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I have to prove $\lim\limits_{n \to \infty} \sum_\limits{i=1}^n \frac{(-1)^{i+1}}{i} = \ln2$.

The hints I have are:

$\lim\limits_{n \to \infty} \sum_\limits{i=1}^n \frac{1}{i} - \ln(n) = c < \infty$

$\sum_\limits{i=1}^{2n} \frac{(-1)^{i+1}}{i} = \sum_\limits{i=1}^{2n} \frac {1}{i} - 2\sum_\limits{i=1}^{2n} \frac{1}{2i} $

I never calculated any limits with sum in the formula, nor I see how the tips are of any help.

Would appreciate a few hints and tips!

Jose M Serra
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Tegernako
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  • See https://math.stackexchange.com/questions/1356517/what-is-the-correct-radius-of-convergence-for-ln1x – lab bhattacharjee Apr 06 '19 at 15:37
  • The second hint should be $\sum_\limits{i=1}^{2n} \frac{(-1)^{i+1}}{i} = \sum_\limits{i=1}^{2n} \frac {1}{i} - 2\sum_\limits{i=1}^{n} \frac{1}{2i}$. – Robert Z Apr 06 '19 at 15:47

1 Answers1

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Let $a_n=\sum_{i=1}^n \frac{1}{i} - \ln(n+1)$. Then the sequence $(a_n)_{n\geq 1}$ is increasing: $$a_{n+1}-a_n=\frac{1}{n+1}-\ln\left(1+\frac{1}{n+1}\right)\geq 0$$ and bounded above (show that part). Then the sequence $(a_n)_n$ converges to a finite limit $c$. Hence, by the second hint, $$\begin{align} \sum_\limits{i=1}^{2n} \frac{(-1)^{i+1}}{i} &= \sum_\limits{i=1}^{2n} \frac {1}{i} - 2\sum_\limits{i=1}^{n} \frac{1}{2i}\\ &=\left(\sum_\limits{i=1}^{2n} \frac {1}{i}-\ln(2n+1)\right) - \left(\sum_\limits{i=1}^{n} \frac{1}{i}-\ln(n+1)\right)+\ln\left(\frac{2n+1}{n+1}\right)\\ &\to c-c+\ln(2)=\ln(2). \end{align}$$ Finally note that $$\sum_\limits{i=1}^{2n+1} \frac{(-1)^{i+1}}{i} =\sum_\limits{i=1}^{2n} \frac{(-1)^{i+1}}{i}+ \frac{1}{2n+1}\to \ln(2)+0=\ln(2).$$

Robert Z
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