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 $\ell=0$

  • $\begingroup$ If $a_n = (-1)^n$ then $\sum_{n=1}^\infty (a_{2n} + a_{2n+1}) = 0$. $\endgroup$
    – Yannik
    Oct 10, 2017 at 14:55
  • $\begingroup$ Perfect, So my answer is right, I think now the teacher wanted to highlight an another method. He starts like that :$u_n=\dfrac{(-1)^n}{n}+\mathcal{O}(n^{-2})$ $\endgroup$
    – Stu
    Oct 10, 2017 at 15:01
  • 1
    $\begingroup$ I'm afraid not, you can't conclude that a series is convergent, just by putting terms together. Note that $\sum_{n=0}^N (-1)^n = 1$ if $N$ is even and $0$ if $N$ is odd. Have you've heard of Riemann's Paradox? $\endgroup$
    – Yannik
    Oct 10, 2017 at 15:07
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    $\begingroup$ You can look at $e_N:=\exp\left(\sum_{n=2}^N u_n\right)$ and use user90369's observation to derive that $(e_N)_N$ converges and so must $(\log(e_N))_N$. $\endgroup$
    – Yannik
    Oct 10, 2017 at 15:14
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    $\begingroup$ See: mathworld.wolfram.com/LeibnizCriterion.html (Leibniz criterion) $\endgroup$
    – user90369
    Oct 10, 2017 at 15:15

1 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 $0$


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