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Find all positive real numbers $p$ such that the series

$$\sum_{n=2}^{\infty}\ln\left[1+\frac{(-1)^n}{n^p}\right]$$

converges conditionally.

First, I show that $\left|\ln\left[1+\frac{(-1)^{n}}{n^p}\right]\right|\leq \frac{1}{n^p}$ for any $n$，then by the comparison test, we can show that the series converges absolutely when $p>1$. For $0<p\leq 1$, Taylor expansion gives:

$$\ln\left[1+\frac{(-1)^n}{n^p}\right]=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}(-1)^{kn}}{kn^{kp}}$$

then our series becomes a double series:

$$\sum_{n=2}^{\infty}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}(-1)^{kn}}{kn^{kp}}$$

then I have trouble handling this.

Kato yu
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2 Answers2

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Let's call $a_n$ the $n$-th term of the series. Then $a_n>0$ if $n$ is even, and $a_n<0$ if $n$ is odd. Clearly $\lim_{n\to\infty}|a_n|=0$. If $p=1$ it is easy to see that $a_n$ is non-increasing. By Leibniz's test, the series converges in this case. However if $0<p<1$, then $|a_{2n}|<|a_{2n+1}|$ and Leibniz's test cannot be applied.

We try a different approach.

Claim. For each $p\in(0,1)$ there exists a constant $C_p>0$ such that $$ C_p\,\frac{(-1)^n}{(n+1)^p}\le a_n\le\frac{(-1)^n}{n^p},\qquad n\ge2.\tag{1} $$

Proof. The right hand side of the inequality holds because $\ln(1+x)\le x$ for all $x>-1$.

A) $n=2\,k$ even. The function $$ f(x)=(x+1)^p\ln(1+x^{-p}) $$ is continuous on $(0,\infty)$, $f(x)>0$, $\lim_{x\to0^+}f(x)=+\infty$, $\lim_{x\to+\infty}f(x)=1$. In particular $$ A_p=\inf_{x\ge2}f(x)>0. $$ Then, letting $x=2\,k$ we get $$ A_p\le(1+2\,k)^p\ln\Bigl(1+\frac{1}{(2\,k)^p}\Bigr),\qquad k\ge1. $$

B) $n=2\,k+1$ odd. The function $$ g(x)=(x+1)^p\ln(1-x^{-p}) $$ is continuous and increasing on $(0,\infty)$. Let $B_p=-g(3)>0$. Then $g(x)\ge -B_p$ for all $x\ge3$. Letting $x=2\,k+1$ we get $$ B_p\le(2\,k+2)^p\ln\Bigl(1-\frac{1}{(2\,k+1)^p}\Bigr),\qquad k\ge1. $$ Taking $C_p=\min(A_p,B_p)$ gives (1).

The series $$ \sum\frac{(-1)^n}{n^p},\qquad \sum\frac{(-1)^n}{(n+1)^p} $$ are both convergent by Leibniz's test. The generalized comparison theorem proves that $\sum a_n$ is also convergent.

Julián Aguirre
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Let $$a_n:=\frac{(-1)^n}{n^p}.$$ For the case of $0<p\leq 1$, note that

$$ \log\left(1+a_n\right)=a_n-\frac12a_n^2+\cdots $$

for large enough $n$ (say, $n\leq M$ for some M>0), since

$$ \log(1+x)=\sum_{n=1}^\infty(-1)^{n+1}\frac{x^n}{n}\quad\text{for}\ \ |x|<1 $$

and $a_n\to 0$.

Let $N>0$ be such that $Np>2$. Then one can divide the Taylor series into two parts:

$$ \log(1+a_n)=\sum_{m=1}^{N-1}(-1)^{m+1}\frac{a_n^m}{m}+f(n),\quad n>M\tag{1} $$

where

$$ f(n)\in O\left(\frac{1}{n^{Np}}\right) $$

Note that $\sum_{n=M}^\infty(-1)^{m+1}\frac{a_n^m}{m}$ converges for each $m$ with $1\leq m\leq N-1$ by the alternating series test. And also $$ \sum_{n=M}^\infty f(n) $$ converges since $Np>2$. Taking the sum on both sides of (1), one can see that $$ \sum_{n=M}^\infty\log(1+a_n) $$ is convergent and so is $$ \sum_{n=1}^\infty\log(1+a_n). $$