Find $\sum_{k=1}^{\infty}a^k \left(\frac{1}{k} - \frac{1}{k+1}\right)$

Question

I want to find $$\sum_{k=1}^{\infty}a^k \left(\frac{1}{k} - \frac{1}{k+1}\right)$$

I'm not sure what to do here. Without the $a^k$ term, it's a simple telescoping series, but that term changes everything. I tried writing out the first few terms but could not come up with anything that doesn't turn right back into the original series.

Answer 1

Hint:

Using What is the correct radius of convergence for $\ln(1+x)$?,

for $-1\le x<1,$

$$\ln(1-x)=-\sum_{k=1}^n\dfrac{x^k}k$$

Now $$a^k\left(\dfrac1k-\dfrac1{k+1}\right)=\dfrac{a^k}k-\dfrac1a\cdot\dfrac{a^{k+1}}{k+1}$$

Answer 2

Hint: write $$\sum_{k=1}^{\infty}a^{k} \left(\frac{1}{k} - \frac{1}{k+1}\right)=\frac{1}{a} \sum_{k=1}^{\infty}a^{k+1} \frac{1}{k(k+1)}$$

Now see that $$\frac{d^2}{da^2} \sum_{k=1}^{\infty}a^{k+1} \frac{1}{k(k+1)}=\sum_{k=1}^{\infty}a^{k-1} =\frac{1}{a(1-a)} = \frac{1}{a-1} -\frac{1}{a}$$

Edit: added about the constants of integration see that $$\sum_{k=1}^{\infty}a^{k+1} \frac{1}{k(k+1)} = 0~~~\text{for} ~~a=0$$ and $$\sum_{k=1}^{\infty}a^{k+1} \frac{1}{k(k+1)} =\sum_{k=1}^{\infty}a^{k+1} \left(\frac{1}{k} - \frac{1}{k+1}\right)= 1 ~~~\text{for} ~~a=1$$