Sum of infinite series $1+\frac22+\frac3{2^2}+\frac4{2^3}+\cdots$

Question

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

Answer 1

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...

Answer 2

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

Answer 3

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

$$S=\boxed{4}$$

Answer 4

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

Answer 5

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

Answer 6

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

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

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