Find the sum of the series $\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\cdots$ My book directly writes-
$$\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\cdots=-\ln 2+1.$$
How do we prove this simply.. I am a high school student. 
 A: How about the following way? Let us prove that
$$\lim_{n\to\infty}\sum_{k=1}^{2n}\frac{(-1)^{k-1}}{k}=\ln 2.$$
Since we have
$$\begin{align}\sum_{k=1}^{2n}\frac{(-1)^{k-1}}{k}&=\left(\sum_{k=1}^{2n}\frac{(-1)^{k-1}}{k}+2\sum_{k=1}^{n}\frac{1}{2k}\right)-2\sum_{k=1}^{n}\frac{1}{2k}\\&=\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^{n}\frac{1}{k}\\&=\sum_{k=1}^{n}\frac{1}{n+k}\end{align}$$
we have
$$\begin{align}\lim_{n\to\infty}\sum_{k=1}^{2n}\frac{(-1)^{k-1}}{k}&=\lim_{n\to \infty}\sum_{k=1}^{n}\frac{1}{n+k}\\&=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{1}{1+(k/n)}\\&=\int_{0}^{1}\frac{1}{1+x}dx\\&=\ln 2.\end{align}$$
So, you'll have
$$\sum_{k=2}^{\infty}\frac{(-1)^k}{k}=1-\ln 2.$$
A: $$\sum_{k=2}^{+\infty}\frac{(-1)^k}{k}=\sum_{n\geq 1}(-1)^{n-1}\int_{0}^{1}x^n\,dx = \int_{0}^{1}\frac{x dx}{1+x} = 1-\int_{0}^{1}\frac{dx}{1+x}=\color{red}{1-\log 2}.$$
A: In calculus there's this famous alternating harmonic series:
$S = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ... + (-1)^{(n+1)} \cdot \frac{1}{n} + ... $
(*) It's convergent and its sum is equal to $ln2$
Your series is equal to exactly $T = -S + 1$ so it's
also convergent and its sum must be exactly $(-ln2+1)$  
I realize that I didn't prove this (*) statement. I am not
aware of an elementary proof but there might be one.  
A: Break up the series into two sub-series, depending on the parity of the denominator, then, after rewriting each of the two in terms of harmonic numbers, use the fact that $H_K~\simeq~\ln K$, and let $N\to\infty$, as follows:

$\begin{align}
\color{green}{S_{2N}}
&~=~\color{blue}{\dfrac11}\color{red}{-\dfrac12}\color{blue}{+\dfrac13}\color{red}{-\dfrac14}\color{blue}{+\dfrac15}\color{red}{-\dfrac16}+~\ldots~\color{blue}{+\dfrac1{2N-1}}\color{red}{-\dfrac1{2N}}~=~
\\\\
&~=~\color{blue}{\dfrac11}\color{red}{+\dfrac12}\color{blue}{+\dfrac13}\color{red}{+\dfrac14}\color{blue}{+\dfrac15}\color{red}{+\dfrac16}+~\ldots~\color{blue}{+\dfrac1{2N-1}}\color{red}{+\dfrac1{2N}}~-~
\\\\
&\color{red}{~\qquad~-\dfrac22\qquad~-\dfrac24\qquad-\dfrac26\qquad-\quad\ldots\qquad~-\frac2{2N}}~=~
\\\\
&~=~\color{blue}{\dfrac11}\color{red}{+\dfrac12}\color{blue}{+\dfrac13}\color{red}{+\dfrac14}\color{blue}{+\dfrac15}\color{red}{+\dfrac16}+~\ldots~\color{blue}{+\dfrac1{2N-1}}\color{red}{+\dfrac1{2N}}~-~
\\\\
&\color{red}{~\qquad~-\dfrac11\qquad~-\dfrac12\qquad-\dfrac13\qquad-\quad\ldots\qquad~-\frac1N}~=~
\\\\
&~=~\color{green}{H_{2N}~-~H_N~\simeq~\ln(2N)-\ln N~=~\ln\dfrac{2N}N~=~\ln2}.
\end{align}$

More rigorously, we could write $H_K~=~\ln K+\gamma_k$ , where $\displaystyle\lim_{k\to\infty}\gamma_k~=~\gamma$, see Euler-Mascheroni constant for more information.
