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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.

Guy Fsone
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Forklift17
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2 Answers2

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

lab bhattacharjee
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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 $$\sum_{k=1}^{\infty}a^{k+1} \frac{1}{k(k+1)} =\sum_{k=1}^{\infty}a^{k+1} \left(\frac{1}{k} - \frac{1}{k+1}\right)= 1 ~~~\text{for} ~~a=1$$

Guy Fsone
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    Amazing (+1) ${}{}{}{}{}{}{}{}{}$ – Jaideep Khare Nov 21 '17 at 07:48
  • I get that result, but then what? Integrate twice and figure out what to do with the constants of integration? – Forklift17 Nov 21 '17 at 08:17
  • For the constants of integration take $a= 0$ and $a=1$ in the initial sum you will fine them. See the edit if it answers your question. – Guy Fsone Nov 21 '17 at 10:07
  • But integrating $\int (\frac{1}{a-1}+\frac{1}{a}), da = \log(a-1)+\log(a) + C_1$, and that is undefined when $a$ is 0 or 1. And why did we not take the derivative of the $1/a$ term in the first place? I am so lost :( – Forklift17 Nov 21 '17 at 18:11
  • That integral is not correct there is Minus instead of plus. also dont forget you have to integrate twice – Guy Fsone Nov 21 '17 at 18:47