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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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    $\begingroup$ 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, 2015 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? $\endgroup$ Jan 6, 2015 at 4:39
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    $\begingroup$ See en.wikipedia.org/wiki/Arithmetico-geometric_sequence $\endgroup$ Jan 6, 2015 at 4:40
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    $\begingroup$ If a ratio depends on $n$, then it is not common. ${}\qquad{}$ $\endgroup$ Jan 6, 2015 at 4:53
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    $\begingroup$ I've posted answers to this question here a number of times. ${}\qquad{}$ $\endgroup$ Jan 6, 2015 at 4:54

6 Answers 6

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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, 2015 at 4:48
  • $\begingroup$ if you see the time i answer that question you may understand that $\endgroup$ Jan 6, 2015 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, 2015 at 4:50
  • $\begingroup$ yes i am slow typer mr fast typer $\endgroup$ Jan 6, 2015 at 4:52
  • $\begingroup$ @erfansoheil I didn't type any answer here at all. $\endgroup$
    – Timbuc
    Jan 6, 2015 at 4:53

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