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4.8: Infinite Series

Difficulty Level: At Grade Created by: CK-12
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Sequences are simply lists of numbers, that’s it. The ones we are most interested in, however, are infinitely long. The only rule is that we must keep the list ordered. For example, the sequence \begin{align*}\left \{1,2,3,\ldots \right \}\end{align*} is not the same as \begin{align*}\left \{2,1,3,4,5,\ldots\right \}\end{align*} because the \begin{align*}1\end{align*} and \begin{align*}2\end{align*} have switched places.

The idea behind the definition for a limit is an intuitive one, although this is somewhat hidden by the terminology. Suppose we have a long list of numbers like \begin{align*}\left \{\frac{1}{1},\frac{1}{2},\frac{1}{3}, \frac{1}{4}, \ldots\right \} \end{align*}. The numbers never actually reach zero, since one over something is never zero. However, we can see at the same time that the numbers get closer and closer to zero. So the limit of the sequence is zero, even though the sequence never quite makes it there. The definition is meant to recognize precisely this kind of situation.

The points keep getting lower and lower and no matter how small a number you can think of, they will eventually get smaller than that number. So the limit is \begin{align*}L=0\end{align*}. In math terms, for any \begin{align*}\epsilon > 0\end{align*}, there is a value \begin{align*}N\end{align*} so that each term is no bigger than \begin{align*}\epsilon\end{align*}.

A variety of techniques for showing that sequences have limits can be found, however they are all based in the geometric idea above. If the terms of the sequence eventually all squish together closer than any imaginable distance, then there is convergence. Otherwise there is not.

In this chapter a much more general kind of sequence is discussed where the elements are not numbers, but functions. These are the sequences of Picard’s Methods. A simpler way to think of a sequence of functions may be as a kind of deforming curve in the \begin{align*}x-y\end{align*} plane, like a chain that is stretched between two points wiggling about until it settles into one final curve. The initial function is like some curve, and the method just wiggles this initial guess so that it more closely solves \begin{align*}\frac{dy}{dx} = f(x ,y)\end{align*} and then we look at the limiting function. In fact, taking a derivative is similar since we also consider a kind of sequence of functions: One for each value of h. As \begin{align*}h \rightarrow 0\end{align*} we look for the limiting function, and this is the derivative.

Infinite Series


An infinite series is actually just a limit of a particular sequence. Given some sequence, we can take try to find its limit like we did in the previous lesson. Or alternatively, we can create new sequences from the individual terms. For example, if we are given an infinite sequence \begin{align*}\left \{a_1,a_2,a_3 \ldots , a_n , \ldots \right \}\end{align*} we could define the so-called infinite sequence of partial sums:

\begin{align*}S = \left \{s_1 = a_1, s_2 = a_1 + a_2 ,\ldots , s_n = \sum_{i=1}^n a_n , \ldots\right \} \end{align*}

Then the limit of this sequence is called the sum:

\begin{align*}\sum_{i=1}^\infty a_n : = limit S\end{align*}

More simply-put, an infinite series is just a summation with an infinite number of terms. This can be illustrated with the nice geometric proof that Zeno’s Sum is \begin{align*}1\end{align*}:

\begin{align*} \sum_{i=1}^\infty \left ( \frac{1}{2} \right ) ^ n = 1\end{align*}


One absolutely indispensible piece of knowledge with regard to series is the ability to quickly recognize and find the sum of a geometric series. In order to give students this ability, the following may be fruitful:

Have students put away all of their materials and listen closely so that they understand every step of what follows. Tell them they will have to do this on their own in a minute, without anything on the board, so they should ask questions if they have any. Then show them the general geometric series:

\begin{align*}\sum_{n=1}^\infty r ^ n = r + r ^ 2 + r ^ 3 + \ldots + r ^ n + \ldots \end{align*}

and ask them to think of a way to solve this. Then explain that there is a very nice little trick that begins with setting the sum equal to some number, say, \begin{align*}S\end{align*}:

\begin{align*}S = r + r ^ 2 + r ^ 3 + \ldots + r ^ n + \ldots\end{align*}

And consider multiplying this equation by the number \begin{align*}r\end{align*}:

\begin{align*} r S = r ^ 2 + r ^ 3 + r ^ 4 + \ldots + r ^ {n + 1} + \ldots \end{align*}

Then we can subtract these:

\begin{align*} S & = r + r ^ 2 + r ^ 3 + \ldots + r ^ n + \ldots \\ -rS & = - r ^ 2 - r ^ 3 - r ^ 4 - \ldots - r ^ {n+1} - \ldots \end{align*}

and we see that every term will cancel except for the very first r:

\begin{align*}S & = \cancel {r + r ^ 2 + r ^ 3 + \ldots + r ^ n + \ldots} \\ -rS & = \cancel{- r ^ 2 - r ^ 3 - r ^ 4 - \ldots - r ^ {n + 1}} - \ldots \end{align*}

to give that:

\begin{align*}S - rS = r\end{align*}

which is readily solved for \begin{align*}S\end{align*}:

\begin{align*}S = \frac{r}{1 - r}\end{align*}

This clearly doesn’t make sense if \begin{align*}r=1\end{align*}, and in fact the sum will not converge for any \begin{align*}r \ge 1\end{align*}. So an implicit assumption in all of this is that we have the strict inequality that \begin{align*}r<1\end{align*}. Then give an example. Ask students how to calculate the sum of the following series:

\begin{align*}\sum_{n=1}^\infty e ^ {-n} \end{align*}

After letting them think a little, tell them the ANSWER:

\begin{align*}\sum_{n=1}^\infty e ^ {-n} = \sum_{n=1}^\infty \frac{1}{e ^ n} = \sum_{n=1}^\infty \left ( \frac{1}{e} \right ) ^ n \end{align*}

So this is just a geometric series with \begin{align*}r = \frac{1}{e} \approx 0.368\end{align*} . The sum then is:

\begin{align*}S = \frac{r}{1 - r} = \frac{\frac{1}{e}}{1 - \frac{1}{e}} = \frac{1}{e - 1} \approx 0.582\end{align*}

Next, the board should be thoroughly erased and students should be asked to reproduce the entire derivation for the sum of a geometric series as well as to solve some similar problem. They can do this individually or in small groups, but the problem should be a little disguised or more complicated such as proving the following result:

\begin{align*}\frac{1}{1 * 5} + \frac{1}{ 2 * 5 ^ 2} + \frac{1}{3 * 5 ^ 3} + \ldots + \frac{1}{n * 5 ^ n} < \frac{1}{4} \end{align*} for all n

Finally, it should be made clear that if a geometric series does not begin with the \begin{align*}n=1\end{align*} power, clearly there are just a finite number of terms that differentiate the two. So we can figure out the exact value of the missing terms and find the sum of one series in terms of the other. A few examples will demonstrate this nicely.  

Series Without Negative Terms


The harmonic series is extremely important and indeed many mathematicians have dedicated their entire life’s work to understanding it. So it is worthwhile to show students why it clearly diverges. Consider the sum:

\begin{align*}\sum_{n=1}^\infty \frac{1}{n} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}+ \ldots\end{align*}

The first term, \begin{align*}s_1=1\end{align*}, clear satisfies \begin{align*}s_1 =\ge \frac{1}{2}\end{align*}. The second term is exactly equal to \begin{align*}\frac{1}{2}\end{align*}, so \begin{align*}s_2 \ge \frac{1}{2}\end{align*};

however, \begin{align*}s_3 = \frac{1}{3} < \frac{1}{2}\end{align*} .On the other hand, \begin{align*}s_3 + s_4 = \frac{1}{3} +\frac{1}{4} = \frac{7}{12} \ge \frac{1}{2}\end{align*} and we can see that we can just keep grouping together parts of the series that are always larger than \begin{align*}\frac{1}{2}\end{align*}:

\begin{align*}\sum_{n=1}^\infty \frac{1}{n} = \left [ \frac{1}{1} \right ] + \left [ \frac{1}{2} \right ] + \left [ \frac{1}{3} + \frac{1}{4} \right ] + \left [ \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \right ] + \left [ \frac{1}{9} + \frac{1}{10} + \ldots \right ] + \ldots \end{align*}

Each bracketed part of the sum is larger than \begin{align*}\frac{1}{2}\end{align*} since, for example:

\begin{align*}\left [ \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \right ] < \left [ \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} \right ] = \frac{1}{2}\end{align*}

And so we see that:

\begin{align*}\sum_{n=1}^\infty \frac{1}{n} = \left [ \frac{1}{1} \right ] + \left [ \frac{1}{2} \right ] + \left [ \frac{1}{3} + \frac{1}{4} \right ] + \left [ \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \right ] + \left [ \frac{1}{9} + \frac{1}{10} + \ldots \right ] + \ldots \\ < \left [ \frac{1}{2} \right ] + \left [ \frac{1}{2} \right ] + \left [ \frac{1}{2} \right ] + \ldots\end{align*}

which is the clearly divergent series \begin{align*}\Sigma \frac{1}{2} \end{align*}.

Some interesting questions may arise as to how fast this series diverges and how to gently nudge it so that it does not diverge. It turns out that the series diverges at the same rate as \begin{align*} \mathrm {Ln} (n)\end{align*}, since it can be shown that:

\begin{align*}\lim_{n \to \infty} \frac{\sum_{k=1}^n \frac{1}{k}}{\text {Ln}(n)} = 1\end{align*}

So how many terms can we remove from the harmonic series and still have it diverge? For example, we can remove every term whose denominator is not prime, leaving behind the famous series:

\begin{align*} \Sigma _ {p \ prime} \frac{1}{p} = \frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{11} + \frac{1}{13} + \dots\end{align*}

and the series will still diverge! In fact, it turns out that while the original harmonic series diverges like \begin{align*} Ln(n)\end{align*}, this series of prime inverses will diverge like \begin{align*}Ln(Ln(n))!\end{align*} On the other hand, if we remove all of the terms with any \begin{align*}9\end{align*} as a digit in the denominator, the resulting series:

\begin{align*}\Sigma_ {no \ 9 \ ts} \frac{1}{k} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{8} + \frac{1}{10} + \ldots + \dots + \frac{1}{18}+ \frac{1}{20} + \ldots = \infty\end{align*}


Teaching this material gives some nice opportunities to have students practice with inequalities and integration. By giving series of complicated rational terms the students can try finding an appropriate comparison or integral. In general this will be a daunting and cumbersome activity for students to do on their own, so it is recommended that the class be divided into small groups. Perhaps the following activity would be exciting:

Split the class into two teams, and within each team for pairs or partners. Provide each team with a treasure map like the one on the following page. If each team has \begin{align*}3\end{align*} pairs, there should be \begin{align*}3\end{align*} forks in the paths, (A, B, and C below). If each team has \begin{align*}4\end{align*} pairs, there should be \begin{align*}4\end{align*} forks lettered as (A, B, C, and D) and correspondingly \begin{align*}16\end{align*} final doors. Then the team should be given a series corresponding to each fork in the path. If the series diverges, the team is to turn left at the fork and if the series converges, the team is to turn right at the fork. The team should divide up the problems among its different pairs so that they can work on the problem and decide whether that series prescribes a left turn or a right turn. The team which arrives at the correct final door first is the winner!

Series With Odd or Even Negative Terms


Series that contain both positive and negative terms should be thought of more likely to converge in a sense. This is because very qualitatively the negative terms will counterbalance the positive ones making the sum more reasonable. For this reason, a series that alternates term-by-term between positive and negative has a very simple test for convergence. If the terms trail off to zero then the series converges. This is clearly not good enough for a strictly positive series, as the harmonic series shows.

The remainder theorem for an alternating series actually tells us that when the terms shrink to zero, we actually have something much stronger than \begin{align*}|a_n| \ge | a_{n + 1} |\end{align*}. We have that:

\begin{align*}|a_n| \ge \sum_{i=n + 1}^\infty a_i\end{align*}

This follows from the fact that the signs alternate so that an arrangement of the terms shows that the tail of the series always has the same sign as its leading term. But then since the tail can be written as a sum of terms that all have the same sign, it must be smaller than the leading term if the sign is to come out unchanged.


To teach this material it is good to get students in the habit of writing the first few terms of a series in order to understand its behavior. To this end, a good start to any class on alternating series is to write some series on the board and select students to come up and write out the first few terms of each. This will get everyone on the same page about how these series alternate and what we are really talking about.

Students will be able to understand the concepts of absolute and conditional convergence for the \begin{align*}p-\end{align*}series. It is useful to provide very nice concrete examples instead of asking the students to memorize abstract rules. For example, for what values of \begin{align*}p\end{align*} is the series:

\begin{align*}\sum_{k=1}^\infty \frac{ (-1) ^ {k + 1}}{k ^ p}\end{align*}

absolutely convergent, conditionally convergent, and divergent?

ANSWER: For \begin{align*}p>1\end{align*}, this series will be absolutely convergent by the \begin{align*}p-\end{align*}test. For \begin{align*} 0 < p \le 1\end{align*} the series will be conditionally convergent since the alternating series

\begin{align*}\sum_{k=1}^\infty \frac{ (-1) ^ {k + 1}}{k ^ p}\end{align*}

is convergent whereas its absolute value, the harmonic series or one that is greater, is divergent. Finally, for any non-positive \begin{align*}p\end{align*} the series will diverge since it will simply be a sum of terms, each greater than or equal to one and alternating in sign. They will bounce back and forth across zero without ever getting any closer than where they start.

Ratio Test, Root Test, and Summary of Tests


This lesson summarizes a number of different tests for convergence of series. Students will be know that if they understood and can recall how to find the sum of a geometric series then the proofs of these tests will follow. For example, consider the ratio test: Let

\begin{align*}\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = r\end{align*}.

Then for large enough \begin{align*}n\end{align*}, this ratio will be arbitrarily close to \begin{align*}r\end{align*}, so suppose we choose such an\begin{align*} N\end{align*} so that \begin{align*}\frac{a_ {n + 1}}{a _n} < r + \epsilon\end{align*} for all \begin{align*}n \ge N\end{align*} . Then this implies that, for example, \begin{align*}a_{N + 1} < r a_N\end{align*}. and therefore that iteratively \begin{align*}a_{N+2}< r ^ 2 a_N\end{align*} or in general that \begin{align*}a_{N+k} < r^k a_N\end{align*}. But then the tail of the series after \begin{align*}N\end{align*} is given by:

\begin{align*} & a_N + a_{N + 1} + a_{N + 2} + \ldots \\ & < a_N + r a_N + r ^ 2 a_N + \ldots \\ & = a_N ( 1 + r + r ^ 2 + \ldots )\end{align*}

Notice that the series in parentheses is a geometric series, and we know that this definitely converges if \begin{align*}r<1\end{align*} and definitely diverges if\begin{align*} r>1\end{align*}. So the series as a whole similarly converges when the ratio \begin{align*}r<1\end{align*} and diverges when \begin{align*}r>1\end{align*}.

The root test has a similar proof except it is in a sense simpler or more direct. If we have that:

\begin{align*}\lim_{n \to \infty} \sqrt[n]{a_n} = r\end{align*}

then we again choose \begin{align*}N\end{align*} so that the two sides here are within \begin{align*}\epsilon\end{align*} for all \begin{align*}n \ge N\end{align*} :

\begin{align*}\sqrt[n]{|a_n|} < r + \epsilon \end{align*}

This implies that:

\begin{align*}|a_n| < ( r + \epsilon ) ^ n \end{align*}


\begin{align*}( r + \epsilon) ^ n = r ^ n + \text {positive terms}\end{align*}

so that:

\begin{align*}|a_n| < ( r + \epsilon ) ^ n < r ^ n \end{align*}

for all \begin{align*}n \ge N\end{align*}. But then of course \begin{align*}|a_N| < r ^ N , | a_{N + 1} | = r ^ N r\end{align*} , or in general \begin{align*}| a_{N + k}| < r ^ N r ^ k \end{align*} and we have that the tail of the series after \begin{align*}N\end{align*} is:

\begin{align*}& a_N + a_{N + 1} + a_{N + 2} + \ldots\\ & < r ^ N + r ^ N r + r ^ N r ^ 2 + \ldots \\ & = r ^ N ( 1 + r + r ^ 2 + \ldots )\end{align*}

which again converges if \begin{align*}r<1\end{align*} and diverges if \begin{align*}r>1\end{align*}. ∎


This could be taught in a nice way too by having students play a game in teams against one-another:

Split the class into two teams, and within each team for pairs or partners. Provide each team with a treasure map like the one on the following page. If each team has \begin{align*}3\end{align*} pairs, there should be \begin{align*}3\end{align*} forks in the paths, (A, B, and C below). If each team has \begin{align*}4\end{align*} pairs, there should be \begin{align*}4\end{align*} forks lettered as (A, B, C, and D) and correspondingly \begin{align*}16\end{align*} final doors. Then the team should be given a series corresponding to each fork in the path. If the series diverges, the team is to turn left at the fork and if the series converges, the team is to turn right at the fork. The team should divide up the problems among its different pairs so that they can work on the problem and decide whether that series prescribes a left turn or a right turn. The team which arrives at the correct final door first is the winner!

The idea with this activity is to generate an environment where students are actively working together to seek tests that will demonstrate whether a given series is convergent or divergent. If their tests work then they will know which way to turn, and if not then they will take a wrong turn.  

Power Series

CONTENT The simplest way to present a power series is as an infinitely long polynomial. Just say: “A power series is just an infinitely long polynomial”

The content is more likely to be readily accepted if motivated properly. Therefore it is worthwhile to let students know ahead of time that almost any function there is that is nice enough (differentiable) can be written as an infinite polynomial like this. Since polynomials are so easy to differentiate and integrate, this has enormous utility throughout math, science, business, and engineering.


It’s a good idea to start with some simple examples of power series that are convergent to familiar functions:

\begin{align*}\sum_{n=0}^\infty x ^ n\end{align*} converges for all \begin{align*}-1 < x < 1\end{align*} to the function \begin{align*} f(x) = \frac{1}{1 - x} \end{align*}

\begin{align*}\sum_{n=0}^\infty \frac{x ^ n}{n!}\end{align*}converges for all \begin{align*}x\end{align*} to the function \begin{align*}f(x)=e^x\end{align*}

\begin{align*}\sum_{n=0}^\infty (-1) ^ n \frac{x ^ {2n}}{(2n)!}\end{align*}converges for all \begin{align*}x\end{align*} to the function \begin{align*}f(x)=\mathrm {Cos}(x)\end{align*}

\begin{align*}\sum_{n=0}^\infty (-1) ^ n \frac{x ^ { 2n + 1}} {(2n + 1)!}\end{align*} converges for all \begin{align*}x\end{align*} to the function \begin{align*}f(x)=\mathrm {Sin}(x)\end{align*}

Here it is worthwhile to point out that using these representations, it is easy to see that:

\begin{align*}\frac{d}{dx} e ^ x & = e ^ x \\ \frac{d}{dx} \text {Sin}(x) = \text {Cos}(x) \end{align*}

Here are some more good examples:

\begin{align*}\sum_{n=0}^\infty (-1) ^ {n + 1} \frac{x ^ n}{n}\end{align*} converges for \begin{align*}x > -1\end{align*} to the function \begin{align*}f(x) = Ln(x + 1)\end{align*}

\begin{align*}\sum_{n=0}^\infty (-1) ^ n \frac{x ^ {2n + 1}}{2n + 1}\end{align*}converges for\begin{align*}-1 < x< 1 \end{align*}to the function \begin{align*}f(x) = \mathrm {Tan} ^ {-1} (x)\end{align*}

Students will start to see why finding the interval is important and how these power series can be used. A fun activity may be to have the class divide into partners and then to give each pair a particular power series. The object will be to find the interval of convergence, and the limit for arbitrary \begin{align*}x\end{align*} in this interval if possible.

Then, each pair should compare their intervals with other pairs. If two intervals overlap, then on the overlap both series should converge and they can be added together. The pairs should join together by adding their series to give what is unlikely to be a simple series. Then together they will be able to find the sum of this more complicated series on the overlapping interval.

As a simple example, one pair could be given the geometric series

\begin{align*} \sum_{n=0}^\infty (2x) ^ n\end{align*}

which will converge for \begin{align*}- \frac{1}{2} < x < \frac{1}{2}\end{align*} to the limit \begin{align*}f(x) = \frac{1}{1 - 2x}\end{align*}. Another pair could be given the geometric series:

\begin{align*}\sum_{n=0}^\infty (3x) ^ n \end{align*}

which will converge for \begin{align*}- \frac{1}{3} < x < \frac{1}{3}\end{align*} to the limit \begin{align*}g(x) = \frac{1}{1 - 3x}\end{align*}.

So we see that on the overlap, when we have that for \begin{align*}- \frac{1}{3} < x < \frac{1}{3}\end{align*}, both of these series converge and we can consider the series:

\begin{align*}\sum_{n=0}^\infty (2x) ^ n + (3x)^ n = \sum_{n=0}^\infty ( 2 ^ n + 3 ^ n) x ^ n\end{align*}

which is not a geometric series. However it is the sum of two simple geometric series, each of which converge, so the limit should be:

\begin{align*}\sum_{n=0}^\infty ( 2 ^ n + 3 ^ n ) x ^ n = \frac{1}{1 -3x} + \frac{1}{1 - 2x} = \frac{2 - 5x}{(1 - 3x)(1 - 2x)} = \frac{2 - 5x}{6x ^ 2 - 5x + 1}\end{align*} PRODUCTS

Students should be asked to find the Taylor and Maclauren Series expansions for a variety of complicated functions using the formulas that:

\begin{align*}f(x) = \sum_{n=0}^\infty a_n ( x - x_0) ^ n\end{align*}


\begin{align*}a_n = \frac{f ^ {(n)}(x_0)}{n!}\end{align*}

The only difficulty in carrying this out is in obtaining a general formula for the n^th derivative of a function. Therefore the best questions lead them towards the answer in the following sort of way:

1. Consider the function \begin{align*}f(x) = \mathrm {Sin}(x)\mathrm {Cos}(x)\end{align*}

a. Find \begin{align*}f ^ {(0)} (0)\end{align*} :

i. ANSWER:\begin{align*}f ^ {(0)} (0) = f(0) = 0\end{align*}

b. Find \begin{align*}f ^ {(1)}(0)\end{align*}:

i. ANSWER: \begin{align*}f ^ {(1)}(0) = 1\end{align*} c. Find \begin{align*}f^{(2)} (0)\end{align*}:

i. ANSWER:\begin{align*} f^{(0)} (0) = 0\end{align*}

d. Find \begin{align*}f^{(3)}(0)\end{align*}: i. ANSWER: \begin{align*}f^{(3)} (0)=-4\end{align*}

e. Find \begin{align*}f^{(4)} (0)\end{align*}:

ANSWER: \begin{align*}f^{(4)} (0)=0\end{align*}

f. Find \begin{align*}f^{(5)}(0)\end{align*}:

i. ANSWER: \begin{align*}f^{(5)}(0)=16\end{align*}

g. Find \begin{align*}f^{(6)}(0)\end{align*}:

i. ANSWER: \begin{align*}f^{(6)}(0)=0\end{align*}

h. Find \begin{align*}f^{(7)}(0)\end{align*}:

i. ANSWER: \begin{align*}f^{(7)}(0)=-64\end{align*}

I. Find \begin{align*}f^{(8)}(0)\end{align*}:

i. ANSWER: \begin{align*}f^{(8)}(0)=0\end{align*}

j. Find \begin{align*}f^{(9)}(0)\end{align*}:

i. ANSWER:\begin{align*} f^{(9)}(0)=256\end{align*}

k. Find \begin{align*}f^{(10)}(0)\end{align*}:

i. ANSWER: \begin{align*}f^{(10)}(0)=0\end{align*}

l. Find \begin{align*}f^{(11)}(0)\end{align*}:

i. ANSWER: \begin{align*}f^{(11)}(0)=-1024\end{align*}

m. Find \begin{align*}f^{(12)}(0)\end{align*}:

i. ANSWER: \begin{align*}f^{(12)}(0)=0\end{align*}

n. Find \begin{align*}f^{(n)}(0)\end{align*}:

i. ANSWER: \begin{align*} f ^ {(n)} (0) = \begin{cases} 0 \ & \mathrm {n even} \\ (-1) ^ {\frac{n -1}{2}} 2 ^ {n -1} \ & \mathrm{ n odd} \end{cases}\end{align*}

o. Find the Taylor Series expansion for \begin{align*}f(x)=\mathrm {Sin}(x)\mathrm {Cos}(x)\end{align*} near \begin{align*}x_0=0\end{align*}: i. ANSWER: \begin{align*}\mathrm {Sin}(x)\mathrm {Cos}(x) = \sum_{n=0}^\infty \frac{f ^ {(m)}(0)}{m!} x ^ m = \sum_{n=0}^\infty (-1) ^ n \frac{4 ^ n}{(2n + 1)!}x ^ {2n + 1}\end{align*}

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