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