Reading a Series from a Partial-Sum Formula

From Partial Sums Back to Terms

A series is often introduced by listing its terms and asking what happens when they are added. But sometimes the information is presented in the reverse direction: a formula gives the sum of the first $n$ terms, and we must recover the terms themselves. This is possible because each new partial sum differs from the previous one by exactly one term.

That simple difference also connects the terms to the value of an infinite series. If the partial sums approach a finite limit, that limit is the series sum. If they do not approach a finite limit, the series does not converge to a finite value.

Definitions and the Consecutive-Difference Rule

Let $a_1,a_2,a_3,\ldots$ be the terms of a series. For $n\in\mathbb{N}$, its $n$th partial sum is the sum of its first $n$ terms. We also define the zeroth partial sum to be zero, representing a sum with no terms.

S_0=0, \qquad S_n=\sum_{k=1}^{n}a_k \quad(n\ge 1).

Subtracting the sum of the first $n-1$ terms from the sum of the first $n$ terms leaves only the $n$th term.

Theorem: For a series with partial sums $S_n$, its terms are recovered by $$ a_n=S_n-S_{n-1}\quad(n\ge 1), $$ where $S_0=0$.

The rule applies to any series whose partial sums are given, whether or not the infinite series converges. Recovering the terms and deciding convergence are separate tasks: the terms come from consecutive differences, while convergence depends on the behavior of the partial sums as $n$ grows.

Take care with the first term. The formula for $a_1$ is $a_1=S_1-S_0=S_1$. If a closed formula for $S_n$ is stated only for $n\ge 1$, use the defined value $S_0=0$; do not assume that substituting $n=0$ into the closed formula gives the zeroth partial sum.

Why the Difference Gives the Term

The rule follows directly from the definition of a partial sum. For $n\ge 2$, write out the two sums and cancel the terms they share. For $n=1$, the result follows from $S_1=a_1$ and $S_0=0$.

\begin{aligned} S_n-S_{n-1} &=(a_1+a_2+\cdots+a_{n-1}+a_n) -(a_1+a_2+\cdots+a_{n-1})\\ &=a_n. \end{aligned}

There is a useful reverse check as well. If we define each term by $a_k=S_k-S_{k-1}$, then summing the differences from $k=1$ to $n$ recovers the original partial sum. Every intermediate partial sum appears once positively and once negatively, except for $S_n$ and $S_0$.

\sum_{k=1}^{n}(S_k-S_{k-1}) =(S_1-S_0)+(S_2-S_1)+\cdots+(S_n-S_{n-1}) =S_n-S_0 =S_n.

This identity checks that the recovered terms really generate the stated partial sums. It also shows why the value of an infinite series, when it exists, is determined by the limit of those partial sums.

Worked Examples

Worked Example: Recover the Terms and Find the Series Sum

Suppose the partial sums are $S_n=\frac{n}{n+1}$ for $n\ge 1$, with $S_0=0$. Find $a_n$ and determine the infinite sum.

For $n=1$, the first term is $a_1=S_1-S_0=\frac{1}{2}-0=\frac{1}{2}$. For $n\ge 2$, apply the consecutive-difference rule:

\begin{aligned} a_n &=\frac{n}{n+1}-\frac{n-1}{n}\\ &=\frac{n^2-(n-1)(n+1)}{n(n+1)}\\ &=\frac{n^2-(n^2-1)}{n(n+1)}\\ &=\frac{1}{n(n+1)}. \end{aligned}

The same expression gives $a_1=\frac{1}{1\cdot2}=\frac12$, so it describes every term for $n\ge1$. To check the recovery, sum its consecutive differences from $k=1$ to $n$:

\sum_{k=1}^{n}a_k =\sum_{k=1}^{n}\left(\frac{k}{k+1}-\frac{k-1}{k}\right) =\frac{n}{n+1}-0 =\frac{n}{n+1} =S_n.

Finally, the given partial sums approach $1$, since $\frac{n}{n+1}=1-\frac{1}{n+1}$ and $\frac{1}{n+1}$ approaches zero. Therefore the series converges and its sum is $1$.

\sum_{n=1}^{\infty}a_n =\lim_{n\to\infty}S_n =\lim_{n\to\infty}\frac{n}{n+1} =1.

The conclusion is checked against the partial sums: their limit is exactly the claimed series sum.

Worked Example: Partial Sums That Grow Without Bound

Suppose $S_n=n^2+2n$ for $n\ge1$, and $S_0=0$. Recover $a_n$ and decide whether the series has a finite sum.

For $n\ge2$, the previous partial sum is

S_{n-1}=(n-1)^2+2(n-1) =n^2-2n+1+2n-2 =n^2-1.

Thus, for $n\ge2$,

a_n=S_n-S_{n-1} =(n^2+2n)-(n^2-1) =2n+1.

For $n=1$, $a_1=S_1-S_0=(1+2)-0=3$, which also agrees with $2(1)+1=3$. Therefore $a_n=2n+1$ for every $n\ge1$.

Check by adding the recovered differences. The sum from $k=1$ to $n$ is

\begin{aligned} \sum_{k=1}^{n}a_k &=\sum_{k=1}^{n}\bigl((k^2+2k)-((k-1)^2+2(k-1))\bigr)\\ &=S_n-S_0\\ &=n^2+2n. \end{aligned}

This reproduces the given partial sums. Since $n^2+2n$ grows without bound as $n$ increases, the partial sums do not approach a finite limit. The infinite series therefore diverges; it has no finite sum.

Worked Example: Oscillating Partial Sums

Suppose $S_n=1-(-1)^n$ for $n\ge1$, and $S_0=0$. Find the terms and determine whether the series converges.

For $n\ge2$, subtract the previous partial sum:

\begin{aligned} a_n &=\bigl(1-(-1)^n\bigr)-\bigl(1-(-1)^{n-1}\bigr)\\ &=(-1)^{n-1}-(-1)^n\\ &=2(-1)^{n-1}. \end{aligned}

For $n=1$, $a_1=S_1-S_0=(1-(-1))-0=2$, also matching $2(-1)^0=2$. Hence $a_n=2(-1)^{n-1}$ for every $n\ge1$.

Summing these consecutive differences verifies the formula: their sum through the $n$th term is $S_n-S_0=1-(-1)^n$. In particular, the partial sums are $2$ for odd $n$ and $0$ for even $n$. They keep alternating between these two values and do not approach a single limit. Thus the series diverges.

This example illustrates why knowing the terms is not enough to claim that an infinite sum exists. The convergence test here is the behavior of the partial sums themselves.

Using the Formula Carefully

A common mistake is to subtract the wrong partial sum or to mishandle the initial index. For example, with $S_n=\frac{n}{n+1}$, the incorrect calculation $a_n=S_n-S_{n+1}$ gives

\frac{n}{n+1}-\frac{n+1}{n+2} =\frac{n(n+2)-(n+1)^2}{(n+1)(n+2)} =\frac{-1}{(n+1)(n+2)}.

This cannot be the $n$th term: it is negative, even though the given partial sums are increasing. The correction is to subtract the previous partial sum, not the next one:

a_n=S_n-S_{n-1}.

Also, when $n=1$, use $S_0=0$. For instance, if a formula is supplied only for positive $n$, extending it to $n=0$ is not part of the definition. The first term must be found from $a_1=S_1$.

Once the terms are recovered, determine convergence from the limit of the partial sums. A finite limit gives the series sum; unbounded or oscillating partial sums do not yield a finite sum. A closed formula for $a_n$ alone does not replace this check.

Why Partial-Sum Formulas Matter

A partial-sum formula packages the result of adding many terms. Taking consecutive differences reverses that packaging and reveals the contribution added at each stage. The method can identify individual terms, check a proposed description of a series, and determine its sum whenever the partial sums have a finite limit.

The essential sequence of reasoning is concise: interpret the given formula as $S_n$, calculate $S_n-S_{n-1}$ to recover $a_n$, and examine $\lim_{n\to\infty}S_n$ to decide whether a finite series sum exists. Keeping these steps distinct helps prevent both indexing errors and unsupported convergence claims.

Check Your Understanding

For each question, pay attention to the index and the role of the partial sums.

If $S_n=2n$ for $n\ge1$ and $S_0=0$, what is $a_n$? Does the series have a finite sum?
Why must the first term be calculated as $a_1=S_1-S_0$?
If $S_n$ approaches $L$, what is the value of the infinite series, provided the series is defined by those partial sums?
What does it imply about convergence if the partial sums alternate between two distinct values?
How can summing the differences $S_k-S_{k-1}$ check that a recovered term formula is correct?