I have to find the sum of :

$$\frac{x^2}{2*1} - \frac{x^3}{3*2} + \frac{x^4}{4*3} - \frac{x^5}{5*4} +\cdots$$

So far I have :

$$\sum_{n=1}^{\infty} \frac{(-1)^{n-1} \, x^{n+1}}{(n+1)(n)}$$

which is very close to $\ln(1+x)$... but I just can't figure out what I have to do from there.

  • $\begingroup$ Hint: Since it's close to the series for $\ln(1+x)$, write out some terms and compare the two series. Do you see how to get from one to the other? $\endgroup$ – rogerl Oct 13 '18 at 20:06


On its domain of convergence, the derivative of the sum of this power series is $$x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\dotsm=\ln(1+x).$$

  • $\begingroup$ Thanks, with that I get that my function is (x+1)ln(x+1) +ln(x+1)-x-1+C? Just wondering what to do with C? Do I just leave it in? $\endgroup$ – Cappuccino Oct 13 '18 at 20:34
  • $\begingroup$ No. The power series is equal to $0$ if $x=0$, so you can deduce a value for $C$. $\endgroup$ – Bernard Oct 13 '18 at 20:38
  • 1
    $\begingroup$ The exact solution is $\; (x+1)\ln(x+1)-x$. $\endgroup$ – Bernard Oct 13 '18 at 20:45

Note that $$\frac {\mathrm d^2}{\mathrm dx^2}\frac{(-1)^nx^n}{n(n-1)}=\frac {\mathrm d}{\mathrm dx}\frac{(-1)^nx^{n-1}}{n-1}=(-1)^nx^{n-2}=(-x)^{n-2}$$ so that we expect $f(x)_=\sum_{n=2}^\infty\frac{(-1)^nx^n}{n(n-1)}$ to be a function with $f''(x)=\sum_{n=0}^\infty (-x)^{n} =\frac1{1+x}$.


\begin{align} \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \, x^{n+1}}{(n+1)(n)} &= \sum_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{n+1}\right) \, (-1)^{n-1} \, x^{n+1} \\ &= x \, \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \, x^{n}}{n} + \sum_{n=1}^{\infty} \frac{(-1)^n \, x^{n+1}}{n+1} \\ &= x \, \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \, x^{n}}{n} + \sum_{n=2}^{\infty} \frac{(-1)^{n-1} \, x^{n}}{n} \\ &= (x+1) \, \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \, x^{n}}{n} - x \\ &= (x+1) \, \ln(1+x) - x. \end{align}


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