Proof of the Necessary Condition

The Proof Comes from Consecutive Partial Sums

For a series $\sum_{n=1}^{\infty}a_n$, write $S_N=\sum_{n=1}^{N}a_n$, with $S_0=0$. The Necessary Condition for Series Convergence, stated in “The Necessary Condition for Series Convergence,” says that if these partial sums converge, then the terms $a_n$ approach zero. The proof rests on a simple observation: each term is the difference between two consecutive partial sums.

a_n=S_n-S_{n-1}.

If both partial sums on the right are close to the same limit, their difference must be small. The main task in making this argument rigorous is to choose the error tolerance and the index threshold so that both $S_n$ and $S_{n-1}$ are covered. We first isolate the sequence fact that makes the reasoning work.

Proposition (Consecutive Differences of a Convergent Sequence): Suppose $(x_n)_{n=0}^{\infty}$ is a sequence of real numbers and $x_n\to L$. Then $x_n-x_{n-1}\to0$ as $n\to\infty$.

Proof. Let $\varepsilon>0$. Since $x_n\to L$, there is an integer $N_0\geq0$ such that whenever $k\geq N_0$,

|x_k-L|<\frac{\varepsilon}{2}.

If $n\geq N_0+1$, then both $n\geq N_0$ and $n-1\geq N_0$. Therefore,

|x_n-x_{n-1}| =|(x_n-L)-(x_{n-1}-L)| \leq |x_n-L|+|x_{n-1}-L| <\frac{\varepsilon}{2}+\frac{\varepsilon}{2} =\varepsilon.

This is exactly the definition of $x_n-x_{n-1}\to0$. The shift from $N_0$ to $N_0+1$ matters: it ensures that the preceding term $x_{n-1}$, as well as $x_n$, is in the range where the convergence estimate applies. $\square$

Applying the Sequence Proposition to a Series

Suppose $\sum_{n=1}^{\infty}a_n$ converges to $S$. By definition, its partial sums satisfy $S_N\to S$. Apply the proposition with $x_N=S_N$. Since $a_n=S_n-S_{n-1}$, it follows that $a_n\to0$.

Proof (Necessary Condition for Series Convergence): Suppose $\sum_{n=1}^{\infty}a_n$ converges, and let its sum be $S$. Given $\varepsilon>0$, choose $N_0$ so that $|S_k-S|<\varepsilon/2$ whenever $k\geq N_0$. For every $n\geq N_0+1$, both $n$ and $n-1$ are at least $N_0$. Hence $$ |a_n|=|S_n-S_{n-1}| =|(S_n-S)-(S_{n-1}-S)| \leq |S_n-S|+|S_{n-1}-S| <\varepsilon. $$ Thus $a_n\to0$.

This proof is an application of the general sequence proposition, not an additional convergence test. Its essential ingredients are the partial-sum identity and the fact that two sufficiently late partial sums are both close to the same limit. The tolerance is split into two parts because the triangle inequality combines two errors.

Worked Example: Partial Sums Approaching One

Define $S_0=0$, and for $N\geq1$ let

S_N=\frac{N}{N+2}.

Then $S_N\to1$, since

1-S_N=1-\frac{N}{N+2}=\frac{2}{N+2}\longrightarrow0.

Define the terms by $a_n=S_n-S_{n-1}$. For $n\geq1$, the calculation is

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

For example, $a_1=2/(2\cdot3)=1/3=S_1$, and $a_2=2/(3\cdot4)=1/6=S_2-S_1=1/2-1/3$. In general, $a_n>0$ and $a_n\leq 2/(n+1)^2$, so $a_n\to0$. The sequence proposition explains this limit from $S_N\to1$, without needing the explicit formula for $a_n$.

Worked Example: Partial Sums with a Small Oscillation

Let $S_0=0$, and define, for $N\geq1$,

S_N=-1+\frac{(-1)^N}{N+1}.

The oscillating part tends to zero because its absolute value is $1/(N+1)$. Thus $S_N\to-1$. The terms obtained by taking consecutive differences are

a_n=S_n-S_{n-1} =(-1)^n\left(\frac{1}{n+1}+\frac{1}{n}\right) =\frac{(-1)^n(2n+1)}{n(n+1)}.

At the first two indices, $a_1=-3/2=S_1-S_0$ and $a_2=5/6=S_2-S_1$, as the formula gives. For every $n\geq1$,

|a_n|=\frac{1}{n}+\frac{1}{n+1}\leq\frac{2}{n},

so $a_n\to0$. The signs alternate, but the proof does not depend on their pattern: it depends on the convergence of the partial sums.

A Consequence for Finite Blocks of Terms

The same reasoning controls more than a single term. A sum of terms over a block is also a difference of partial sums. When the block has a fixed length and moves farther out, both endpoints of that difference approach the same series sum.

Proposition (Fixed-Length Blocks Tend to Zero): Suppose $\sum_{n=1}^{\infty}a_n=S$ converges. For each fixed positive integer $r$, $$ \sum_{k=n}^{n+r-1}a_k\longrightarrow0 \quad\text{as }n\to\infty. $$

Proof. By the block-sum identity for partial sums,

\sum_{k=n}^{n+r-1}a_k=S_{n+r-1}-S_{n-1}.

Let $\varepsilon>0$. Since $S_N\to S$, choose $N_0$ so that $|S_j-S|<\varepsilon/2$ whenever $j\geq N_0$. If $n\geq N_0+1$, then both $n-1\geq N_0$ and $n+r-1\geq N_0$. Consequently,

\left|\sum_{k=n}^{n+r-1}a_k\right| =|(S_{n+r-1}-S)-(S_{n-1}-S)| \leq |S_{n+r-1}-S|+|S_{n-1}-S| <\varepsilon.

This proves the assertion. The length $r$ is fixed, although the argument also shows directly why the location of the block is what matters: its two partial-sum indices both move to infinity. $\square$

Worked Example: Terms from a Prescribed Convergent Sequence

Set $S_0=0$ and, for $N\geq1$, let

S_N=1-\frac{1}{\sqrt{N+1}}.

Since $1/\sqrt{N+1}\to0$, the partial sums converge to $1$. The corresponding terms are

a_n=S_n-S_{n-1} =\left(1-\frac{1}{\sqrt{n+1}}\right) -\left(1-\frac{1}{\sqrt n}\right) =\frac{1}{\sqrt n}-\frac{1}{\sqrt{n+1}}.

For $n=1$, this gives $a_1=1-1/\sqrt2=S_1$. For every $n\geq1$, $a_n$ is positive and satisfies

0<a_n=\frac{1}{\sqrt n}-\frac{1}{\sqrt{n+1}} \leq\frac{1}{\sqrt n}.

Because $1/\sqrt n\to0$, the terms tend to zero. Moreover, for any fixed positive integer $r$,

\sum_{k=n}^{n+r-1}a_k =S_{n+r-1}-S_{n-1} =\frac{1}{\sqrt n}-\frac{1}{\sqrt{n+r}},

which also tends to zero as $n\to\infty$. This verifies the fixed-length block conclusion directly in this example.

What the Proof Does—and Does Not—Say

The proof uses convergence of the partial sums to control the difference between neighboring partial sums. It does not assume that the terms are positive, decreasing, or otherwise regular. The terms may change sign or behave irregularly; if the partial sums converge, the difference identity still forces each term to become small.

A common mistake is to reverse the implication. The proof establishes that convergence of the series implies $a_n\to0$; it does not establish that $a_n\to0$ implies convergence. For instance, the terms $1/n$ approach zero, but this necessary-condition proof alone does not decide the convergence of their series. The condition is decisive when it fails, and inconclusive when it holds.

Another proof-writing pitfall is to say only that “the difference of two convergent sequences tends to zero.” In this setting the two sequences are $S_n$ and $S_{n-1}$, and the limit calculation is valid, but an epsilon proof makes the index shift explicit. One first requires both indices to lie beyond the convergence threshold, which is why $n\geq N_0+1$ is used. Likewise, the error $\varepsilon$ must be divided between the two partial-sum estimates.

Write the term as a difference.
Use $a_n=S_n-S_{n-1}$, so the series question becomes a statement about consecutive partial sums.

Use the common limit.
If $S_N\to S$, choose a threshold beyond which each partial sum is within $\varepsilon/2$ of $S$.

Cover both indices.
Take $n$ at least one greater than the threshold, ensuring that both $S_n$ and $S_{n-1}$ satisfy the estimate.

Apply the triangle inequality.
The two errors add to less than $\varepsilon$, proving that $|a_n|$ tends to zero.

Takeaway: A term of a series is the change between two consecutive partial sums. If the partial sums converge, both are eventually close to the same limit, so their difference tends to zero. The same argument shows that any fixed-length block of terms, moved far enough into the series, has sum tending to zero.

Check Your Understanding

Use the consecutive-difference proof and its block-sum consequence to answer the following questions.

Why is it enough to require $n\geq N_0+1$ when estimating $S_n-S_{n-1}$?
In the proof, why is the tolerance for each partial sum chosen to be $\varepsilon/2$?
If $x_n\to L$, what can be concluded about $x_n-x_{n-1}$, and what inequality proves it?
For a convergent series, express the sum from $a_n$ through $a_{n+r-1}$ as a difference of partial sums.
Does $a_n\to0$ by itself prove that $\sum_{n=1}^{\infty}a_n$ converges? Explain what the necessary-condition proof establishes.

Proof of the Necessary Condition

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