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# 5.8: Infinite Series

Difficulty Level: At Grade Created by: CK-12

## Sequences

Zeno of Elea was a Greek philosopher who's most famous for the paradoxes that have been attributed to his name. While Zeno proposed his paradoxes to support, or discredit, various philosophical viewpoints, the paradox is frequently “solved” with a little bit of analysis.

The most famous of Zeno's Paradoxes is about Achilles and the Tortoise. Taken from Aristotle: “In a race the quickest runner can never overtake the slowest, since the pursuer must first reach the point whence the pursued started, so that the slower must always hold a lead.” Put in numerical terms, if the tortoise has a 80\begin{align*}80\end{align*} meter lead, Achilles must first endeavor to make up that deficit. But by the time Achilles makes it to 80\begin{align*}80\end{align*} meters, the half as fast tortoise is now at 120\begin{align*}120\end{align*} meters, covering 40\begin{align*}40\end{align*} meters in the same time it took Achilles to travel 80\begin{align*}80\end{align*}. Now Achilles must make it to the 120\begin{align*}120\end{align*} meter mark, but when he gets there, the tortoise is now at the 140\begin{align*}140\end{align*} meter point. This continues on, making the point that Achilles will always be some distance behind the tortoise.

This is very similar to another paradox about motion: “That which is in locomotion must arrive at the half-way stage before it arrives at the goal.” This is the paradox that should be looked at as a sequence. The paradox states that this makes it impossible to actually reach a goal, as you must pass through the half way point, and then you are at a new location yet to reach your new half way point and that the distance left is always going to be half the distance you are currently away from the destination. For ease of work, lets say the goal is 10m\begin{align*}10m\end{align*} away. Make a list of the half way locations: {5,2.5,1.25,.625,.3125,}\begin{align*}\left \{5,2.5,1.25,.625,.3125,\ldots \right \}\end{align*}which may be better expressed as fractions of the original distance: {102,104,108,1016,}\begin{align*} \left \{ \frac{10}{2},\frac{10}{4}, \frac{10}{8},\frac{10}{16},\ldots \right \}\end{align*}. Now it should become clear that we can express the sequence of locations with an expression: Sn=102n\begin{align*}S_n = \frac{10}{2 ^ n}\end{align*} . If we want to know if we will ever get to the end, we need to know where this sequence will end up, which is another way of saying, what is the limit of this sequence. We can see that limn102n=0\begin{align*} \lim_{n \to \infty} \frac{10}{2 ^ n} = 0 \end{align*}. This means that even though by taking half of each quantity, this is a sequence that gets to zero in an effective way.

There are many other paradoxes, some with more mathematical involvement than others, that can be fun to consider. It is also a fun exercise to try to create new ones, or modify those from Zeno to new situations.

## Infinite Series

Sometimes some interesting accounting techniques can provide the opportunity for banks to lend more money out then they strictly have possession of. If we assume that only 8%\begin{align*}8\% \end{align*}of the amount deposited is in use, the rest remain in the account, then the bank is free to loan out the other 92%\begin{align*}92\%\end{align*}. If the bank then assumes that the cash they loan out will be coming back in the form of income deposited by another party, they can then lend 92%\begin{align*}92 \% \end{align*} of that quantity and so on. If we try a model with the first deposit being \$1000, what is the total amount of money that is deposited back into the bank?

This is an infinite series question, as 92%\begin{align*}92 \%\end{align*} of the previous deposit is never going to be exactly 0\begin{align*}0\end{align*}, and we are adding the amount each time. Listing out some partial sums may give us some insight to the correct way to write the summation.

S0=1000\begin{align*}S_0=1000\end{align*}

S1=1000+1000(.92)=1920\begin{align*}S_1=1000+1000(.92)=1920\end{align*}

S2=1000+1000(.92)+920(.92)=2766.4\begin{align*}S_2=1000+1000(.92)+920(.92)=2766.4\end{align*}

S3=1000+1000(.92)+920(.92)+846.4(.92)=3545.09\begin{align*}S_3=1000+1000(.92)+920(.92)+846.4(.92)=3545.09\end{align*}

I nearly always start out with writing out partial sums if I was not supplied the summation by the problem. The process of writing out the sums, and finding the answers, often gives me clues, such as each term being able to be written as 1000\begin{align*}1000\end{align*} times some multiple of .92\begin{align*}.92\end{align*}:

S0=1000(.92)0\begin{align*}S_0=1000(.92)^0\end{align*}

S1=1000(.92)0+1000(.92)=1920\begin{align*}S_1=1000(.92)^0+1000(.92)=1920\end{align*}

S2=1000(.92)0+1000(.92)+920(.92)2=2766.4\begin{align*}S_2=1000(.92)^0+1000(.92)+920(.92)^2=2766.4\end{align*}

S3=1000(.92)0+1000(.92)+920(.92)2+846.4(.92)3=3545.09\begin{align*}S_3=1000(.92)^0+1000(.92)+920(.92)^2+846.4(.92)^3=3545.09\end{align*}

Now I have a clear idea that the summation will be:

i=0n1000(.92)t\begin{align*} \sum_{i=0}^n 1000 (.92) ^ t \end{align*}

Which is a geometric series that converges.

i=0n1000(.92)i1=10001.92=12500\begin{align*}\sum_{i=0}^n 1000 (.92) ^ {i -1} = \frac{1000}{1 - .92} = 12500 \end{align*}

Taking that quantity and dividing by the original deposit is a quantity called the credit multiplier. Many different fields in the study of economics look at multipliers as a method of analysis or comparison.

## Series Without Negative Terms

The text omits the proof that the harmonic series is a divergent series, but it is not out of the scope of capability for a first year student to accomplish. The harmonic series is interesting to look at because it can trick you at first with how slowly it grows. A good question to ask students to get a feel for the rate of growth is to find how many terms in the partial sum to get to 10\begin{align*}10\end{align*}? To 50\begin{align*}50\end{align*}? To 100\begin{align*}100\end{align*}? (And please don't do the latter two by hand... it will take a very long time! Use a computer or calculator to help.) Also, as the next chapter will illustrate, the alternating harmonic series does converge. Therefore, it may not seem obvious that the harmonic series diverges.

One way, and probably the most obvious way, to prove divergence is with the integral test. The function 1x\begin{align*}\frac{1}{x}\end{align*} is clearly decreasing and the starting value is greater than 0\begin{align*}0\end{align*} so:

11xdx=limb11xdx=limbln(x)b1=limbln(b)\begin{align*}\int\limits_{1}^{\infty} \frac{1}{x} dx = \lim_{b \to \infty} \int\limits_{1}^{\infty} \frac{1}{x} dx = \lim_{b \to \infty} \text {ln} (x) \big |_{1} ^ {b} = \lim_{b \to \infty} \text {ln} (b) \end{align*}

Which is divergent.

Another, slightly more elementary and crafty method, is the one that is briefly outlined in the text. It is a process like the comparison test, but the comparison test requires the inequality to hold term by term. Here we are going to group a set of terms to compare to a series that is divergent. If we list out the first 20\begin{align*}20\end{align*} terms of the sequence:

1+12+13+13+14+15+16+17+18+19+110+111+112+113+114+115+116+117+118+119+120+\begin{align*}1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}+ \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12} + \frac{1}{13} + \frac{1}{14} + \frac{1}{15}+ \frac{1}{16} + \frac{1}{17} + \frac{1}{18} + \frac{1}{19} + \frac{1}{20}+ \end{align*}

We can group the terms such that each group will be greater than \begin{align*}\frac{1}{2}\end{align*}.

\begin{align*} [1] + \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} + \frac{1}{11} + \frac{1}{12} + \frac{1}{13} + \frac{1}{14} + \frac{1}{15} + \frac{1}{16} \right ] + \big [ \frac{1}{17} + \frac{1}{18} + \frac{1}{19} + \frac{1}{20} + \ldots \end{align*}

Since there will be infinitely many groupings, we find that this sequence will be larger than an infinite sum of \begin{align*}\frac{1}{2}\end{align*}, which is clearly divergent.

## Series With Odd or Even Negative Terms

The methods of having alternating signs in the terms of a series introduces some puzzles for writing series in summation notation. As is clear from the text, the way of writing an alternating sign is to have a factor of \begin{align*}-1 \end{align*}to an exponent. While this is clear enough, another consideration must be the index, as a series:

\begin{align*}1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} - \ldots \end{align*}

Will not be written the same way as the series:

\begin{align*}- 1 + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \ldots\end{align*}

Therefore, some tricky work must be done with the indexing. Often times there will be numbers added or subtracted to the indexing of the summation, the exponent or anywhere else to get signs and numbers to agree. In the case of the above series, each are an alternating harmonic series, so we know we will start out with:

\begin{align*}\sum_{i=1}^\infty (-1) ^ 7 \frac{1}{i} \end{align*}

In the first case, the first term is positive, so we need the exponent to be even for the first term. Therefore we need to add one. No such addition is needed for the second series, as the negatives work with the regular indexing.

Other interesting places to get tripped up include use of all even, or all odd numbers. For instance, how would you write the series of all even numbers added with alternating signs, where the first term is positive? Writing even sequences is a little trick that many students learn and become comfortable with around the first year of calculus. The same way that the trick to alternating signs is the exponent being odd or even, the trick to getting all even numbers is to multiply by \begin{align*}2\end{align*}. Odd numbers will be handled by multiplying by \begin{align*}2\end{align*} and then adding or subtracting one depending on what the starting value needs to be. Therefore our series ask above is:

\begin{align*}\sum_{i=1}^\infty (-1) ^{i + 1} (2i) \end{align*}

It is a good challenge for students to try to think up series that skip terms, alternate signs and other tricks that may require a bit of puzzle solving to write out.

## Ratio Test, Root Test and Summary of Tests

An added challenge for students can using some of the techniques of calculus to not only determine convergence, but find the sum of the series.

\begin{align*}\sum_{i=2}^\infty \frac{1}{i ^ 3 - i} = \end{align*}

The first thing to do is to show that this series converges. This is easily done by the comparison test, which is nearly always my first attempt, especially for expressions with polynomials in the denominator. Here we can compare it to \begin{align*}\frac{1}{x ^ 2} \end{align*} which is easy to show convergence with the integral test.

Now finding the value of the sum is a little bit tricky. This is a nice application of the method of partial fractions outside of integrals, as wel will need to split up that denominator to find a solution.]

\begin{align*} \frac{1}{i ^ 3 - i} = \frac{1}{i(i - 1)(i + 1)} = \frac{A}{i} + \frac{B}{i - 1} + \frac{C}{i + 1} \\ 1 = A(i - 1) (i + 1) + Bi(i + 1) + C(i -1) = (A+B+C) t ^ 2 + (B - C)i - A\\ A + B + C = 0 && B - C = 0 && -A = 1\end{align*}

Substituting in and then splitting up the summation:

\begin{align*} \sum_{i=2}^\infty \frac{-1}{i} + \sum_{i=2}^\infty \frac{1}{2(i - 1)} + \sum_{i=2}^\infty \frac{1}{(2 + i)}\end{align*}

Now we can change the index of each to eliminate the terms in the denominator.

\begin{align*} - \sum_{i=2}^\infty \frac{1}{i} + \frac{1}{2} \sum_{i=1}^\infty \frac{1}{i} + \frac{1}{2} \sum_{i=3}^\infty\frac{1}{i}\end{align*}

We need one more change of index now to get compare the sums. By taking the first two terms from the middle sum and the first term from the first sum we can start each of them at an index of \begin{align*}3\end{align*}:

\begin{align*} - \frac{1}{2} - \sum_{i=3}^\infty \frac{1}{i} + \frac{1}{2} \times 1 + \frac{1}{2} \times \frac{1}{2} + \frac{1}{2} \sum_{i=3}^\infty \frac{1}{i} + \frac{1}{2} \sum_{i=3}^\infty\frac{1}{i}\end{align*} The summations all cancel, adding to zero, so the sum is equal to the evaluation of the constants = \begin{align*}\frac{1}{4}\end{align*}.

## Power Series

Finding ways to approximate functions with power series is a tough task for students. Here is some additional reinforcement with another standard problem.

Find the power series representation for the function \begin{align*} f(x) = \text {ln} (x + 1)\end{align*} with center zero.

There are two tricks here. First of all we want to try to convert to a series at some point and usually the easiest way is to use a geometric series. Also, a common trick to get logarithms into the form of a geometric series is to use the derivative. This gives a fraction that can be manipulated into the correct form:

\begin{align*} f'(x) = \frac{1}{1 - (-x)} dx = (-x) ^ 0 + (-x) ^ 1 + (-x) ^ 2 + (-x) ^ 3 + \ldots \end{align*}

Integrate both sides:

\begin{align*}\int f'(x) dx = \int 1 -x + x ^ 2 - x ^ 3 + \ldots dx\\ \text {ln} (x+1) = x - \frac{x ^2}{2} + \frac{x ^ 3}{3} - \frac{x ^ 4}{4} + \ldots = \sum_{n=0}^\infty\frac{(-1) ^ n x ^ n}{n} \end{align*}

Checking for the radius of convergence:

\begin{align*}a_n = \frac{(-1) ^ n x ^ n}{n} \rightarrow \left |\frac{ (-1) ^ {n + 1} x ^ {n + 1}}{n + 1} \times \frac{n}{ (-1) ^ {n + 1} x ^n} \right | = \left | \frac{-xn}{n + 1} \right | \end{align*}

So taking the limit:

\begin{align*}\lim_{n \to \infty} \left | \frac{-n}{n + 1} \times x \right | = |x| \end{align*}

Therefore the radius of convergence is \begin{align*}|x|<1\end{align*}.

## Taylor and MacLaurin Series

John Machin was a \begin{align*}17th\end{align*} century mathematician who is probably most famous for developing a formula for to approximate pi:

\begin{align*}\frac{\pi}{4} = 4 tan ^ {-1} \left ( \frac{1}{5} \right ) - tan ^ {-1} \left ( \frac{1}{239} \right ) \end{align*}

We can examine why this was important with the following questions. Remember, the whole advantage of Taylor series is that it allows nearly any function to be calculated as a polynomial. This has two implications; first, this is how computers and calculators compute transcendental functions. Second, if you do not have a caculator, or you are attempting to find a value that is previously unknown so it does not appear in a table, the first number of terms in a taylor sum will allow you to find that value.

First we need to find the taylor sum for \begin{align*}tan ^ {-x} (x) \end{align*}. Here we are going to take a roundabout approach. The first thing to do is to look at the binomial expansion for the function:

\begin{align*}( 1 + u ) ^ {-1} = 1 - u + u ^ 2 - u ^ 3 + u ^ 4 - \ldots \end{align*}

Substituting \begin{align*}u=x^2\end{align*}:

\begin{align*} (1 + x ^ 2) ^ {-1} = 1 - x ^ 2 + x ^ 4 - x ^6 + x ^ 8 - \ldots\end{align*}

Now you should recognize that this function is the derivative of \begin{align*}tan ^ {-x} (x)\end{align*} . We can then integrate both sides to get the taylor series:

\begin{align*}tan ^ {-1} (x) = x - \frac{x ^3}{3} + \frac{x ^ 5}{5} - \frac{x ^ 7}{7} + \frac{x ^ 9}{9} - \ldots\end{align*}

A student may ask how we knew to take the binomial expansion of that particular function. There is no really good answer, as all the time mathematicians are asserting that something is true, and then proving it later, seemingly picking ideas out of thin air. In fact, we will make a doozy of an assumption later. Sometimes guess and check can tell us where we need to go. Here, we are taking a function and that is close to some form of our original function.

Now the temptation is to say that since \begin{align*}tan ^ {-1} (1) = \frac{\pi}{4}\end{align*}, why not use this expansion to calculate pi? You can, and it will converge to the correct number, but the \begin{align*}3rd\end{align*} decimal place is correct after \begin{align*}1000\end{align*} terms. If you notice Machin’s formula uses fractions that when put into the taylor sum, it converges very quickly. In fact, you only need about \begin{align*}5\end{align*} or \begin{align*}6\end{align*} terms to get a very accurate approximation for pi.

But we still need to show that Machin’s formula is correct. We will start by making the assertion that:

\begin{align*}tan ^ {-1} \left ( \frac{120}{119} \right ) - tan ^ {-1} \left ( \frac{1}{239} \right ) = tan ^ {-1} (1) \end{align*}

To show that this is the case, use the angle sum formula for tangent:

\begin{align*}tan(A + B) = \frac{tan(A) + tan(B)}{1 - tan(A)tan(B)} \end{align*}

If you use angle \begin{align*}A = tan ^ {-1} \left ( \frac{120}{119} \right )\end{align*} and \begin{align*}B = tan ^ {-1} \left ( \frac{-1}{239} \right )\end{align*} the assertion above is proven. All that is required is a little bit of arithmetic as all of the tangent and tangent inverses cancel each other.

Much the same way, we now need to show:

\begin{align*}4tan ^ {-1} \left ( \frac{1}{5} \right ) = tan ^ {-1} \left ( \frac{120}{119} \right ) \end{align*}

This is easiest to show in two steps. First show:

\begin{align*}2tan ^ {-1} \left ( \frac{1}{5} \right ) = tan ^ {-1} \left ( \frac{5}{12} \right ) \end{align*}

Again by using the angle addition rule with\begin{align*}A =B = tan ^ {-1} \left ( \frac{1}{5} \right ) \end{align*} Then show that:

\begin{align*}2tan ^ {-1} \left ( \frac{5}{12} \right ) = tan ^ {-1} \left ( \frac{120}{119} \right )\end{align*}

With the same arithmetic techniques for a third time with \begin{align*} A =B = tan ^ {-1} \left ( \frac{5}{12} \right ) \end{align*}

It should all come together now substituting back to the top. It is also useful to remember that negatives inside of a tangent become negatives outside due to symmetry. It is common for students to believe that taylor series are antiquated, made obsolete by the calculator. As it actually stands, someone has to program all of those functions into the calculator, and the most common technique is to use the equivalent taylor series. Our calculators would not know how to take the tangent of an angle otherwise. This is an elegant way to compute many digits of pi without extreme computer power.

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