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This question already has an answer here:

How do I find the sum of $\displaystyle 1+{2\over2} + {3\over2^2} + {4\over2^3} +\cdots$

I know the sum is $\sum_{n=0}^\infty (\frac{n+1}{2^n})$ and the common ratio is $(n+2)\over2(n+1)$ but i dont know how to continue from here

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marked as duplicate by user147263, heropup, TrueDefault, Claude Leibovici, hardmath Jan 6 '15 at 6:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ You're wrong about the common ratio; only a geometric series has a common ratio, and this isn't a geometric series. $\endgroup$ – MJD Jan 6 '15 at 4:33
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    $\begingroup$ @Fundamental the question in the title is not even close to this question, how can this be a duplicate? $\endgroup$ – Irrational Person Jan 6 '15 at 4:39
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    $\begingroup$ See en.wikipedia.org/wiki/Arithmetico-geometric_sequence $\endgroup$ – lab bhattacharjee Jan 6 '15 at 4:40
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    $\begingroup$ If a ratio depends on $n$, then it is not common. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jan 6 '15 at 4:53
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    $\begingroup$ I've posted answers to this question here a number of times. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jan 6 '15 at 4:54
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After establishing convergence, you could do the following: $$S = 1 + \frac22 + \frac3{2^2}+\frac4{2^3}+\dots$$ $$\implies \frac12S = \frac12 + \frac2{2^2} + \frac3{2^3}+\frac4{2^4}+\dots$$ $$\implies S - \frac12S = 1+\frac12 + \frac1{2^2} + \frac1{2^3}+\dots$$ which is probably something you can recognise easily...

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Consider the geometric series $$\sum_{n = 0}^\infty x^n = \frac{1}{1 - x}, \quad |x| < 1.$$

Differentiating both sides with respect to $x$,

$$\sum_{n = 1}^\infty nx^{n-1} = \frac{1}{(1 - x)^2},$$

or

$$\sum_{n = 0}^\infty (n+1)x^n = \frac{1}{(1 - x)^2}.$$

Now substitute $x = \frac{1}{2}$.

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Once you've determined that the sum converges (you can do this by the ratio test or by integration as other users have pointed out), you can find the value quite nicely. Let

$$S=1+\frac{2}{2}+\frac{3}{4}+\frac{4}{8}+\frac{5}{16}+\cdots$$

Then

$$2S=2+\frac{2}{1}+\frac{3}{2}+\frac{4}{4}+\frac{5}{8}+\frac{6}{16}+\cdots$$

From the second term on, this shares denominators with $S$ itself, so we can write

$$2S-S=2+1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots$$

But this is just a geometric series, so we get

$$S=\boxed{4}$$

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HINT: Consider the series $\sum_{n=0}^\infty x^n$ and differentiate (do you know this theorem?).

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$$1+\frac22+\frac34+\frac48+\frac5{16}+\cdots$$
$$=1+\frac12+\frac14+\frac18+\frac1{16}+\cdots$$ $$\ \ \ \ \ \ +\frac12+\frac14+\frac18+\frac1{16}+\cdots$$ $$\ \ \ \ \ \ +\frac14+\frac18+\frac1{16}$$ $$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\frac18+\frac1{16}+\cdots$$ $$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\cdots$$
$$=2+1+\frac12+\frac14+\cdots=\boxed4.$$

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$\sum x^n=\frac{1}{1-x}$ if $|x|<1 $ so

$$\frac{x}{1-x}=\sum x^{n+1}$$, then $$\frac{1}{(1-x)^{2}}=\sum (n+1)x^n$$

put $x=\frac{1}{2}$

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    $\begingroup$ There is not much point to repeat an answer, is there? $\endgroup$ – Timbuc Jan 6 '15 at 4:48
  • $\begingroup$ if you see the time i answer that question you may understand that $\endgroup$ – erfan soheil Jan 6 '15 at 4:49
  • $\begingroup$ #erfan About 7 minutes after the other answer was posted. Perhaps you're a slow typer, though. $\endgroup$ – Timbuc Jan 6 '15 at 4:50
  • $\begingroup$ yes i am slow typer mr fast typer $\endgroup$ – erfan soheil Jan 6 '15 at 4:52
  • $\begingroup$ @erfansoheil I didn't type any answer here at all. $\endgroup$ – Timbuc Jan 6 '15 at 4:53

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