3.1 Numerical series and convergence tests

Partial sums and tails

Definition

Series convergence

The series $\sum_{n=1}^\infty a_n$ converges to $s$ if its partial sums $s_N=\sum_{n=1}^N a_n$ converge to $s$. It converges absolutely if $\sum|a_n|$ converges, and conditionally if it converges but not absolutely.

The Cauchy criterion for partial sums says convergence is equivalent to: for every $\varepsilon>0$ there exists $N$ such that $|\sum_{k=m}^n a_k|<\varepsilon$ for all $n\ge m\ge N$. In particular, convergence forces $a_n\to0$, but this necessary condition is not sufficient: the harmonic series diverges.

Theorem

Comparison and absolute convergence

For $0\le a_n\le b_n$ eventually, convergence of $\sum b_n$ implies convergence of $\sum a_n$; divergence of $\sum a_n$ implies divergence of $\sum b_n$. If $\sum|a_n|$ converges, then $\sum a_n$ converges.

Proof

Tail control proves absolute convergence

If $\sum|a_n|$ converges, its nonnegative tails are arbitrarily small. By the triangle inequality, $|\sum_{k=m}^n a_k|\le\sum_{k=m}^n|a_k|$, so the signed partial sums are Cauchy and converge in $\mathbb R$. The comparison statements follow from monotone bounded partial sums and, for the divergence conclusion, contraposition.

Worked example

Geometric, harmonic and p-series

For $|r|<1$, $\sum_{n=0}^N r^n=(1-r^{N+1})/(1-r)$ tends to $1/(1-r)$. For the harmonic series, grouping terms from $1/2$ onward as $1/2$, then $1/3+1/4$, then $1/5+\cdots+1/8$, and so on gives each group at least $1/2$. Partial sums are unbounded, so it diverges although its terms tend to zero. The standard $p$-series $\sum n^{-p}$ converges for $p>1$ and diverges for $p\le1$; the integral or dyadic comparison test establishes this threshold.

Ratio and alternating tests

If $a_n\ne0$ eventually and $\limsup|a_{n+1}/a_n|<1$, the series converges absolutely by comparison with a geometric series. If $|a_{n+1}/a_n|\ge1$ eventually, the terms cannot tend to zero, so the series diverges. A limiting ratio of $1$ is inconclusive: both $\sum1/n$ and $\sum1/n^2$ have ratio tending to $1$.

Proof of the convergent ratio case. Suppose $a_n\ne0$ for all sufficiently large $n$ and $\limsup |a_{n+1}/a_n|<1$. Choose $0<q<1$ strictly larger than this limsup. For some $N$, every $n\ge N$ satisfies $|a_{n+1}/a_n|\le q$. Iterating gives

$$|a_{N+k}|\le |a_N|q^k\qquad(k\ge0).$$

The geometric series $\sum_{k\ge0}|a_N|q^k$ converges, so comparison proves convergence of the absolute tail. A finite prefix does not affect convergence; hence $\sum a_n$ converges absolutely. A uniform conclusion for a family of series requires a bound valid uniformly in the input.

If $b_n\ge0$ is nonincreasing and tends to zero, then $\sum(-1)^{n-1}b_n$ converges, and the error after $N$ terms has magnitude at most $b_{N+1}$. The even partial sums are nondecreasing, the odd partial sums are nonincreasing, and $S_{2n+1}-S_{2n}=b_{2n+1}\to0$. Thus the alternating harmonic series converges but not absolutely.

Worked example

Choose the right test

For $\sum_{n=1}^\infty n/3^n$, the ratio of successive positive terms is $(n+1)/(3n)\to1/3<1$, so the series converges. For $\sum(-1)^{n-1}/n$, the alternating test gives convergence while harmonic divergence rules out absolute convergence. A ratio tending to $1$ in the latter case cannot decide it.

Method comparison

First test whether $a_n\to0$; failure proves divergence, success proves nothing alone. Compare nonnegative series with geometric or $p$-series. Use the ratio test for factorials and exponentials when the ratio settles away from $1$. Use alternating monotone decrease to zero for conditional convergence and an error bound. Absolute convergence permits stronger operations than conditional convergence; arbitrary rearrangement of a conditionally convergent series requires special care.

Proof of the convergent ratio case. Suppose $a_n\ne0$ for all sufficiently large $n$ and $\limsup |a_{n+1}/a_n|<1$. Choose $0<q<1$ strictly larger than this limsup. For some $N$, every $n\ge N$ satisfies $|a_{n+1}/a_n|\le q$. Iterating gives

$$|a_{N+k}|\le |a_N|q^k\qquad(k\ge0).$$

The geometric series $\sum_{k\ge0}|a_N|q^k$ converges, so comparison proves convergence of the absolute tail. A finite prefix does not affect convergence; hence $\sum a_n$ converges absolutely. A uniform conclusion for a family of series requires a bound valid uniformly in the input.

Practice

1. Determine convergence of $\sum 1/2^n$.
2. Determine convergence of $\sum 1/n$.
3. Determine absolute or conditional convergence of $\sum(-1)^{n-1}/n$.
4. Explain why $a_n\to0$ does not prove $\sum a_n$ converges.

Full solutions

1. It is geometric with ratio $1/2$, so converges to $1$ when indexed from $n=1$.
2. The dyadic grouping argument above shows the partial sums grow without bound, so it diverges.
3. The alternating test gives convergence, and its absolute series is the divergent harmonic series. It is conditionally convergent.
4. The harmonic terms $1/n$ tend to zero while their series diverges. The term test is necessary, not sufficient.

Synthesis

A series is a limit of partial sums. Tests control tails using comparison or cancellation; checking the individual terms alone cannot establish convergence.