In the 18th century, mathematician Leonhard Euler solved one of the foremost infinite series problems of his day by examining the series \begin{align*}\sum \limits_{n=1}^{\infty}\frac{2}{n(n+1)}\end{align*}

Find the first five partial sums of this series and make an observation about the sum of the infinite series.

Source: *Plus Magazine*

### Partial Sums

An **infinite series** is a series with an infinite number of terms. In other words, the value of \begin{align*}n\end{align*}

\begin{align*}\sum \limits_{n=1}^{\infty}3n+1 & =4+7+10+13+ \ldots \\
\sum \limits_{n=1}^{\infty}4(2)^{n-1} & =4+8+16+32+ \ldots\\
\sum \limits_{n=1}^{\infty}8 \left(\frac{1}{2}\right)^{n-1}&=8+4+2+1+ \frac{1}{2}+ \ldots\end{align*}

These sums continue forever and can increase without bound.

Since we cannot find the sums of these series by adding all the terms, we can analyze their behavior by observing patterns within their **partial sums**. A partial sum is a sum of a finite number of terms in the series. We can look at a series of these sums to observe the behavior of the infinite sum. Each of these partial sums is denoted by \begin{align*}S_n\end{align*}

Let's find the first five partial sums of \begin{align*}\sum \limits_{n=1}^{\infty} 2n-1\end{align*}

The first five partial sums are \begin{align*}S_1, S_2, S_3, S_4\end{align*}

\begin{align*}S_1&=a_1=1 \\
S_2&=a_1+a_2=1+3=4 \\
S_3&=a_1+a_2+a_3=1+3+5=9 \\
S_4&=a_1+a_2+a_3+a_4=1+3+5+7=16 \\
S_5&=a_1+a_2+a_3+a_4+a_5=1+3+5+7+9=25\end{align*}

Notice that each sum can also be found by adding the \begin{align*}n^{th}\end{align*}

For example: \begin{align*}S_5=S_4+a_4=16+9=25\end{align*}

The sequence of the first five partial sums is 1, 4, 9, 16, 25. This pattern will continue and the terms will continue to grow without bound. In other words, the partial sums continue to grow and the infinite sum cannot be determined as it is infinitely large.

Now, let's find the first five partial sums of \begin{align*}\sum \limits_{n=1}^{\infty} \left(\frac{1}{2}\right)^{n-1}\end{align*}

The first five terms of this sequence are: \begin{align*}1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}\end{align*}

\begin{align*}S_1&=1 \\
S_2&=1.5 \\
S_3&=1.75 \\
S_4&=1.875 \\
S_5&=1.9375\end{align*}

Consider what happens with each subsequent term: We start with 1 and add \begin{align*}\frac{1}{2}\end{align*}

To support our conjecture further, we can use the calculator to find the \begin{align*}50^{th}\end{align*}

Finally, let's find the first five partial sums of \begin{align*}\sum \limits_{n=1}^{\infty} \frac{1}{n}\end{align*}

Use the calculator to find the following sums:

\begin{align*}S_1&=sum(seq(1/x,x,1,1))=1 \\
S_2&=sum(seq(1/x,x,1,2))=1.5 \\
S_3&=sum(seq(1/x,x,1,3))=1.833 \\
S_4&=sum(seq(1/x,x,1,4))=2.083 \\
S_5&=sum(seq(1/x,x,1,5))=2.283\end{align*}

In this series, the behavior is not quite as clear. Consider some additional partial sums:

\begin{align*}S_{50}&=4.499 \\
S_{100}&=5.187 \\
S_{500}&=6.793\end{align*}

In this case, the partial sums don’t seem to have a bound. They will continue to grow and therefore there is no finite sum.

### Examples

#### Example 1

Earlier, you were asked to find the first five partial sums of the series \begin{align*}\sum \limits_{n=1}^{\infty}\frac{2}{n(n+1)}\end{align*}

The first five terms of this sequence are: \begin{align*}1, \frac{1}{3}, \frac{1}{6}, \frac{1}{10}, \frac{1}{15}\end{align*}

\begin{align*}S_1&=1 \\
S_2&=1.333 \\
S_3&=1.5 \\
S_4&=1.6 \\
S_5&=1.6666\end{align*}

If this pattern is continued, we will get ever closer to 2 but never actually reach two. Therefore, the sum is said to “converge to” or “approach” 2.

**Find the first five partial sums of the infinite series below and additional partial sums if needed to determine the behavior of the infinite series. Use the calculator to find the partial sums.**

#### Example 2

\begin{align*}\sum \limits_{n=1}^{\infty} 4 \left(\frac{3}{2}\right)^{n-1}\end{align*}

\begin{align*}S_1=4; \ S_2=10; \ S_3=19; \ S_4=32.5; \ S_5=52.75;\end{align*}

#### Example 3

\begin{align*}\sum \limits_{n=1}^{\infty} 500 \left(\frac{2}{3}\right)^{n-1}\end{align*}

\begin{align*}S_1=500; \ S_2=833.333; \ S_3=1055.556; \ S_4=1203.704; \ S_5=1302.469;\end{align*}

#### Example 4

\begin{align*}\sum \limits_{n=1}^{\infty} \frac{5}{6n}\end{align*}

\begin{align*}S_1=0.833; \ S_2=1.25; \ S_3=1.528; \ S_4=1.736; \ S_5=1.903;\end{align*}

### Review

Find the first five partial sums and additional partial sums as needed to discuss the behavior of each infinite series. Use your calculator to find the partial sums.

- \begin{align*}\sum \limits_{n=1}^{\infty}5 \left(\frac{1}{2} \right)^{n-1}\end{align*}
∑n=1∞5(12)n−1 - \begin{align*}\sum \limits_{n=1}^{\infty}2 \left(\frac{3}{4} \right)^{n-1}\end{align*}
∑n=1∞2(34)n−1 - \begin{align*}\sum \limits_{n=1}^{\infty}10(0.9)^{n-1}\end{align*}
∑n=1∞10(0.9)n−1 - \begin{align*}\sum \limits_{n=1}^{\infty}8(1.03)^{n-1}\end{align*}
∑n=1∞8(1.03)n−1 - \begin{align*}\sum \limits_{n=1}^{\infty}\frac{1}{2}n\end{align*}
∑n=1∞12n - \begin{align*}\sum \limits_{n=1}^{\infty}\frac{10}{n}\end{align*}
∑n=1∞10n - \begin{align*}\sum \limits_{n=1}^{\infty}\frac{1}{2} \left(\frac{3}{4}\right)^{n-1}\end{align*}
∑n=1∞12(34)n−1 - \begin{align*}\sum \limits_{n=1}^{\infty}\frac{1}{n^2}\end{align*}
∑n=1∞1n2 - \begin{align*}\sum \limits_{n=1}^{\infty}6(0.1)^{n-1}\end{align*}
∑n=1∞6(0.1)n−1 - \begin{align*}\sum \limits_{n=1}^{\infty}0.01n+5\end{align*}
∑n=1∞0.01n+5 - \begin{align*}\sum \limits_{n=1}^{\infty}2 \left(\frac{7}{8}\right)^{n-1}\end{align*}
∑n=1∞2(78)n−1 - Which of the series above are arithmetic? Do any of them have a finite sum? Can you explain why?
- Which of the series above are geometric? Do any of them have a finite sum? Can you explain why?
- What is the difference between the series that converge and diverge?
- Make a conjecture about the converging series in this problem set.

### Answers for Review Problems

To see the Review answers, open this PDF file and look for section 11.11.