Series Convergence & Tests

Given a sequence $(a_n)_{n=1}^\infty$ in $\mathbb{R}$, the infinite series $\sum_{n=1}^\infty a_n$ is the sequence of partial sums $(S_n)_{n=1}^\infty$ defined by

$S_n = \sum_{k=1}^n a_k = a_1 + a_2 + \cdots + a_n.$

The terms $a_n$ are the terms of the series and $S_n$ is the $n$th partial sum.

The series $\sum_{n=1}^\infty a_n$ converges if the sequence of partial sums $(S_n)$ converges — that is, if $\lim_{n \to \infty} S_n = S$ exists as a finite real number. We write $\sum_{n=1}^\infty a_n = S$ and call $S$ the sum of the series. If $(S_n)$ does not converge, the series diverges.

Definitions 1 and 2 convert every series problem into a sequence problem. The Monotone Convergence Theorem, the Cauchy criterion, and the Algebra of Limits — all from Sequences, Limits & Convergence — now apply to partial sums. In particular:

(a) If all $a_n \geq 0$, then $(S_n)$ is increasing, so $\sum a_n$ converges iff $(S_n)$ is bounded (Monotone Convergence).

(b) $\sum a_n$ converges iff for every $\varepsilon > 0$ there exists $N$ such that $|S_m - S_n| = \left|\sum_{k=n+1}^m a_k\right| < \varepsilon$ for all $m > n \geq N$ (Cauchy criterion for series).

If $\sum_{n=1}^\infty a_n$ converges, then $\lim_{n \to \infty} a_n = 0$.

Equivalently (contrapositive): if $a_n \not\to 0$, then $\sum a_n$ diverges.

If the series converges with sum $S$, then $S_n \to S$ and $S_{n-1} \to S$. Since $a_n = S_n - S_{n-1}$, the Algebra of Limits gives $a_n \to S - S = 0$.

The harmonic series $\sum 1/n$ has $a_n = 1/n \to 0$, but it diverges (we prove this in the next section). Having $a_n \to 0$ is necessary for convergence, but far from sufficient. The convergence tests in the following sections provide stronger criteria.

For $r \in \mathbb{R}$:

$\sum_{n=0}^\infty r^n = \frac{1}{1-r} \quad \text{if } |r| < 1, \qquad \text{diverges if } |r| \geq 1.$

The partial sum has the closed form $S_n = \sum_{k=0}^{n} r^k = \frac{1 - r^{n+1}}{1 - r}$ (verified by multiplying both sides by $1 - r$ and telescoping). If $|r| < 1$, then $r^{n+1} \to 0$ (from Sequences, Limits & Convergence, the sequence $|r|^n \to 0$ for $|r| < 1$), so $S_n \to \frac{1}{1-r}$. If $|r| \geq 1$, then $r^n \not\to 0$, so $S_n$ does not converge.

The geometric series is the most important series in mathematics and in ML. In reinforcement learning, the discounted return $G_t = \sum_{k=0}^\infty \gamma^k R_{t+k+1}$ is a geometric series weighted by rewards — convergence requires $|\gamma| < 1$, which is why discount factors satisfy $\gamma \in [0, 1)$. In momentum-based optimization (e.g., Adam), the bias correction factor $1/(1 - \beta^t) \to 1/(1 - \beta) = \sum_{k=0}^\infty \beta^k$ is the geometric series sum. In the ratio test (Theorem 6), convergence is established by showing the terms eventually decay faster than a geometric series.

(a) $\sum_{n=0}^\infty (1/2)^n = 1/(1-1/2) = 2$.

(b) $\sum_{n=0}^\infty (-1/3)^n = 1/(1+1/3) = 3/4$.

(c) $\sum_{n=1}^\infty 3 \cdot (0.9)^n = 3 \cdot \frac{0.9}{1-0.9} = 27$.

For $p > 0$, the $p$-series is $\sum_{n=1}^\infty \frac{1}{n^p}$.

$\sum_{n=1}^\infty \frac{1}{n^p}$ converges if and only if $p > 1$.

We use the Cauchy condensation test. Consider the “condensed” series $\sum_{k=0}^\infty 2^k \cdot a_{2^k} = \sum_{k=0}^\infty 2^k \cdot \frac{1}{2^{kp}} = \sum_{k=0}^\infty 2^{k(1-p)}$. This is a geometric series with ratio $2^{1-p}$, which converges iff $2^{1-p} < 1$, i.e., $1 - p < 0$, i.e., $p > 1$.

The Cauchy condensation theorem states: for a positive decreasing sequence $(a_n)$, $\sum a_n$ converges iff $\sum 2^k a_{2^k}$ converges. The key idea is that the block of terms $a_{2^k+1}, a_{2^k+2}, \ldots, a_{2^{k+1}}$ contains $2^k$ terms, each between $a_{2^{k+1}}$ and $a_{2^k}$ (since the sequence is decreasing). So the block sum is bounded between $2^k a_{2^{k+1}}$ and $2^k a_{2^k}$, and the condensed series captures the growth rate exactly.

Setting $p = 1$: $\sum 1/n = \infty$.

Alternative proof (Oresme’s grouping):

$1 + \frac{1}{2} + \left(\frac{1}{3} + \frac{1}{4}\right) + \left(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}\right) + \cdots \geq 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \cdots = \infty.$

Each group of $2^k$ terms sums to at least $1/2$, so the partial sums grow without bound. This is perhaps the most important divergence result in analysis — it says that even though $1/n \to 0$, the terms don’t shrink fast enough to produce a finite sum.

• $\sum 1/n^2 = \pi^2/6$ (Basel problem, converges — $p = 2 > 1$).
• $\sum 1/n^{3/2}$ converges ($p = 3/2 > 1$).
• $\sum 1/\sqrt{n}$ diverges ($p = 1/2 \leq 1$).

Let $0 \leq a_n \leq b_n$ for all $n \geq N_0$. Then:

1. If $\sum b_n$ converges, then $\sum a_n$ converges.
2. If $\sum a_n$ diverges, then $\sum b_n$ diverges.

(1) The partial sums $S_n^{(a)} = \sum_{k=1}^n a_k$ are increasing (since $a_k \geq 0$) and bounded above by $S_{N_0}^{(a)} + \sum_{k=N_0+1}^\infty b_k < \infty$. By the Monotone Convergence Theorem (Sequences, Limits & Convergence, Theorem 1), $S_n^{(a)}$ converges.

(2) Contrapositive of (1).

(a) $\sum \frac{1}{n^2 + n}$ converges: $\frac{1}{n^2+n} < \frac{1}{n^2}$ and $\sum 1/n^2$ converges.

(b) $\sum \frac{\ln n}{n}$ diverges: $\frac{\ln n}{n} \geq \frac{1}{n}$ for $n \geq 3$ and $\sum 1/n$ diverges.

Let $a_n > 0$ and $b_n > 0$ for all $n$. If $\lim_{n \to \infty} \frac{a_n}{b_n} = L$ where $0 < L < \infty$, then $\sum a_n$ and $\sum b_n$ either both converge or both diverge.

Since $a_n/b_n \to L > 0$, for $\varepsilon = L/2$ there exists $N$ such that $n \geq N \Rightarrow L/2 < a_n/b_n < 3L/2$. Thus $(L/2)b_n < a_n < (3L/2)b_n$ for all $n \geq N$. If $\sum b_n$ converges, the direct comparison gives $\sum_{n \geq N} a_n \leq (3L/2)\sum_{n \geq N} b_n < \infty$, so $\sum a_n$ converges. If $\sum b_n$ diverges, then $\sum_{n \geq N} a_n \geq (L/2)\sum_{n \geq N} b_n = \infty$.

$\sum \frac{n}{n^3 + 1}$ converges: compare with $b_n = 1/n^2$, since $\frac{n/(n^3+1)}{1/n^2} = \frac{n^3}{n^3+1} \to 1$, and $\sum 1/n^2$ converges.

The direct comparison test requires an explicit inequality $a_n \leq b_n$ or $a_n \geq b_n$, which can be algebraically demanding. The limit comparison test requires only that $a_n \sim L \cdot b_n$ asymptotically, which is usually easier to verify. In practice, the limit comparison test with $p$-series benchmarks handles most polynomial-type series.

Let $(a_n)$ be a sequence with $a_n \neq 0$ for all $n$. Define $L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|$ (if the limit exists, or use $\limsup$).

1. If $L < 1$, the series $\sum a_n$ converges absolutely.
2. If $L > 1$ (or $L = \infty$), the series diverges.
3. If $L = 1$, the test is inconclusive.

(1) If $L < 1$, choose $r$ with $L < r < 1$. By definition of the limit, there exists $N$ such that $|a_{n+1}/a_n| < r$ for all $n \geq N$. Then $|a_n| < |a_N| \cdot r^{n-N}$ for $n > N$ (by induction). Since $\sum r^{n-N}$ is a convergent geometric series ($r < 1$), the comparison test gives $\sum |a_n| < \infty$.

(2) If $L > 1$, eventually $|a_{n+1}| > |a_n|$, so $|a_n|$ is eventually increasing and $a_n \not\to 0$. The series diverges by the divergence test.

(a) $\sum \frac{n!}{n^n}$: $\frac{a_{n+1}}{a_n} = \frac{(n+1)!/(n+1)^{n+1}}{n!/n^n} = \frac{n^n}{(n+1)^n} = \left(\frac{n}{n+1}\right)^n \to 1/e < 1$. Converges.

(b) $\sum \frac{2^n}{n!}$: $\frac{a_{n+1}}{a_n} = \frac{2}{n+1} \to 0 < 1$. Converges.

(c) $\sum \frac{1}{n}$: $\frac{a_{n+1}}{a_n} = \frac{n}{n+1} \to 1$. Inconclusive (we know it diverges by other means).

Let $L = \limsup_{n \to \infty} |a_n|^{1/n}$.

1. If $L < 1$, the series $\sum a_n$ converges absolutely.
2. If $L > 1$, the series diverges.
3. If $L = 1$, the test is inconclusive.

(1) If $L < 1$, choose $r$ with $L < r < 1$. By definition of $\limsup$, there exists $N$ such that $|a_n|^{1/n} < r$ for all $n \geq N$ (this is where $\limsup$ is needed rather than $\lim$ — there may be finitely many exceptions). Then $|a_n| < r^n$ for all $n \geq N$, and $\sum r^n < \infty$ (convergent geometric series).

(2) If $L > 1$, then $|a_n|^{1/n} > 1$ for infinitely many $n$, so $|a_n| > 1$ for infinitely many $n$, hence $a_n \not\to 0$.

(a) $\sum \left(\frac{n}{2n+1}\right)^n$: $|a_n|^{1/n} = \frac{n}{2n+1} \to 1/2 < 1$. Converges.

(b) $\sum \left(\frac{1}{1 + 1/n}\right)^{n^2}$: $|a_n|^{1/n} = \left(\frac{n}{n+1}\right)^n \to 1/e < 1$. Converges.

The root test is strictly stronger than the ratio test: whenever the ratio test gives a verdict, the root test gives the same verdict, but there exist series where the root test succeeds and the ratio test is inconclusive. This is because $\liminf |a_{n+1}/a_n| \leq \liminf |a_n|^{1/n} \leq \limsup |a_n|^{1/n} \leq \limsup |a_{n+1}/a_n|$. In practice, the ratio test is more convenient for factorials and exponentials, while the root test is better for $n$th powers.

Let $f: [1, \infty) \to \mathbb{R}$ be positive, continuous, and decreasing. Then $\sum_{n=1}^\infty f(n)$ converges if and only if $\int_1^\infty f(x)\,dx$ converges.

Since $f$ is decreasing, for all $k \geq 1$: $f(k+1) \leq \int_k^{k+1} f(x)\,dx \leq f(k)$. Summing from $k = 1$ to $n$:

$\sum_{k=2}^{n+1} f(k) \leq \int_1^{n+1} f(x)\,dx \leq \sum_{k=1}^n f(k).$

If $\int_1^\infty f(x)\,dx$ converges, the right inequality gives $S_n \leq f(1) + \int_1^\infty f(x)\,dx < \infty$, so $(S_n)$ is bounded and increasing, hence convergent by the Monotone Convergence Theorem. If $\int_1^\infty f(x)\,dx = \infty$, the left inequality gives $S_{n+1} - f(1) \geq \int_1^{n+1} f(x)\,dx \to \infty$, so the partial sums diverge.

For $f(x) = 1/x^p$: $\int_1^\infty x^{-p}\,dx = \frac{x^{1-p}}{1-p}\big|_1^\infty$, which converges iff $1-p < 0$, i.e., $p > 1$. This recovers Theorem 3 via the integral test.

$\sum \frac{1}{n(\ln n)^2}$ for $n \geq 2$: the ratio and root tests both give $L = 1$ (inconclusive). The integral test with $f(x) = 1/(x(\ln x)^2)$: substituting $u = \ln x$,

$\int_2^\infty \frac{dx}{x(\ln x)^2} = \int_{\ln 2}^\infty u^{-2}\,du = \frac{1}{\ln 2} < \infty.$

Converges. This is a series that only the integral test can handle among our toolkit.

The integral test also provides error bounds: if $R_n = \sum_{k=n+1}^\infty f(k)$ is the remainder after $n$ terms, then

$\int_{n+1}^\infty f(x)\,dx \leq R_n \leq \int_n^\infty f(x)\,dx.$

This tells you how many terms you need for a given accuracy — exactly the kind of bound used in numerical analysis and in bounding truncation error in series approximations for ML (e.g., truncating a Taylor expansion or a Fourier series).

A series $\sum_{n=1}^\infty (-1)^{n+1} b_n$ where $b_n > 0$ for all $n$ is an alternating series.

If $(b_n)$ is decreasing ($b_{n+1} \leq b_n$) and $\lim_{n \to \infty} b_n = 0$, then the alternating series $\sum_{n=1}^\infty (-1)^{n+1} b_n$ converges.

Consider the even partial sums:

$S_{2n} = (b_1 - b_2) + (b_3 - b_4) + \cdots + (b_{2n-1} - b_{2n}).$

Each parenthesized pair is $\geq 0$ (since $b_k$ is decreasing), so $(S_{2n})$ is increasing. Also,

$S_{2n} = b_1 - (b_2 - b_3) - \cdots - (b_{2n-2} - b_{2n-1}) - b_{2n} \leq b_1,$

so $(S_{2n})$ is bounded above by $b_1$.

By the Monotone Convergence Theorem, $S_{2n} \to S$ for some $S \leq b_1$. Now $S_{2n+1} = S_{2n} + b_{2n+1} \to S + 0 = S$. Since both the even and odd subsequences converge to $S$, the full sequence $S_n \to S$.

$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots = \ln 2$. The terms $b_n = 1/n$ are decreasing and tend to $0$, so the Leibniz test applies. The sum is $\ln 2$ (provable via the Taylor series of $\ln(1+x)$ at $x=1$, which we establish in Power Series & Taylor Series).

Note: $\sum 1/n = \infty$ diverges, so this series converges conditionally but not absolutely. We formalize this distinction in the next section.

If the alternating series converges to $S$, the error after $n$ terms satisfies $|S - S_n| \leq b_{n+1}$ — the error is bounded by the first omitted term. This is a remarkably tight bound, much better than what most convergence tests provide.

A series $\sum a_n$ converges absolutely if $\sum |a_n|$ converges.

A series $\sum a_n$ converges conditionally if it converges but does not converge absolutely — that is, $\sum a_n$ converges but $\sum |a_n| = \infty$.

If $\sum |a_n|$ converges, then $\sum a_n$ converges, and $\left|\sum a_n\right| \leq \sum |a_n|$.

We use the Cauchy criterion. Since $\sum |a_n|$ converges, for any $\varepsilon > 0$ there exists $N$ such that $\sum_{k=n+1}^m |a_k| < \varepsilon$ for all $m > n \geq N$. By the triangle inequality:

$\left|\sum_{k=n+1}^m a_k\right| \leq \sum_{k=n+1}^m |a_k| < \varepsilon.$

This is the Cauchy criterion for $\sum a_n$, so $\sum a_n$ converges.

The inequality $|\sum a_n| \leq \sum |a_n|$ follows by taking limits of $|S_n| \leq T_n$ where $T_n = \sum_{k=1}^n |a_k|$.

Note: This proof uses the completeness of $\mathbb{R}$ — the Cauchy criterion requires completeness. In an incomplete space (like $\mathbb{Q}$), absolute convergence need not imply convergence. See Completeness & Compactness for why completeness matters.

(a) $\sum (-1)^{n+1}/n^2$ converges absolutely ($\sum 1/n^2 < \infty$).

(b) $\sum (-1)^{n+1}/n$ converges conditionally ($\sum 1/n = \infty$, but the Leibniz test gives convergence).

(c) $\sum (-1)^n$ diverges (not even conditional convergence — the $n$th-term test kills it).

Let $\sum a_n$ be a conditionally convergent series. Then for any $S \in \mathbb{R} \cup \{-\infty, +\infty\}$, there exists a rearrangement $\sum a_{\sigma(n)}$ (where $\sigma: \mathbb{N} \to \mathbb{N}$ is a bijection) that converges to $S$.

Let $p_k$ be the positive terms (in their original order) and $q_k$ the negative terms. Since $\sum a_n$ is conditionally convergent, both $\sum p_k = +\infty$ and $\sum |q_k| = +\infty$ — if either were finite, the series would converge absolutely.

To reach a target $S$: add positive terms $p_1, p_2, \ldots$ until the partial sum first exceeds $S$; then add negative terms $q_1, q_2, \ldots$ until it drops below $S$; then add more positive terms until it exceeds $S$ again; continue alternating.

The key insight: because $p_k \to 0$ and $q_k \to 0$ (since $a_n \to 0$ by the divergence test applied to the original convergent series), each overshoot/undershoot gets smaller. The partial sums oscillate around $S$ with decreasing amplitude, converging to $S$ by the squeeze theorem.

If $\sum a_n$ converges absolutely, then every rearrangement converges to the same sum. This is why absolute convergence is the “safe” mode: it is invariant under permutation. Conditional convergence is inherently order-dependent.

In computational settings, finite-precision arithmetic implicitly rearranges series (due to rounding and the order of evaluation), so absolute convergence guarantees reproducibility, whereas conditional convergence does not.

The standard ordering gives $\sum (-1)^{n+1}/n = \ln 2 \approx 0.693$.

The rearrangement $1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} + \frac{1}{7} - \frac{1}{4} + \cdots$ (two positive terms, then one negative term) converges to $\frac{3}{2}\ln 2 \approx 1.040$.

1. Always check the divergence test first. If $a_n \not\to 0$, the series diverges — done.
2. Recognize geometric series ($a_n = cr^n$) and $p$-series ($a_n = 1/n^p$) immediately. These are the benchmarks.
3. Factorials or $n$th powers of constants: ratio test.
4. $n$th powers of $n$-dependent expressions: root test.
5. Series resembling $1/n^p$ asymptotically: limit comparison with the $p$-series.
6. $f(n)$ where $f$ has an elementary antiderivative: integral test.
7. Alternating series: Leibniz test.
8. If you need absolute convergence: apply tests 3–6 to $\sum |a_n|$.