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could anyone help with this question?

$$\sum_{n=1}^{\infty}\frac{(x-\frac{1}{2})^{n+1}}{n(n+1)}$$

I have to find sum of this power series using differentiation or integration. Thanks a lot!

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  • $\begingroup$ The operation is called "differentiation". A "derivation" is a calculation or proof of a result. (Or an object from algebraic geometry, which is also not what you mean.) $\endgroup$ – Chappers Apr 27 '15 at 22:47
  • $\begingroup$ Thanks, I will edit question. $\endgroup$ – Krop Apr 27 '15 at 22:51
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Recall that: $$\frac{\left(x-\frac{1}{2}\right)^{n+1}}{n(n+1)}=\int_{0}^{x-\frac{1}{2}}\frac{y^n}{n}dy.$$ Then:

$$\sum_{n=1}^{\infty}\frac{(x-\frac{1}{2})^{n+1}}{n(n+1)} = \sum_{n=1}^{\infty}\int_{0}^{x-\frac{1}{2}}\frac{y^n}{n}dy = \\ = \int_{0}^{x-\frac{1}{2}}\sum_{n=1}^{\infty}\frac{y^n}{n}dy = -\int_{0}^{x-\frac{1}{2}}\log(1-y)dy = \\ = -\left((y-1)(\log(1-y)-1)\right)_{0}^{x-\frac{1}{2}} = \\=-\left(\log\left(x-\frac{3}{2}\right)-1\right)\left(x - \frac{3}{2}\right)+1.$$

References:

http://en.wikipedia.org/wiki/Taylor_series#List_of_Maclaurin_series_of_some_common_functions

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  • $\begingroup$ There appears to be a sign error in the last expression. $\endgroup$ – Mark Viola Apr 27 '15 at 23:09
  • $\begingroup$ Nice answer! +1 $\endgroup$ – jm324354 Apr 27 '15 at 23:24
  • $\begingroup$ @Dr.MV thanks for the advice, I fixed it. $\endgroup$ – the_candyman Apr 28 '15 at 7:24
  • $\begingroup$ You're welcome. My pleasure. $\endgroup$ – Mark Viola Apr 28 '15 at 13:29
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Differentiate with respect to $x$ twice: $$ S''(x) = \sum_{n=1}^{\infty} (x-1/2)^{n-1} = \sum_{k=0}^{\infty} (x-1/2)^{k}, $$ geometric series formula gives $$ S''(x) = \frac{1}{1-(x-1/2)} = \frac{1}{3/2-x}. $$ Now you have to integrate this twice to get the answer, using (which you should check) $S(1/2)=S'(1/2)=0$.

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Differentiate the sum $S(x)$ once reveals that

$$S'(x)=\sum_{n=1}^{\infty}\, \frac{(x-1/2)^n}{n}$$

which we recognize as the series representation of $\log(1-(x-1/2))$.

Integrate once and apply the condition that $S(1/2)=0$.

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