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I have tried to compute the first few terms to try to find a pattern but I got

$$\frac{1}{2}+\frac{1}{2}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\frac{6}{64}$$

but I still don't see any obvious pattern(s). I also tried to look for a pattern in the question, but I cannot see any pattern (possibly because I'm overthinking it?) Please help me with this problem.

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  • $\begingroup$ It's a arithmetico geometric series. $\endgroup$ Jul 22, 2016 at 14:07
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    $\begingroup$ See this. $\endgroup$ Jul 22, 2016 at 14:07
  • $\begingroup$ @DavidMitra Please how to turn an answer as a 'community answer'? $\endgroup$ Jul 22, 2016 at 14:10
  • $\begingroup$ @OlivierOloa There's a button for it when you write it; I don't know if it can be changed on a submitted answer. $\endgroup$ Jul 22, 2016 at 14:12
  • $\begingroup$ @DavidMitra Ok, thank you. I'll delete my answer. $\endgroup$ Jul 22, 2016 at 14:13

5 Answers 5

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$$I=\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\frac{6}{64}+\cdots$$ $$2I=1+1+\frac{3}{4}+\frac{4}{8}+\frac{5}{16}+\frac{6}{32}+\cdots$$ $$2I-I=1+\left(1-\frac 12 \right)+\left(\frac 34 -\frac 24 \right)+\left(\frac 48 -\frac 38 \right)+\left(\frac {5}{16} -\frac {4}{16} \right)+\cdots$$ $$I=1+\frac 12+\frac 14+\frac 18+\cdots=2$$

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$$\sum_{n=1}^\infty nx^n=x\sum_{n=1}^\infty nx^{n-1}=x\sum_{n=0}^\infty (x^n)'=x\Bigl(\sum_{n=0}^\infty x^n\Bigr)'=x\Bigl(\frac1{1-x}\Bigr)'=\frac x{(1-x)^2}.$$

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Summation by parts gives: $$ \sum_{n=1}^{N}\frac{n}{2^n} = N\left(1-\frac{1}{2^N}\right)-\sum_{n=1}^{N-1}\left(1-\frac{1}{2^k}\right)=1-\frac{N}{2^N}+\left(1-\frac{1}{2^{N-1}}\right)=\color{red}{2-\frac{N+2}{2^N}}. $$

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Let

$$ L = \sum_{n=1}^\infty \frac{n}{2^n} $$

Then,

$$ L = \frac{1}{2} + \sum_{n=2}^\infty \frac{n}{2^n} \\ L = \frac{1}{2} + \sum_{n=1}^\infty \frac{n+1}{2^{n+1}} \\ L = \frac{1}{2} + \frac{1}{2} \sum_{n=1}^\infty \frac{n}{2^n} + \frac{1}{2} \sum_{n=1}^\infty \frac{1}{2^n} \\ L = \frac{1}{2} + \frac{L}{2} + \frac{1}{2} \sum_{n=1}^\infty \frac{1}{2^n} \\ \frac{L}{2} = \frac{1}{2} + \frac{1}{2} \sum_{n=1}^\infty \frac{1}{2^n} \\ \frac{L}{2} = \frac{1}{2} + \frac{1}{2} \\ \boxed{L = 2} $$

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First consider the partial sum by showing $$ \sum_{n=1}^k \frac{n}{2^n} = 2^{-k}(-k+2^{k+1}-2). $$ Now $k\to \infty$ gives the result $2$.


Edit: Proof of the partial sum by induction

$k=1$: $$\sum_{n=1}^1 \frac{1}{2} = 1/2 = 2^{-1}(-1+2^{2}-2) \quad \checkmark$$

Let $\sum_{n=1}^k \frac{n}{2^n} = 2^{-k}(-k+2^{k+1}-2)$ be true for any $k\geq1$.

Induction step: $$ \sum_{n=1}^{k+1} \frac{n}{2^n} = \sum_{n=1}^k \frac{n}{2^n} + \frac{k+1}{2^{k+1}} = 2^{-k}(-k+2^{k+1}-2) + \frac{k+1}{2^{k+1}} = \frac{2(-k+2^{k+1}-2)+k+1}{2^{k+1}} = \frac{-k+2^{k+2}-3}{2^{k+1}} = 2^{-(k+1)}(-(k+1)+2^{k+2}-2) $$

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