$\displaystyle\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n(n+1)}=2\ln 2-1$ I have evaluated this sum and found that it is equal to $2\log2-1$. However, my friend found $\log2-1$. When we expanded, we got both the answers are same. So I am confused which one is correct answer.
Thank you.
 A: $$
\sum_{k=1}^{n}\frac{(-1)^{k+1}}{k(k+1)}=\sum_{k=1}^{n}(-1)^{k+1}\left(\frac1k-\frac1{k+1}\right)
=\sum_{k=1}^n\frac{(-1)^{k+1}}{k}+\sum_{k=1}^n\frac{(-1)^{k+2}}{k+1}
\\=2\sum_{k=1}^n\frac{(-1)^{k+1}}{k}-1-\frac{(-1)^{n+1}}{n+1}.
$$
Hence, 
$$
\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(n+1)}=2\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}-1=2\ln 2-1,
$$
since
$$
\sum_{n=1}^n\frac{(-1)^{n+1}}{n}=\ln 2.
$$
Proof. If $\lvert x\rvert<1$, then
$$
\frac{1}{1+x}=\sum_{k=0}^n(-1)^{k}x^k+\frac{(-1)^{n+1}x^{n+1}}{1+x},
$$
and hence
$$
\log(1+x)=\int_0^x\frac{dt}{1+t}=\sum_{k=0}^n(-1)^{k}\int_0^tt^k\,dt+\int_0^x\frac{(-1)^{n+1}t^{n+1}\,dt}{1+t} \\=\sum_{k=0}^n\frac{(-1)^kx^{n+1}}{n+1}+R_n(x).
$$
Clearly, for $x\in [0,1]$ 
$$
\lvert R_n(x)\rvert = \int_0^x\frac{t^{n+1}\,dt}{1+t}\le\int_0^xt^{n+1}\,dt\le \frac{1}{n+2}.
$$
Hence
$$
\ln 2=\lim_{x\to 1^-}\ln(1+x)=\lim_{x\to 1^-}\sum_{k=0}^n\frac{(-1)^kx^{n+1}}{n+1}+\lim_{x\to 1^-}R_n(x)=\sum_{k=0}^n\frac{(-1)^k}{n+1}+R_n(1),
$$
and hence
$$
\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}=\lim_{n\to\infty}\sum_{k=0}^n\frac{(-1)^k}{n+1}=\lim_{n\to\infty}\big(\ln 2+R_n(1)\big)=\ln 2.
$$
A: Using falling factorials and finite calculus we can write
$$\begin{align}\sum\frac{(-1)^{n+1}}{n(n+1)}\delta n&=\sum (-1)^{n+1}(n-1)^{\underline{-2}}\delta n\\&=(-1)^{n}(n-1)^{\underline{-1}}-\sum\frac{(-1)^{n+2}-(-1)^{n+1}}{(-1)(n+1)}\delta n\\&=\frac{(-1)^n}n+2\sum\frac{(-1)^{n}}{n+1}\delta n\end{align}$$
where the second step is summation by parts (check this, by example), and the indefinite sum of a falling factorial is defined as
$$\sum n^{\underline m}\delta n=\frac{n^{\underline{m+1}}}{m+1}+ K$$
Now taking limits above we have
$$\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n(n+1)}=\sum\nolimits_1^\infty\frac{(-1)^{n+1}}{n(n+1)}\delta n=\frac{(-1)^n}n\bigg|_1^\infty+2\sum\nolimits_1^\infty\frac{(-1)^{n}}{n+1}\delta n=\\=1+2(\ln (2)-1)=2\ln(2)-1$$
because the Taylor expansion of $\ln (1+x)$ is
$$\ln (1+x)=\sum_{n=1}^\infty(-1)^{n+1}\frac{x^k}{k},\quad\text{whenever }x\in(-1,1]$$
A: I got my answer in this way:
$\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n(n+1)}=\sum_{n=1}^{\infty}(-1)^{n+1}\left(\frac1n-\frac1{n+1}\right)=\sum_{n=1}^{\infty}\frac {(-1)^{n+1}}{n}-\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n+1}=\\log2+\log2-1=2\log2-1$
A: Let 
$$ f(x)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n(n+1)}x^{n+1} $$
and hence 
$$ f'(x)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}x^{n},  f''(x)=\sum_{n=1}^\infty(-1)^{n+1}x^{n-1}=\frac{1}{1+x}.$$
Note that
$$ f(1)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n(n+1)}, f'(0)=0.$$
So
\begin{eqnarray}
f(1)&=&\int_0^1\int_0^x\frac{1}{1+t}dtdx\\
&=&\int_0^1\int_t^1\frac{1}{1+t}dxdt\\
&=&\int_0^1\frac{1-t}{1+t}dt\\
&=&2\ln2-1.
\end{eqnarray}
