Hi! I need help in finding the sum of the series:
I would appreciate any kind of help here, since I don't even know how to start.
I think that x=0,1 means x=1/10 which gives a convergent series. Anyway I'll solve it for general.
$\displaystyle S = \sum_{n=0}^{\infty} \frac{x^n}{n+1}$
$\displaystyle x\cdot S = \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1}$
$\displaystyle x\cdot S = \int \sum_{n=0}^{\infty} \frac{\partial}{\partial x} \frac{x^{n+1}}{n+1}~dx$
$\displaystyle x\cdot S = \int \sum_{n=0}^{\infty} x^n~dx$
$\displaystyle x\cdot S = \int \frac{1}{1-x} ~dx$
$\displaystyle x\cdot S = - \ln (1-x) + C$
We can find C using,
$\displaystyle x\cdot S = \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1} = - \ln (1-x) + C$
Use $\displaystyle x = 0$, then
$\displaystyle 0=-\ln (1-0)+C$
$\displaystyle C=0$
So finally,
$\displaystyle x\cdot S = - \ln (1-x)$
$\displaystyle S = - \frac{\ln (1-x)}{x}$
$\displaystyle \sum_{n=0}^{\infty} \frac{x^n}{n+1} = - \frac{\ln (1-x)}{x}$
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Assuming $\displaystyle x=0.1$ (this is the x in the original question, so 1-6x makes 0.4),
$\displaystyle S = - \frac{\ln (1-0.4)}{0.4} \approx 1.27706$
$\displaystyle \sum\limits_{n\,=\,1}^{\infty }{\frac{1}{2^{n}n(n+1)}}=\sum\limits_{n\,=\,1}^{\i nfty }{\frac{(n-1)!}{2^{n}(n+1)!}}=\sum\limits_{n\,=\,1}^{\infty }{\frac{\Gamma (n)\Gamma (2)}{2^{n}\Gamma (n+2)}}.$
Hence the sum equals,
$\displaystyle \sum\limits_{n\,=\,1}^{\infty }{\int_{0}^{1}{\frac{x^{n-1}(1-x)}{2^{n}}\,dx}}=\int_{0}^{1}{\left\{ \frac{1-x}{x}\sum\limits_{n\,=\,1}^{\infty }{\left( \frac{x}{2} \right)^{n}} \right\}\,dx}=\int_{0}^{1}{\frac{1-x}{2-x}\,dx}$
and the conclusion follows. (This sum can be tackled from many ways.)
Since $\displaystyle \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$:
$\displaystyle S = \sum_{n=1}^{\infty} \frac{1}{n(n+1)2^n} = \sum_{n=1}^{\infty} \frac{1}{n \, 2^n} - \sum_{n=1}^{\infty} \frac{1}{(n+1) \, 2^n}$
$\displaystyle = \sum_{n=1}^{\infty} \frac{1}{n \, 2^n} - 2 \sum_{n=1}^{\infty} \frac{1}{(n+1) \, 2^{n+1}}$ ........
*Ahem*
Plato remarked that it was.
I doubt Krizalid's approach was what Plato had in mind (which in no way is meant to diminish K's solution).
My post is what Plato might have had in mind (I won't presume that it was). I think it's evident that Plato's remark was both reasonable and accurate.
True, since most people don't study to the level that this question is contexted within.