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# Partial Sums

## Evaluate partial sums of various types of series

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Partial Sums

In the 18th century, mathematician Leonhard Euler solved one of the foremost infinite series problems of his day by examining the series n=12n(n+1)\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 n\begin{align*}n\end{align*} increases without bound as shown in the series below.

n=13n+1n=14(2)n1n=18(12)n1=4+7+10+13+=4+8+16+32+=8+4+2+1+12+\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 Sn\begin{align*}S_n\end{align*} where n\begin{align*}n\end{align*} denotes the index of the last term in the sum. For example, S6\begin{align*}S_6\end{align*} is the sum of the first 6 terms in an infinite series.

Let's find the first five partial sums of n=12n1\begin{align*}\sum \limits_{n=1}^{\infty} 2n-1\end{align*} and make an observation about the sum of the infinite series.

The first five partial sums are S1,S2,S3,S4\begin{align*}S_1, S_2, S_3, S_4\end{align*} and S5\begin{align*}S_5\end{align*}. To find each of these sums we will need the first five terms of the sequence: 1, 3, 5, 7, 9. No we can find the partial sums as shown:

S1S2S3S4S5=a1=1=a1+a2=1+3=4=a1+a2+a3=1+3+5=9=a1+a2+a3+a4=1+3+5+7=16=a1+a2+a3+a4+a5=1+3+5+7+9=25\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 nth\begin{align*}n^{th}\end{align*} term to the previous sum: Sn=Sn1+an\begin{align*}S_n=S_{n-1}+a_n\end{align*}.

For example: S5=S4+a4=16+9=25\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 n=1(12)n1\begin{align*}\sum \limits_{n=1}^{\infty} \left(\frac{1}{2}\right)^{n-1}\end{align*} and make an observation about the sum of the infinite series.

The first five terms of this sequence are: 1,12,14,18,116\begin{align*}1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}\end{align*}. The partial sums are thus:

S1S2S3S4S5=1=1.5=1.75=1.875=1.9375\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 12\begin{align*}\frac{1}{2}\end{align*} putting us halfway between 1 and 2. Then we add 14\begin{align*}\frac{1}{4}\end{align*}, putting us halfway between 1.5 and 2. Each time we add another term, we are cutting the distance between our current sum and 2 in half. 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.

To support our conjecture further, we can use the calculator to find the 50th\begin{align*}50^{th}\end{align*} partial sum: S50=2\begin{align*}S_{50}=2\end{align*}. Eventually, if we sum enough terms, the calculator will give us the value to which the sum approaches due to rounding.

Finally, let's find the first five partial sums of n=11n\begin{align*}\sum \limits_{n=1}^{\infty} \frac{1}{n}\end{align*}, the “harmonic series” and make an observation about the sum of the infinite series. (You may need to find addition partial sums to see the behavior of the infinite series.)

Use the calculator to find the following sums:

S1S2S3S4S5=sum(seq(1/x,x,1,1))=1=sum(seq(1/x,x,1,2))=1.5=sum(seq(1/x,x,1,3))=1.833=sum(seq(1/x,x,1,4))=2.083=sum(seq(1/x,x,1,5))=2.283\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:

S50S100S500=4.499=5.187=6.793\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 n=12n(n+1)\begin{align*}\sum \limits_{n=1}^{\infty}\frac{2}{n(n+1)}\end{align*}, and make an observation about the sum of the infinite series.

The first five terms of this sequence are: 1,13,16,110,115\begin{align*}1, \frac{1}{3}, \frac{1}{6}, \frac{1}{10}, \frac{1}{15}\end{align*}. The partial sums are thus:

S1S2S3S4S5=1=1.333=1.5=1.6=1.6666\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

n=14(32)n1\begin{align*}\sum \limits_{n=1}^{\infty} 4 \left(\frac{3}{2}\right)^{n-1}\end{align*}

S1=4; S2=10; S3=19; S4=32.5; S5=52.75;\begin{align*}S_1=4; \ S_2=10; \ S_3=19; \ S_4=32.5; \ S_5=52.75;\end{align*} The partial sums are growing with increasing speed and thus the infinite series will have no bound.

#### Example 3

n=1500(23)n1\begin{align*}\sum \limits_{n=1}^{\infty} 500 \left(\frac{2}{3}\right)^{n-1}\end{align*}

S1=500; S2=833.333; S3=1055.556; S4=1203.704; S5=1302.469;\begin{align*}S_1=500; \ S_2=833.333; \ S_3=1055.556; \ S_4=1203.704; \ S_5=1302.469;\end{align*} Here the sums seems to be growing by smaller amounts each time. Look at the sum additional partial sums to see if there is an apparent upper limit to their growth. S50=1499.9999=1500; S100=1500\begin{align*}S_{50}=1499.9999 \ldots=1500; \ S_{100}=1500\end{align*}. The sum is clearly approaching 1500 and thus the infinite series has a finite sum.

#### Example 4

n=156n\begin{align*}\sum \limits_{n=1}^{\infty} \frac{5}{6n}\end{align*}

S1=0.833; S2=1.25; S3=1.528; S4=1.736; S5=1.903;\begin{align*}S_1=0.833; \ S_2=1.25; \ S_3=1.528; \ S_4=1.736; \ S_5=1.903;\end{align*} This sequence of partial sums is growing slowly, but will it approach a finite value or continue to grow? Look at additional partial sums: S50=3.749; S100=4.323; S500=5.661\begin{align*}S_{50}=3.749; \ S_{100}=4.323; \ S_{500}=5.661\end{align*}. In this case, the sums continue to grow without bound so the infinite series will have no bound.

### 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.

1. n=15(12)n1\begin{align*}\sum \limits_{n=1}^{\infty}5 \left(\frac{1}{2} \right)^{n-1}\end{align*}
2. n=12(34)n1\begin{align*}\sum \limits_{n=1}^{\infty}2 \left(\frac{3}{4} \right)^{n-1}\end{align*}
3. n=110(0.9)n1\begin{align*}\sum \limits_{n=1}^{\infty}10(0.9)^{n-1}\end{align*}
4. n=18(1.03)n1\begin{align*}\sum \limits_{n=1}^{\infty}8(1.03)^{n-1}\end{align*}
5. n=112n\begin{align*}\sum \limits_{n=1}^{\infty}\frac{1}{2}n\end{align*}
6. n=110n\begin{align*}\sum \limits_{n=1}^{\infty}\frac{10}{n}\end{align*}
7. n=112(34)n1\begin{align*}\sum \limits_{n=1}^{\infty}\frac{1}{2} \left(\frac{3}{4}\right)^{n-1}\end{align*}
8. n=11n2\begin{align*}\sum \limits_{n=1}^{\infty}\frac{1}{n^2}\end{align*}
9. n=16(0.1)n1\begin{align*}\sum \limits_{n=1}^{\infty}6(0.1)^{n-1}\end{align*}
10. n=10.01n+5\begin{align*}\sum \limits_{n=1}^{\infty}0.01n+5\end{align*}
11. n=12(78)n1\begin{align*}\sum \limits_{n=1}^{\infty}2 \left(\frac{7}{8}\right)^{n-1}\end{align*}
12. Which of the series above are arithmetic? Do any of them have a finite sum? Can you explain why?
13. Which of the series above are geometric? Do any of them have a finite sum? Can you explain why?
14. What is the difference between the series that converge and diverge?
15. Make a conjecture about the converging series in this problem set.

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

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