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Not exactly sure how to approach. Cannot do telescoping, or break up the sum into two parts either, cannot find a way to express division as the sum here. Although, if it was something like $\frac{1}{n(n+1)}$ we could write $\frac{1}{n}-\frac{1}{n+1}$. Would like to solve this with simple/clever arithmetic.

Krzysztof Myśliwiec
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Naz
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    I think there is a mistake in your profile: $i^i=e^{-\pi/2}\ne -i$ – Franklin Pezzuti Dyer Oct 05 '18 at 20:58
  • :) $i^2 = -1$ :) – Naz Oct 05 '18 at 21:00
  • But your profile says that $i^i=-1$, not $i^2=-1$. – Franklin Pezzuti Dyer Oct 05 '18 at 21:02
  • oh dear me, thanks for pointing out. I looked like uninitiated all this time – Naz Oct 05 '18 at 21:05
  • No problem! :) $\space$ – Franklin Pezzuti Dyer Oct 05 '18 at 21:09
  • $$\displaystyle \lim_{n \to \infty} \sum_{k=1}^n \frac{3k}{2^k}=3\displaystyle \lim_{n \to \infty} \sum_{k=1}^n \frac{k}{2^k} $$

    OK- for $\displaystyle \sum_{k=1}^{\infty} \frac{k}{2^k}$ I needed a bit help from WolframAlpha: $$ \displaystyle \sum_{k=1}^{\infty} \frac{k}{2^k}=2 $$

     – Krzysztof Myśliwiec Oct 05 '18 at 21:18
  • @KrzysztofMyśliwiec that is the content of the problem, and it is generally solved by differentiating the series below – operatorerror Oct 05 '18 at 22:18
  • @qbert I got an idea today: $\displaystyle \sum_{k=1}^{\infty} \frac{k}{2^k}=\frac{1}{2^1}+\frac{2}{2^2}+\frac{1+2}{2^3}+\frac{2^2}{2^4}+\frac{1+2^2}{2^5}+\frac{2+2^2}{2^6}+\frac{1+2+2^2}{2^7}+\frac{2^3}{2^8}+\frac{1+2^3}{2^9}+...=$ $\frac{1}{2^1}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\frac{1}{2^9}+...+\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^4}+\frac{1}{2^6}+\frac{1}{2^5}+\frac{1}{2^5}+\frac{1}{2^6}+...=> ? 2 \displaystyle \sum_{k=1}^{\infty} \frac{1}{2^k}$ . – Krzysztof Myśliwiec Oct 06 '18 at 17:44

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hint

Consider the series of functions

$$\sum \frac{x^k}{2^k}$$

and its derivative.

hamam_Abdallah
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  • I know I can use L'Hopitals here, but this is the school programme, so should be solvable without derivatives :) – Naz Oct 05 '18 at 20:59