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I found this question and found it very interesting, $\lim_{k\to\infty\ }4\sum_{n=1}^k\frac{\left(-1\right)^{n+1}}{2n+1}$ How to solve it?

projector israel
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    First I noticed that it's very similar to a well known formula for pi. (Leibniz series) – David Lui Mar 10 '23 at 18:51
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    Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Another User Mar 10 '23 at 18:55
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    @coudy the OP's bounds are from $1$ to $k$ while the link you provide is for $0$ to $\infty$ – Kamal Saleh Mar 10 '23 at 18:58
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    @coudy I didn't see the limit part. So I guess it is a duplcate. – Kamal Saleh Mar 10 '23 at 19:01
  • @KamalSaleh is there is a closed form if it is not infinite series ? I know this make a function $f(x) = \sum_{k=1}^k\frac{\left(-1\right)^{k+1} x^{k+1}}{2k+1}$ and differentiate it then sum the alternating geometric series then integrate it back but does the integral can be done without special functions – pie Oct 24 '23 at 16:55

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Start with $$\log (1-x) = -x -\frac{x^2}{2}-\frac{x^3}{3} - \cdots$$

$$\log (1+x) = x -\frac{x^2}{2}+\frac{x^3}{3} - \cdots$$

so that $$\frac{1}{2ix}\left[\log(1+ix)-\log(1-ix) \right] = 1-\frac{x^2}{3} + \frac{x^4}{5} - \cdots$$

$$ 4\left( 1-\frac{1}{3} + \frac{1}{5} - \cdots \right) = \frac{2}{i}\left[\log(1+i)-\log(1-i) \right]$$

$$\phantom{ 4\left( 1-\frac{1}{3} + \frac{1}{5} - \cdots \right)} = \frac{2}{i}\log\frac{1+i}{1-i} $$

$$\phantom{ 4\left( 1-\frac{1}{3} + \frac{1}{5} - \cdots \right)} = \frac{2}{i}\log i $$

$$\phantom{ 4\left( 1-\frac{1}{3} + \frac{1}{5} - \cdots \right)} = \pi.$$

mjw
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