Infinite Series $a_k=\frac{(-1)^{k+1}}{2k-1}$ Let $$a_k=\frac{(-1)^{k+1}}{2k-1}.$$
Compute the sum $\sum_{k=1}^{\infty} {a_k}$.
Then compute $\lim _ { n \to \infty}\left(\sum_{k=n}^{2n}|a_k|\right)$.
The sum I want to compute is $1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}. . .$. I know the sum should come out to be $\frac{\pi}{4}$, but I don't know how to get to that point. I don't know where to start on computing the limit. 
 A: For the first one, recall that
$$\arctan(x)=\int_0^x\frac1{1+t^2}\ dt$$
Then use the geometric series to see that for $t\in(-1,1)$,
$$\frac1{1+t^2}=\sum_{k=1}^\infty(-1)^{k+1}t^{2k-2}$$
Then rigorously show that when $x=1$, we can interchange the integral and series to get
$$\frac\pi4=\arctan(1)=\int_0^1\frac1{1+t^2}\ dt=\sum_{k=1}^\infty\int_0^1(-1)^{k+1}t^{2k-2}\ dt=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{2k-1}$$

One may then note that $|a_k|=\frac1{2k-1}$, thus, we want to compute the following limit:
$$L=\lim_{n\to\infty}\sum_{k=n}^{2n}\frac1{2k-1}$$
This may easily be done by noticing that
$$0\le\sum_{k=n}^{2n}\frac1{2k-1}-\int_n^{2n}\frac1{2x-1}\ dx\le\frac1{2n-1}+\frac1{4n-1}$$
And by seeing that
$$\int_n^{2n}\frac1{2x-1}\ dx=\frac12\ln\left(\frac{4n-1}{2n-1}\right)\stackrel{n\to\infty}\longrightarrow\frac12\ln(2)$$
We conclude by the squeeze theorem that
$$L=\frac12\ln(2)$$
A: HINT:
Let $f(x)=\arctan(x)$.  Then $f'(x)=\frac{1}{1+x^2}=\sum_{k=0}^{\infty} (-x^2)^k$.
Now integrate term by term and set $x=1$

$$\begin{align}
\sum_{k=n}^{2n}\frac{1}{2k-1}&=\sum_{k=0}^{n}\frac1{2n+2k-1}\\\\
&=\frac1{2n}\sum_{k=0}^n\frac{1}{1+\frac{k-1/2}{n}}\tag 1
\end{align}$$
Now note that $(1)$ is the Riemann Sum for $\frac12\int_0^1 \frac{1}{1+x}\,dx$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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*

*\begin{align}
\sum_{k = 1}^{\infty}{\pars{-1}^{k + 1} \over 2k - 1} & =
\sum_{k = 1}^{\infty}{\ic^{\pars{2k - 1} + 3} \over 2k - 1} =
-\ic\sum_{k = 1}^{\infty}{\ic^{2k - 1} \over 2k - 1} =
-\ic\sum_{k = 1}^{\infty}{\ic^{k} \over k} +
\ic\sum_{k = 1}^{\infty}{\ic^{2k} \over 2k}
\\[5mm] & =
\ic\ln\pars{1 - \ic} - {1 \over 2}\,\ic\ln\pars{1 - \ic^{2}} =
\ic\bracks{{1 \over 2}\,\ln\pars{2} - {\pi \over 4}\ic} -
{1 \over 2}\,\ic\ln\pars{2} = \bbx{\ds{\pi \over 4}}
\end{align}



*\begin{align}
\lim_{n \to \infty}\sum_{k = n}^{2n}{1 \over 2k - 1} & =
{1 \over 2}\,\lim_{n \to \infty}\pars{%
\sum_{k = 1}^{2n}{1 \over k - 1/2} - \sum_{k = 1}^{n - 1}{1 \over k - 1/2}}
\end{align}


Note that

\begin{align}
\sum_{k = 1}^{N}{1 \over k - 1/2} & = \sum_{k = 0}^{N - 1}{1 \over k + 1/2} =
\sum_{k = 0}^{\infty}\pars{{1 \over k + 1/2} - {1 \over k + N + 1/2}} =
\Psi\pars{N + {1 \over 2}} - \Psi\pars{1 \over 2}
\\[5mm] & \sim
-\Psi\pars{1 \over 2} + \ln\pars{N + {1 \over 2}}\quad \mbox{as}
\quad N \to \infty
\end{align}

\begin{align}
\mbox{Then,}\quad \lim_{n \to \infty}\sum_{k = n}^{2n}{1 \over 2k - 1} & =
{1 \over 2}\,\lim_{n \to \infty}\bracks{%
\ln\pars{2n + {1 \over 2}} - \ln\pars{n - {1 \over 2}}} =
\bbx{\ds{{1 \over 2}\,\ln\pars{2}}}
\end{align}
A: Let $\varphi , \psi : [- \pi , \pi] \to \mathbb{R}$ given by $\varphi = \frac{1}{2} Id_{[- \pi , \pi]}$ and $\psi = {\chi}_{[- \pi , 0]} - {\chi}_{(0 , \pi)}$. Let's expand $\varphi$ and $\psi$ to $\mathbb{R}$ such that $\varphi$ and $\psi$ are both $2 \pi$-periodic functions. Two of a lot of forms to compute these sum are computing Fourier series of $\varphi$ and $\psi$.
