Evaluate $\displaystyle{\lim_{n\to \infty}}(1-\frac 12 +\frac 13 - \frac 14 + \cdots + \frac{1}{2n-1}-\frac{1}{2n}) $ I want to calculate this limit.
$$\lim_{n\to \infty}(1-\frac 12 +\frac 13 - \frac 14 + \cdots + \frac{1}{2n-1}-\frac{1}{2n}) $$
I tried to pair the terms of the sum in order to reduce each other but without any succes. By writing the sum:
$$\frac{1}{1*2} + \frac{1}{3*4}+\frac{1}{5*6}+\cdots+\frac{1}{2n(2n-1)}$$
I have not reached anything useful. Could you help me?
 A: We can write the given sequence by
$$u_n=\sum_{k=1}^{2n}\frac{(-1)^{k-1}}{k}$$
Using that
$$\frac1k=\int_0^1 t^{k-1}dt$$
we get
$$u_n=\int_0^1\sum_{k=1}^{2n}(-t)^{k-1}dt=\int_0^1\frac{dt}{1+t}-\int_0^1\frac{(-t)^{2n}}{1+t}dt$$
The first integral in the RHS is $\ln 2$ and for the second integral
$$\left\vert\int_0^1\frac{(-t)^{2n}}{1+t}dt\right\vert\le \int_0^1t^{2n}dt=\frac1{2n+1}\xrightarrow{n\to\infty}0$$
Hence we can conclude that $\lim u_n=\ln 2$.
A: Solution 1:
Prove by induction that 
$$1-\frac 12 +\frac 13 - \frac 14 + \cdots + \frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+...+\frac{1}{2n}$$
Then use the fact that $\frac{1}{n+1}+...+\frac{1}{2n}=\frac{1}{n}\sum_{k=1}^n \frac{1}{1+\frac{k}{n}}$ is a Riemann sum.
Solution 2
$$\begin{align}
u_n &=1-\frac 12 +\frac 13 - \frac 14 + \cdots + \frac{1}{2n-1}-\frac{1}{2n}\\&=1+\frac 12 +\frac 13 + \frac 14 + \cdots + \frac{1}{2n-1}+\frac{1}{2n}-2 (\frac{1}{2}+...+\frac{1}{2n})\\&=\left(1+\frac 12 +\frac 13 + \frac 14 + \cdots + \frac{1}{2n-1}+\frac{1}{2n} -\ln (2n)\right)\\&\qquad-\left(\frac{1}{1}+\frac{1}{2}+\dots+\frac{1}{n}-\ln(n)\right)+\ln(2)
\end{align}$$
A: Hint.
$$
x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots=\int_0^x\left(1-t+t^2-t^3-t^4+\cdots\right)\,dt=\int_0^x\frac{dt}{1+t}=\log(1+x)
$$
From this, one can obtain that
$$
1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots=\ln 2
$$
A: If I remember correctly, Ramanujan made several contributions with these types of infinite series. More precisely, he used series of the form$$\varphi(a,n)=1+2\sum\limits_{k=1}^n\frac 1{(ak)^3-ak}$$But that's enough of a history lesson.

Let's first perform a few manipulations to the summation at hand$$\begin{align*}\sum\limits_{k=1}^{2n}\frac {(-1)^{k+1}}k & =1-\frac 12+\frac 13-\frac 14+\frac 15-\ldots+\frac 1{2n-1}-\frac 1{2n}\\ & =\left(1+\frac 12+\frac 13+\ldots+\frac 1{2n}\right)-\left(\frac 22+\frac 24+\ldots+\frac 2{2n}\right)\\ & =\left(1+\frac 12+\frac 13+\ldots+\frac 1{2n}\right)-\left(1+\frac 12+\ldots+\frac 1n\right)\\ & =\sum\limits_{k=1}^{2n}\frac 1k-\sum\limits_{k=1}^n\frac 1k\end{align*}$$Now, we're ready. Using the definition of the Euler Mascheroni constant$$\lim\limits_{x\to\infty}\left\{\sum\limits_{k\leq x}\frac 1k-\log x\right\}=\gamma,$$we have$$\begin{align*}\lim\limits_{n\to\infty}\left\{\sum\limits_{k=1}^{2n}\frac {(-1)^{k+1}}k\right\} & =\lim\limits_{n\to\infty}\left\{\sum\limits_{k=1}^{2n}\frac 1k-\sum\limits_{k=1}^n\frac 1k\right\}\\ & =\lim\limits_{n\to\infty}\left\{\sum\limits_{k=1}^{2n}\frac 1k-\log 2n\right\}-\lim\limits_{n\to\infty}\left\{\sum\limits_{k=1}^n\frac 1k-\log n\right\}+\log2\\ & =\gamma-\gamma+\log2\\ & =\log2\end{align*}$$
P.S This was how Ramanujan found the limits of certain series in some of his early works. Other forms are$$\tfrac 43\log 2=1+\sum\limits_{k\geq1}\frac {2(-1)^k}{(3k)^3-3k}$$
A: Hint: We can write it as $S_n = (H_{2n}-\ln(2n))-(H_n-\ln n)+\ln2$ and use Mascheroni constant result. Here $H_n = 1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots + \dfrac{1}{n}$. Can you proceed?
