I want to show that the series whose nth term is $a_n=\ln(1+\frac{(-1)^n}{n+1})$ is convergent. I wanted to use the limit comparison test to compare it to the $p$ series but $a_n$ is not positive. I thought of writing the power series representation of $a_n$ using the power series representation of $\ln(1+x)$ with $x=b_n=\frac{(-1)^n}{n+1}$ we find that $$a_n=b_n-\frac{1}{2}b_n^2+\frac{1}{3}b_n^3-\frac{1}{4}b_n^4+\cdots$$ Now the seris $\sum b_n$ is convergent by the alternating series test and the other terms are all terms of absolutely convergent series but it is an infinte sum, can I say so ? I mean is the infinite sum of convergent series a convergent series ? Is this correct and is there any other way to do it ?

  • $\begingroup$ Use $\lvert \ln (1+x) - x \rvert \leqslant \lvert x\rvert^2$ for $\lvert x\rvert \leqslant \frac{1}{2}$. $\endgroup$ – Daniel Fischer Feb 14 '16 at 21:40
  • $\begingroup$ The alternating series test also applies to $a_n$, which alternates in sign and converges to 0. $\endgroup$ – Matt Samuel Feb 14 '16 at 21:42
  • $\begingroup$ @Matt Samuel How I write $a_n=(-1)^nc_n$ what is $c_n$ ?Do you mean that I write $c_n=|\ln (1+\frac{(-1)^n}{n+1}|$ ? $\endgroup$ – palio Feb 14 '16 at 21:53
  • $\begingroup$ @MattSamuel It needs to have an absolute value decreasing (non-increasing, rather) to zero, convergence is not enough. $\endgroup$ – Clement C. Feb 14 '16 at 21:53
  • 2
    $\begingroup$ Let $a_n=\ln\left(1+\frac{(-1)^n}{n+1}\right)$ for every $n\geqslant1$, then $a_n\to0$ and $a_{2n-1}+a_{2n}=0$ hence the series $\sum\limits_{n\geqslant1}a_n$ converges to $a_1=-\ln2$. If one starts at $n=0$, the sum is $0$. $\endgroup$ – Did Feb 14 '16 at 22:36

One may use the Taylor series expansion, as $x\to 0$, $$ \log(1+x)=x+O(x^2) $$ giving, for some great $n_0$ and all $N$ greater than $n_0$, $$ \sum_{n_0 \leq n\leq N}\ln(1+\frac{(-1)^n}{n+1})=\sum_{n_0 \leq n\leq N}\frac{(-1)^n}{n+1}+\sum_{n_0 \leq n\leq N} O\left(\frac1{(n+1)^2}\right) $$ then conclude to the convergence of the initial series.

  • $\begingroup$ That's exactly what i want to use but without the big O notation what should be the procedure ? Thank you for your help! $\endgroup$ – palio Feb 14 '16 at 21:48
  • $\begingroup$ @OlivierOloa: it's technically correct, but seeing $\frac{1}{n+1}$ instead of $\frac{1}{n}$ seems... morally wrong. $\frac{(-1)^n}{n} + O(\frac{1}{n^2})$? $\endgroup$ – Clement C. Feb 14 '16 at 21:51
  • $\begingroup$ @palio You have $\displaystyle \left|O\left(\frac1{(n+1)^2}\right)\right|\leq \frac{C}{(n+1)^2}$ giving the sought convergence. Thanks. $\endgroup$ – Olivier Oloa Feb 14 '16 at 21:52
  • $\begingroup$ @ClementC. I see what you mean... $\endgroup$ – Olivier Oloa Feb 14 '16 at 21:54
  • $\begingroup$ It does not really matter here... actually, not at all here :) It's really more a question of habit and not "letting one's guard down" (otherwise, sneaky things based on differences of expressions like $e^{\frac{1}{n}}$ and $e^{\frac{1}{n+1}}$ can appear). $\endgroup$ – Clement C. Feb 14 '16 at 21:56

This series is alternating and the positive parts of the terms decrease to 0. It therefore converges. Note this little experiment and you will see why.


import math
def f(n):
    return math.log(1 + (-1)**n/n)

for k in range(2,20):
    print ("%s:%7.8f"% (k, f(k)))


  • $\begingroup$ How do you prove the second point elegantly, though? $\endgroup$ – Clement C. Feb 14 '16 at 21:52
  • $\begingroup$ $t\mapsto 1/(1+t)$ is increasing on (-1, 1).. $\endgroup$ – ncmathsadist Feb 14 '16 at 21:54
  • $\begingroup$ $\lvert a_n\rvert = \left\lvert \ln\left(1+\frac{(-1)^n}{n+1}\right)\right\rvert$. The jump to monotonicity of $t\mapsto \frac{1}{t+1}$ is not that straightforward, and may require at least a couple lines? $\endgroup$ – Clement C. Feb 14 '16 at 21:58
  • $\begingroup$ separate the even and odd terms first. It is a bit of delicate affair. $\endgroup$ – ncmathsadist Feb 14 '16 at 22:10
  • $\begingroup$ As pointed out by Did in another comment, the simplest option to show monotonicity of $\lvert a_n\rvert$ is probably to note that $$0 < a_{2n} = \ln\left(1+\frac{1}{2n}\right)=\ln\left(\frac{2n+1}{2n}\right)= -\ln\left(\frac{2n}{2n+1}\right) = -\ln\left(1-\frac{1}{2n+1}\right) = -a_{2n+1}.$$ $\endgroup$ – Clement C. Feb 14 '16 at 23:10

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