Series of Nonnegative Terms

Why Nonnegative Terms Change the Picture

For a general series, terms of opposite signs can cancel: partial sums may rise and fall while still approaching a finite limit. When every term is nonnegative, cancellation is impossible. Each new term either leaves the partial sum unchanged or increases it. This simple fact has strong consequences: convergence becomes a question of bounded growth, and controlling one long tail controls every shorter block within it.

Let $a_n\geq0$ for every $n$, and define the partial sums by $S_0=0$ and $S_N=\sum_{n=1}^{N}a_n$ for $N\geq1$. The Block-Sum Identity gives, whenever $M>N$,

S_M-S_N=\sum_{n=N+1}^{M}a_n\geq0.

Thus $(S_N)$ is nondecreasing. The earlier theorem Bounded Increasing Partial Sums says that a nondecreasing sequence of partial sums converges if it is bounded above. If it is unbounded above, it tends to $+\infty$, by Unbounded Increasing Partial Sums Tend to Infinity. Consequently, a series of nonnegative terms has only two possibilities: its partial sums converge to a finite real number, or they grow without bound and tend to $+\infty$.

Definition: A series $\sum_{n=1}^{\infty}a_n$ is a series of nonnegative terms if $a_n\geq0$ for every $n$. Its partial sums $S_N=\sum_{n=1}^{N}a_n$ form a nondecreasing sequence. The series converges precisely when these partial sums are bounded above.

The last equivalence uses the earlier theorem; it is not true for arbitrary series. For example, bounded partial sums that repeatedly rise and fall need not converge. Here, monotonicity rules out that behavior.

A Tail Criterion for Nonnegative Series

The Cauchy Criterion for Series requires every sufficiently late finite block to have small sum. For nonnegative terms, it is enough to control the entire tail starting from one fixed index: each later block is no larger than a tail that contains it. This gives a particularly useful form of the criterion.

Theorem (Tail Criterion for Nonnegative Series): Suppose $a_n\geq0$ for every $n$. Then $\sum_{n=1}^{\infty}a_n$ converges if and only if, for every $\varepsilon>0$, there is an integer $N\geq1$ such that $$ \sum_{n=N}^{M}a_n<\varepsilon $$ for every integer $M\geq N$.

Proof. Suppose first that the series converges. By the Cauchy Criterion for Series, for the given $\varepsilon>0$, there is an integer $N$ such that

\left|\sum_{n=p}^{q}a_n\right|<\varepsilon

whenever $q\geq p\geq N$. Taking $p=N$ and $q=M$ gives the required bound. The absolute value can be removed because the sum is nonnegative.

Conversely, suppose the stated tail bound holds for every $\varepsilon>0$. Fix $\varepsilon>0$, and choose $N$ for that bound. For any $q\geq p\geq N$, nonnegativity gives

0\leq\sum_{n=p}^{q}a_n \leq\sum_{n=N}^{q}a_n <\varepsilon.

The second sum contains every term in the first, as well as any terms from $N$ through $p-1$; all those additional terms are nonnegative. Hence every sufficiently late finite block has absolute value less than $\varepsilon$. The Cauchy Criterion for Series proves convergence. $\square$

The uniformity in $M$ is essential. It is not enough to find, for each endpoint $M$, some starting index that makes the sum small. One starting index $N$, chosen from $\varepsilon$, must work for every finite endpoint $M\geq N$. The criterion is useful precisely because it captures that uniform control.

Corollary (Remainders Decrease to Zero): If $\sum_{n=1}^{\infty}a_n=S$ converges and $a_n\geq0$, then $R_N=S-S_N$ is nonnegative, nonincreasing, and tends to zero.

Proof. Since $S_N\to S$, $R_N=S-S_N\to0$. Also, $S_N\leq S$: the partial sums are nondecreasing and converge to $S$, so none can exceed their limit. Thus $R_N\geq0$. Finally,

R_{N+1}=S-S_{N+1}=S-S_N-a_{N+1}=R_N-a_{N+1}\leq R_N.

This proves all three claims. In particular, for $M>N$, the finite tail satisfies

\sum_{n=N+1}^{M}a_n=S_M-S_N\leq S-S_N=R_N.

So a remainder estimate bounds every finite block that begins after $N$.

Worked Applications

Worked Example: An Exact Sum from a Finite Telescope

Consider the nonnegative terms $a_n=1/(n(n+2))$. For every $n\geq1$,

\frac{1}{n(n+2)} =\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right),

because $\frac{1}{n}-\frac{1}{n+2}=\frac{(n+2)-n}{n(n+2)}=\frac{2}{n(n+2)}$. The finite sum therefore telescopes:

S_N =\frac{1}{2}\sum_{n=1}^{N}\left(\frac{1}{n}-\frac{1}{n+2}\right) =\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{N+1}-\frac{1}{N+2}\right).

The last two terms tend to zero as $N\to\infty$. Hence $S_N\to\frac{3}{4}$, and the series converges with sum $\frac{3}{4}$. The remainder is also explicit:

R_N=\frac{1}{2}\left(\frac{1}{N+1}+\frac{1}{N+2}\right).

This is nonnegative and tends to zero, in agreement with the Remainders Decrease to Zero corollary.

Worked Example: Convergence by Comparison

Consider $a_n=n/(n^3+1)$. Each term is positive. For $n\geq1$, multiplying by the positive quantity $n^2(n^3+1)$ shows that

\frac{n}{n^3+1}\leq\frac{1}{n^2} \quad\Longleftrightarrow\quad n^3\leq n^3+1,

which is true. The $p$-Series Convergence Criterion says that $\sum_{n=1}^{\infty}1/n^2$ converges. The Comparison Test for Nonnegative Series therefore gives convergence of $\sum_{n=1}^{\infty}n/(n^3+1)$.

The same comparison also gives a tail estimate. For $M\geq N$,

0\leq\sum_{n=N}^{M}\frac{n}{n^3+1} \leq\sum_{n=N}^{M}\frac{1}{n^2}.

Because the comparison series converges, its tails can be made arbitrarily small. The Tail Criterion then gives another way to see why the original series converges.

Worked Example: The Harmonic Series Has Unbounded Partial Sums

Consider $\sum_{n=1}^{\infty}1/n$. For each integer $k\geq0$, take the block from $2^k$ through $2^{k+1}-1$. It contains $2^k$ terms. Every index $n$ in this block satisfies $n<2^{k+1}$, so $1/n>1/2^{k+1}$. Consequently,

\sum_{n=2^k}^{2^{k+1}-1}\frac{1}{n} > 2^k\frac{1}{2^{k+1}} =\frac{1}{2}.

Adding the blocks for $k=0,1,\ldots,K$ shows that

S_{2^{K+1}-1} =\sum_{k=0}^{K}\sum_{n=2^k}^{2^{k+1}-1}\frac{1}{n} >\frac{K+1}{2}.

These partial sums are unbounded as $K$ increases. Since they are nondecreasing, the theorem Unbounded Increasing Partial Sums Tend to Infinity shows that they tend to $+\infty$. Thus the harmonic series diverges. This argument controls whole blocks, rather than relying only on the fact that the individual terms tend to zero.

Finite Subsets and Subseries

Nonnegativity lets us compare not only consecutive blocks, but also sums over arbitrary finite collections of indices. This gives a useful characterization of the total sum and a direct result about subseries. A subseries is obtained by keeping terms whose indices belong to a chosen subset of the positive integers, in their original order.

Theorem (Finite-Subset Characterization): Suppose $a_n\geq0$ for every $n$. The series $\sum_{n=1}^{\infty}a_n$ converges if and only if the sums $\sum_{n\in F}a_n$, over all finite subsets $F$ of the positive integers, are bounded above by one finite constant. When the series converges to $S$, $$ S=\sup\left\{\sum_{n\in F}a_n:F\subseteq\mathbb{N}\text{ is finite}\right\}. $$

Proof. Suppose the series converges to $S$. Let $F$ be any finite subset of the positive integers. If $F$ is empty, its sum is zero and is at most $S$, since $S\geq0$. Otherwise, let $K$ be the largest element of $F$. Since all terms are nonnegative,

0\leq\sum_{n\in F}a_n\leq\sum_{n=1}^{K}a_n=S_K\leq S.

Thus all finite-subset sums are bounded above by $S$. Their supremum is at least $S$ as well: each initial segment $\{1,\ldots,N\}$ is a finite subset, and its sum $S_N$ tends to $S$. Therefore the supremum is exactly $S$.

Conversely, suppose there is a finite constant $C$ such that every finite-subset sum is at most $C$. In particular, for each $N$, the initial segment is a finite subset, so $S_N\leq C$. The partial sums are nondecreasing because the terms are nonnegative. By the earlier theorem Bounded Increasing Partial Sums, $(S_N)$ converges, and hence the series converges. This proves both directions and the supremum formula. $\square$

Corollary (Subseries of a Convergent Nonnegative Series): Every subseries of a convergent series of nonnegative terms converges, and its sum is at most the sum of the original series.

Proof. Let $E$ be the set of indices retained. If $E$ is finite, the subseries is a finite sum and converges. If $E$ is infinite, list its indices in increasing order as $e_1<e_2<\cdots$. Every partial sum of the subseries,

\sum_{j=1}^{r}a_{e_j},

is a sum over a finite subset of the original indices. By the Finite-Subset Characterization, it is at most the sum $S$ of the original series. The subseries partial sums are nondecreasing, since each added term is nonnegative, and they are bounded above by $S$. The theorem Bounded Increasing Partial Sums implies that they converge. Their limit, the sum of the subseries, is at most $S$. $\square$

Worked Example: Keeping Only Even-Indexed Terms

Consider $a_n=2^{-n}$, whose terms are nonnegative. Keeping only the even-indexed terms gives the subseries

\sum_{k=1}^{\infty}a_{2k} =\sum_{k=1}^{\infty}4^{-k}.

Its $M$-th partial sum, by the finite geometric-sum formula, is

\sum_{k=1}^{M}4^{-k} =\frac{1}{3}\left(1-4^{-M}\right).

Since $4^{-M}\to0$, the subseries converges to $1/3$. The original series $\sum_{n=1}^{\infty}2^{-n}$ sums to $1$, so the subseries sum is no larger, as the corollary guarantees. More generally, the corollary does not require the retained indices to follow a simple pattern: any chosen subset gives a convergent subseries.

What to Watch For

The defining advantage of nonnegative terms is monotonicity, not merely the smallness of individual terms. The Necessary Condition for Series Convergence says that convergence forces $a_n\to0$, but that condition alone does not control the accumulation of many small positive terms. The harmonic-series example shows how blocks can keep adding a fixed positive amount even though their individual terms become small.

There is also a useful distinction between controlling one block and controlling all blocks. For general terms, a small sum over one particular grouping does not guarantee that every late block is small. For nonnegative terms, the Tail Criterion works because a block from $p$ through $q$ is bounded by the larger sum from $N$ through $q$, whenever $p\geq N$. Without nonnegativity, that inequality can fail because added terms may cancel.

Takeaway: For a series of nonnegative terms, partial sums increase, so boundedness is equivalent to convergence. Its tails can be tested by one uniform cutoff, and its sum bounds every finite sum—and therefore every subseries—formed from its terms.

Check Your Understanding

Use the results in this tutorial to answer the following questions.

Why does nonnegativity make the partial sums nondecreasing?
In the Tail Criterion, why does a bound on $\sum_{n=N}^{M}a_n$ control every block from $p$ through $q$ with $q\geq p\geq N$?
What additional fact, beyond $a_n\to0$, is needed to establish convergence of a nonnegative series?
Why is the supremum of finite-subset sums equal to the sum of a convergent nonnegative series?
How does boundedness of the original sum bound the partial sums of any subseries?

What You'll Learn