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. 

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.
