# By applying term-wise differentiation and integration find the sum of the series $\sum_{k=1}^{\infty}\frac{x^k}{k}$

I need to find the sum of the following series: $$\sum_{k=1}^{\infty}\frac{x^k}{k}$$ on the interval $x\in[a,b], -1<a<0<b<1$ using term-wise differentiation and integration.

Can anyone give me a hint on how to start this using the method described because it seems fairly simple but I'm stumped.

• You have $$x+\frac{x^2}2=\frac{x^3}3+\ldots\;;$$ what series do you get when you differentiate this term by term, just as if it were a polynomial? Once you have that, you should recognize the result as a very familiar series for which you know a closed form. Integrate that closed form to get the original function. – Brian M. Scott Nov 10 '14 at 14:16
• Oh I think I see. Differentiating everything gives $1+x+x^2+...=\frac{1}{1-x}$. So then do I just integrate back and that's my answer? – user184036 Nov 10 '14 at 14:19
• That’s right. You could write it up and post it as an answer to your own question. – Brian M. Scott Nov 10 '14 at 14:19
• The only reason I didn't accept the answer on this question was because Brian's comments helped me reach the answer before you posted your answer. So because I didn't use your answer I didn't accept it. – user184036 Nov 26 '14 at 11:23
• @john.smith That's fine, but The difference was just 3 minutes, Interestingly, It takes more than $3$ minutes to write an answer of this length! – Aditya Hase Dec 1 '14 at 12:37

$$f(x)=\sum_{k=1}^{\infty}\frac{x^k}{k}=x+\frac{x^2}2+\frac{x^3}3+\ldots$$ By differentiang w.r.t. $x$ $$f'(x)=\sum_{k=1}^{\infty} x^{k-1}=1+x+x^2+\cdots$$
for$|x|\lt1$we have
$$f'(x)=1+x+x^2+\cdots=\frac{1}{1-x}$$ by integrating we get $$f(x)=-\ln (1-x)+C$$ and can determine the unique value of $C$ to get
$$\sum_{k=1}^{\infty}\frac{x^k}{k}=-\ln (1-x)$$
• ... and can determine the unique correct value of $C$ – Hagen von Eitzen Nov 10 '14 at 14:26