Compute the sum of the infinite series $\sum_{k=0}^\infty \frac{(-1)^k}{2k+1}$ 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.
 A: 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.
A: 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.
A: 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 $\dfrac{i\pi}2$
A: $$
\begin{aligned}
&\sum_{k=0}^{\infty}\frac{(-1)^k}{2k+1}\\
&=\frac{1}{2}\sum_{k=0}^{\infty}\frac{(-1)^k}{k+\frac{1}{2}}\\
&=\frac{1}{2}\sum_{k=0}^{\infty}(-1)^k\int_0^1t^{k+1/2-1}dt\\
&=\frac{1}{2}\int_0^1t^{-1/2}\left(\sum_{k=0}^{\infty}(-1)^kt^k\right)dt\\
&=\frac{1}{2}\int_0^1\frac{t^{-1/2}}{1+t}dt\\
&=\frac{1}{4}\left(\psi\left( \frac{3}{4}\right)-\psi\left( \frac{1}{4}\right) \right)\\
&=\frac{1}{4}\left(\pi \cot\left( \frac{\pi}{4}\right) \right)\\
&=\frac{\pi}{4}\\
\end{aligned}
$$
I used the fact that
$$\int_0^1\frac{t^{x-1}}{1+t}dt=\frac{1}{2}\left(\psi\left( \frac{x+1}{2}\right)-\psi\left( \frac{x}{2}\right) \right)$$
and that
$$\psi(1-x)-\psi(x)=\pi \cot(\pi x)$$
A: As shown in this answer,
$$
\sum_{k\in\mathbb{Z}}\frac{(-1)^k}{k+z}=\pi\csc(\pi z)\tag1
$$
Setting $z=\frac12$ gives
$$
\begin{align}
\pi
&=\sum_{k\in\mathbb{Z}}\frac{(-1)^k}{k+\frac12}\tag{2a}\\
&=2\sum_{k=0}^\infty\frac{(-1)^k}{k+\frac12}\tag{2b}\\
&=4\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}\tag{2c}\\
\end{align}
$$
Explanation:
$\text{(2a)}$: set $z=\frac12$ in $(1)$
$\text{(2b)}$: pair $k=-n-1$ and $k=n$
$\text{(2c)}$: multiply by $\frac22$
Therefore,
$$
\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}=\frac\pi4\tag3
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\bbox[5px,#ffd]{\sum_{k = 0}^{\infty}{\pars{-1}^{k} \over 2k + 1}} = -\ic\sum_{k = 0}^{\infty}{\ic^{2k + 1} \over 2k + 1} =
-\ic\sum_{k = 1}^{\infty}{\ic^{k} \over k}\,
{1 - \pars{-1}^{k} \over 2}
\\[5mm] = & \
-\ic\pars{2\ic\,\Im\sum_{k = 1}^{\infty}
{\ic^{k} \over k}{1 \over 2}} =
\Im\sum_{k = 1}^{\infty}{\ic^{k} \over k} =
\Im\bracks{-\ln\pars{1 - \ic}}
\\[5mm] = & \ -\arctan\pars{-1} =
\bbx{\color{#44f}{\pi \over 4}} \approx 0.7854
\end{align}
