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Is it possible to directly show $$ \sum_{k=0}^\infty \frac{(-1)^k}{2k+1}=\frac{\pi}4 $$ using partial sums?

I think I was able to do something else to get this result as follows, which I posted separately as my own answer to this question.

However, I was unable to compute the sum using $n$-th partial sums, but I would like to ask if someone knows how to do that here. Writing out the first four terms $$ \sum_{k=0}^\infty \frac{(-1)^k}{2k+1}=1-\frac 13+\frac 15-\frac 17+\cdots $$ the series is not telescoping (wishful thinking...), so that would not help us here. It also seems that we cannot use partial fractions for this series. But I hope what I posted as an answer is not the only way to compute the sum.

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Starting from the geometric sum formula $$ \frac 1{1+x^2}=\sum_{k=0}^\infty (-1)^kx^{2k} $$ for all $x \in (-1,1)$, we integrate both sides to get $$ \tan^{-1}x=\sum_{k=0}^\infty \frac{(-1)^k}{2k+1}x^{2k+1}. $$ We send $x \to 1^-$ of both sides to conclude $$ \frac{\pi}4=\sum_{k=0}^\infty \frac{(-1)^k}{2k+1}. $$

I also verified this sum on WolframAlpha.

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  • $\begingroup$ Note: You need Abel's theorem when you take the limit. $\endgroup$ – Botond Sep 12 at 19:05
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For $-1\le x<1,$ $$\ln(1-x)=-\sum_{r=1}^\infty\dfrac{x^r}r$$

So, $\ln\dfrac{1+x}{1-x}=\ln(1+x)-\ln(1-x)=?$

If $S=\sum_{k=0}^\infty\dfrac{(-1)^k}{2k+1},$

$2iS=\ln\dfrac{1+i}{1-i}=\ln(i)$ whose principal value is $\dfrac{i\pi}2$

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Considering that $$\frac{1}{2k+1}=\int_0^1{x^{2k}dx},$$ we have that $$\sum_{k=0}^n{\frac{(-1)^k}{2k+1}}=\int_0^1{\frac{dx}{1+x^2}}-\int_0^1{\frac{(-x^2)^{n+1}}{1+x^2}dx}=\frac{\pi}{4}+(-1)^{n}\int_0^1{\frac{x^{2n+2}}{1+x^2}dx}.$$ However $$0\leqslant\int_0^1{\frac{x^{2n+2}}{1+x^2}dx}\leqslant\int_0^1{x^{2n+2}dx}=\frac{1}{2n+3}\underset{n\rightarrow +\infty}{\longrightarrow}0$$ since $\forall x\in[0,1],\,\frac{1}{1+x^2}\leqslant 1$. Letting $n\rightarrow +\infty$ gives the result.

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