Express the following in terms of elementary functions, without integrals or series: $\sum_{n=0}^\infty \frac{(−1)^nx^{n+1}}{n(n+1)}$ Not sure how to proceed. 
Do I play with the algebra, and knowing the fact that $\frac{1}x = \sum_{n=0}^\infty$ ? 
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\mathsf{No\ Integrals\ !!!}}$.

\begin{align}
\sum_{n = 1}^{\infty}\pars{-1}^{n}\,{x^{n + 1} \over n\pars{n + 1}} & =
x\sum_{n = 1}^{\infty}{\pars{-x}^{n} \over n} -
x\sum_{n = 1}^{\infty}{\pars{-x}^{n} \over n + 1} =
x\sum_{n = 1}^{\infty}{\pars{-x}^{n} \over n} -
x\sum_{n = 2}^{\infty}{\pars{-x}^{n - 1} \over n}
\\[5mm] & =
x\sum_{n = 1}^{\infty}{\pars{-x}^{n} \over n} +
\sum_{n = 2}^{\infty}{\pars{-x}^{n} \over n} =
x\sum_{n = 1}^{\infty}{\pars{-x}^{n} \over n} + x +
\sum_{n = 1}^{\infty}{\pars{-x}^{n} \over n}
\\[5mm] & = 
\pars{x + 1}\sum_{n = 1}^{\infty}{\pars{-x}^{n} \over n} + x =
\bbx{x - \pars{x + 1}\ln\pars{1 + x}}
\end{align}
A: Probably there is a typo in $$\sum_{n=0}^\infty \frac{(−1)^nx^{n+1}}{n(n+1)}$$
because the first term at $n=0$ would be infinite.
Supposing that the correct series is :
$$f(x)=\sum_{n=1}^\infty \frac{(−1)^nx^{n+1}}{n(n+1)}$$
$$f'(x)=\sum_{n=1}^\infty \frac{(−1)^nx^{n}}{n}=-\ln(1+x)$$
$$f(x)=-\int\ln(1+x)dx=-(1+x)\ln(1+x)+x+C$$
$f(0)=0\quad\to\quad C=0$
$$\sum_{n=1}^\infty \frac{(−1)^nx^{n+1}}{n(n+1)}=-(1+x)\ln(1+x)+x$$
In the result there is no integral and no series as requested. 
