Let $u_n=\ln\left(1+\frac {(-1)^n}{n}\right)$, $n\ge2$, show that the series $\sum u_n$ converges.

Question

Let $u_n=\ln\left(1+\dfrac {(-1)^n}{n}\right)\qquad n\ge2,\quad$, show that the series $\displaystyle \sum u_n$ converges and its limit.

I found the result, but the correction gives me something different and more complicated. So I'm wondering if my solution is correct.

Let $u_{2p}=\ln\left(1+\dfrac {1}{2p}\right)$ and $u_{2p+1}=\ln\left(1-\dfrac {1}{2p+1}\right)$

$\begin{array}l \displaystyle \sum_{n\ge2} u_n &=\displaystyle \sum_{p\ge1}(u_{2p}+u_{2p+1})\\ &=\displaystyle \sum_{p\ge1}\bigg[\ln\left(1+\dfrac {1}{2p}\right)+\ln\left(1-\dfrac {1}{2p+1}\right)\bigg] \\&=\displaystyle\sum_{p\ge1} \bigg(\ln(2p+1)-\ln2p+\ln 2p-\ln(2p+1)\bigg)\\ &=\displaystyle\sum_{p\ge1} 0=0 \end{array}$

So $\displaystyle \sum_{n\ge2} u_n$ converges to $\ell=0$

Answer 1

I won't use at the moment the Leibniz criterion .

For $n>0$

$S_{2n+1}=\displaystyle \sum_{k=2}^{2n+1} \ln\left(1+\dfrac {(-1)^k}{k}\right)$ as $\forall p\in \mathbb{N}$ we have $u_{2p}+u_{2p+1} =0$ so $S_{2n+1}=0$

$S_{2n}=\displaystyle \sum_{k=2}^{2n} \ln\left(1+\dfrac {(-1)^k}{k}\right)=\ln\left(1+\dfrac{1}{2n}\right)$

Thus $S_{2n+1}-S_{2n}=-\ln\left(1+\dfrac{1}{2n}\right)=\ln\left(\dfrac{2n}{2n+1}\right)\underset{n\to +\infty}{\longrightarrow}0$ so $(S_n)$ converges

With the same argument ($u_{2p}+u_{2p+1} =0$), we can say that $S_{2n-1}=\displaystyle \sum_{k=2}^{2n-1} \ln\left(1+\dfrac {(-1)^k}{k}\right)=0$

So we can conclude this series converges to $0$