Use a partial-sum formula
A running total that levels off
The partial sum \(S_n\) of a series is the total of its first \(n\) terms. Here \(S_n=\frac{7n}{n+4}\), and the bar at \(n\) has height \(S_n\).
Substitute values of \(n\) to find partial sums: \(S_1=\frac75\), \(S_3=3\) and \(S_{10}=5\).
Move the slider to the right. The bars rise, but each rise is smaller than the one before, and at \(n=40\) the partial sum is \(\frac{70}{11}\), about 6.36.
The orange segment is the gap \(7-S_n\) between the bar and the dashed line at height 7. This gap equals \(\frac{28}{n+4}\), which shrinks toward 0, so the partial sums approach 7 and the series adds up to 7.
The sum is the height where the bars level off
The bars show any partial sums \(S_1, S_2, S_3,\dots\). These partial sums form a sequence, so you find its limit as in the last skill.
If the bars level off at a height \(L\), the series has the sum \(L\).
If the bars rise past every height or jump between two heights, the partial sums have no limit, so the series has no sum.
Whether the series has a sum depends on the limit of the partial sums, not on the size of any one of them.
Notation
A formula for \(S_n\) gives every partial sum. Substitute the number of terms for \(n\).
\[\begin{aligned}\cb{S_n}&=\dfrac{7n}{n+4}\\ \cb{S_3}&=\dfrac{7\cdot3}{3+4}=3\end{aligned}\]The sum of a series with terms \(a_n\) is the limit of its partial sums.
\[\sum_{n=1}^{\infty}a_n=\lim_{n\to\infty}\cb{S_n}=\cy{L}\]If the limit is a number, the series converges to it. If the limit is infinite or does not exist, the series diverges and has no sum.
When it fails
The limit of \(S_n\) must be a number. For \(S_n=\frac{n^2}{n+4}\), the top grows faster than the bottom, so the partial sums pass every height.
The bars rise for both formulas, so rising alone does not decide the question. The series has a sum only if the bars level off at a height.
Example 1: A closed form that levels off
A series has the partial sums \(S_n=12\left(1-\dfrac{1}{3^n}\right)\). Find \(S_1\), \(S_2\) and the sum of the series.
Substitute \(n=1\) and \(n=2\) into the formula.
\[\begin{aligned}S_1&=12\left(1-\tfrac13\right)=8\\ S_2&=12\left(1-\tfrac19\right)=\tfrac{32}{3}\end{aligned}\]Because 3 is greater than 1, \(3^n\) grows without bound, so \(\frac1{3^n}\) approaches 0.
\[\lim_{n\to\infty}S_n=12(1-0)=12\]The bars approach the dashed line at 12, so the series converges and its sum is 12.
Example 2: One partial sum is not the sum
A series has the partial sums \(S_n=\dfrac{6n}{n+9}\). Find the sum of the series.
Substitute a large number first, to see where the bars go. For \(n=99\) the partial sum is 5.5.
\[S_{99}=\dfrac{6\cdot99}{99+9}=5.5\]A tempting wrong move: The partial sum at \(n=99\) is 5.5, so the sum of the series is 5.5.
\[S=S_{99}=5.5\]The orange bracket shows that the bar at \(n=99\) is still 0.5 below the dashed line at 6. The bars \(S_{100}\), \(S_{101}\) and so on are higher.
The sum is the limit of the partial sums, not one of them. Divide the top and the bottom by \(n\). Then let \(n\) grow.
\[\lim_{n\to\infty}\frac{6n}{n+9}=\lim_{n\to\infty}\frac{6}{1+\frac9n}=6\]
Practice
A series has the partial sums \(S_n=\dfrac{5n^2+2}{2n^2+3n}\). Does the series converge? If it does, find its sum.
Move the slider to large \(n\). Read the height at which the bars level off.
Solution
Divide the top and the bottom by \(n^2\), the highest power of \(n\).
\[S_n=\frac{5+\frac{2}{n^2}}{2+\frac3n}\]As \(n\) grows, \(\frac2{n^2}\) and \(\frac3n\) approach 0. The limit is a number, so the series converges.
\[\lim_{n\to\infty}S_n=\frac52\]The series has the sum \(\frac52\).
Questions
Questions 1 and 2 use the figure. It shows the partial sums \(S_n=\frac{8n}{n+3}\) as bars, for \(n=1\) to 12. The vertical axis is marked from 1 to 8.
What is the sum of the first 5 terms, \(S_5\)?
- 5
- 8
- 4
Answer
5
What is the sum of the series?
- 8
- 6.4
- The series has no sum, because the bars rise
Answer
8
A series has the partial sums \(S_n=\dfrac{n^2+4}{2n+1}\). Which statement is correct?
- The series diverges, because \(S_n\) grows without bound.
- The series converges to \(\frac12\).
- The series converges to 4.
Answer
The series diverges, because \(S_n\) grows without bound.
Summary
You decide whether a series converges by taking the limit of its partial sums.
This works because the sum of a series is defined as the number that the partial sums approach.
You use it when a problem gives a formula for the running total \(S_n\), to find the sum or to show that there is none.
This is one of five skills in topic 10.1 of Which & Why: AP Calculus.