Compute the series Compute the series
$$1)\space\sum_{n=1}^{\infty}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n+1}\right)\frac{1}{n(n+1)}$$
$$2)\space\sum_{n=1}^{\infty}\left(1-\frac{1}{2}+\frac{1}{3}-\cdots(-1)^{n}\frac{1}{n+1}\right)\frac{1}{n(n+1)}$$
 A: Hint: Consider the partial fraction expansion of $\frac{1}{n(n+1)}$
A: In both of these, we will use the telescoping sum
$$
\begin{align}
\sum_{k=n}^\infty\frac1{k(k+1)}
&=\sum_{k=n}^\infty\frac1k-\frac1{k+1}\\
&=\frac1n
\end{align}
$$
$1)$ Here is one way:
$$
\begin{align}
\sum_{n=1}^\infty\sum_{k=0}^n\frac1{k+1}\frac1{n(n+1)}
&=1+\sum_{n=1}^\infty\sum_{k=1}^n\frac1{k+1}\frac1{n(n+1)}\\
&=1+\sum_{k=1}^\infty\sum_{n=k}^\infty\frac1{k+1}\frac1{n(n+1)}\\
&=1+\sum_{k=1}^\infty\frac1{k+1}\frac1k\\[6pt]
&=1+1\\[12pt]
&=2
\end{align}
$$
$2)$ Here is a similar way:
$$
\begin{align}
\sum_{n=1}^\infty\sum_{k=0}^n(-1)^k\frac1{k+1}\frac1{n(n+1)}
&=1+\sum_{n=1}^\infty\sum_{k=1}^n(-1)^k\frac1{k+1}\frac1{n(n+1)}\\
&=1+\sum_{k=1}^\infty\sum_{n=k}^\infty(-1)^k\frac1{k+1}\frac1{n(n+1)}\\
&=1+\sum_{k=1}^\infty(-1)^k\frac1{k+1}\frac1k\\
&=1+\sum_{k=1}^\infty(-1)^k\frac1k-\sum_{k=1}^\infty(-1)^k\frac1{k+1}\\
&=1+\sum_{k=1}^\infty(-1)^k\frac1k+\sum_{k=1}^\infty(-1)^{k+1}\frac1{k+1}\\[6pt]
&=1-\log(2)+(1-\log(2))\\[12pt]
&=2-2\log(2)
\end{align}
$$
A: For the first part, recognize that $\frac {1}{n(n+1)} = \frac {1}{n} - \frac {1}{n+1}$. So the sum can we written as 
$$ \sum_{n=1}^{\infty} ( 1 + \frac {1}{2} + \frac {1}{3}+ \ldots + \frac {1}{n+1} ) ( \frac {1}{n} - \frac {1}{n+1}) $$
Break this up into the separate terms of the right parenthesis, and we find that consecutive terms cancel out. Note that we are canceling terms that are directly next to each other, so this step is justified. We'd obtain
$$\sum_{n=1}^{\infty} \big[ (1 + \frac {1}{2} + \ldots + \frac  {1}{n+1}) - (1 + \frac {1}{2} + \ldots + \frac  {1}{n}) \big] (\frac {1}{n}) = \sum_{n=1}^{\infty} \frac {1}{(n+1)n}$$
Finally, evaluate this as telescoping series $\sum_{n=1}^{\infty} \frac {1}{(n+1)n} = \sum_{i=1}^{\infty} \frac {1}{n} - \frac {1}{n+1} = 1$.

With these ideas, you should try the second part by yourself. At the most, you will need to evaluate $\sum_{n=1}^\infty \frac {(-1)^n}{n+1}$, which you can do by using the MacLaurin expansion $\ln(1+x) = x - \frac {x^2}{2} + \frac {x^3}{3} - \frac {x^4}{4}$ for a suitable value of $x$.
