Sum of power series using derivation or integration

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!

• 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.) – Chappers Apr 27 '15 at 22:47
• Thanks, I will edit question. – Krop Apr 27 '15 at 22:51

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

• There appears to be a sign error in the last expression. – Mark Viola Apr 27 '15 at 23:09
• Nice answer! +1 – jm324354 Apr 27 '15 at 23:24
• @Dr.MV thanks for the advice, I fixed it. – the_candyman Apr 28 '15 at 7:24
• You're welcome. My pleasure. – Mark Viola Apr 28 '15 at 13:29

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) $S(1/2)=S'(1/2)=0$.

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 $S(1/2)=0$.