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Exercise 37 in Apostol $10.20$ asks to find all complex $z$ such that $$\sum_{n=1}^{\infty} \frac{(-1)^n}{z+n}$$ converges.

I suspect this requires the use of either Abel's Test or Dirichlet's Test. My attempt so far has been to set $\{b_n\}=\frac{1}{n}$, which is a decreasing sequence of real numbers that converges to $0$. Then, I set $\{a_n\}=(-1)^n\frac{n}{z+n}$.

As $n\to\infty$, $$\{a_n\} \to (-1)^ne^{i\arg(\frac{n}{z+n})}$$ since $$|\frac{n}{z+n}|=\frac{n}{|z+n|}\to 1$$

In order to prove that this converges for all complex $z\not=-1,-2,\dots$, I must show that $A_n=\sum_{k=1}^{n} a_n$ is a bounded sequence (not necessarily that it converges). This would satisfy the hypotheses for Dirichlet's test.

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$\frac{1}{z+n}\to0$ for $n\to\infty$. so the oscillating series converges. – akkkk Jun 19 '12 at 17:26
@Auke: Alternating series test requires terms to (eventually) decrease to 0, not just converge. You can easily construct a counterexamples. – Erick Wong Jun 19 '12 at 17:34
up vote 3 down vote accepted

$$ c_n:=\frac{1}{z+n}=\frac{x+n}{|z+n|^2}-i\frac{y}{|z+n|^2}=:a_n-ib_n \quad \forall \ z=x+iy \ne -n. $$ For every $z \in \mathbb{C}\setminus(-\mathbb{N})$ the sequences $a_n, \ b_n$ decrease to $0$, therefore the series $\sum_{n=1}^\infty(-1)^nc_n$ converges for every $z \in \mathbb{C}\setminus(-\mathbb{N})$.

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Clear and concise! – Andrew Salmon Jun 19 '12 at 22:55
Very nice answer! – albmiz-mth Feb 9 '13 at 3:25

Use Cauchy property of convergent sequences and write $$ \sum_{k=m}^n \frac{(-1)^k}{z+k} $$ Then write (suppose $m$ even) $$ \frac{1}{z+m} - \frac{1}{z+m+1} = \frac{1}{(z+m)(z+m+1)} \sim m^{-2} $$ This should persuade you that the series converges for $z$ that makes any denominator different from zero.

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We can use Abel transform. Let $s_n:=\sum_{j=0}^n(-1)^j$. We have for $n$ and $m$ integers, and $z\notin -\Bbb N$, \begin{align} \sum_{j=n+1}^{n+m}\frac{(-1)^j}{z+j}&=\sum_{j=n+1}^{n+m}\frac{s_j-s_{j-1}}{z+j}\\ &=\sum_{k=n+1}^{n+m}\frac{s_k}{z+k}-\sum_{k=n}^{m+n-1}\frac{s_k}{z+k+1}\\ &=-\frac{s_n}{z+n+1}+\frac{s_{n+m}}{z+m+n}+\sum_{k=n+1}^{n+m}\frac{s_k}{(z+k)(z+k+1)}. \end{align} This gives, for $n,m> |z|$, that $$\left|\sum_{j=n+1}^{n+m}\frac{(-1)^j}{z+j}\right|\leq \frac 1{n+1-|z|}+\frac 1{m+n-|z|}+\sum_{k=n+1}^{n+m}\frac 1{(k-|z|)(k+1-|z|)}.$$ Since the series $\sum_{k=1}^{+\infty}\frac 1{(k-|z|)(k+1-|z|)}$ is convergent, we get that the sequence $\{\sum_{k=1}^n\frac{(-1)^k}{z+k}\}$ is Cauchy if $z\neq -k$ for each integer $k$.

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