sum of the series of real numbers of following.

Question

Find the value of the series $$\sum_{n=0}^\infty \frac{n^{2}}{2^{n}}.$$ I tried the problem but not getting the sum. Please help.

Answer 1

Hint: Let $$f(x)=\sum_02^{-nx}~~~~~;~~x>0$$ and consider $f''(1)$.

Edit. Let's know: \begin{eqnarray} f(x)&=&\sum_0(2^{-x})^n\\ &=&\frac{1}{1-2^{-x}}\\ f''(x)=\sum_0(-n\ln2)^22^{-nx}&=&2^{-x}(\ln2)^2\frac{1-2^{-x}+2^{1-x}}{(1-2^{-x})^3}\\ f''(1)=(\ln2)^2\sum_0n^22^{-n}&=&6(\ln2)^2\\ \sum_0\frac{n^2}{2^n}&=&\color{blue}{6} \end{eqnarray}

Answer 2

$$S = \sum_{i=0}^n\frac{i^2}{2^i}$$

$$S -\frac{S}{2} = \sum_{i=0}^n\frac{i^2}{2^i} -\sum_{i=0}^n\frac{i^2}{2^{(i+1)}}$$

$$\frac{S}{2} = \sum_{i=0}^n\frac{i^2}{2^i} -\sum_{i=0}^n\frac{i^2}{2^{(i+1)}}$$

$$\frac{S}{2} = (\frac{1}{2}+\frac{4}{4}+\frac{9}{8}+\frac{16}{16}+ ... +\frac{n^2}{2^n}) - (\frac{1}{4}+\frac{4}{8}+\frac{9}{16}+\frac{16}{32}+ ... +\frac{n^2}{2^{(n+1)}})$$

$$\frac{S}{2} = (\frac{1}{2}+\frac{3}{4}+\frac{5}{8}+\frac{7}{16}+ ... +\frac{n^2-({n-1}^2)}{2^n}) - \frac{n^2}{2^{(n+1)}}$$

$$\frac{S}{2} = (\frac{1}{2}+\frac{3}{4}+\frac{5}{8}+\frac{7}{16}+ ... +\frac{2n-1}{2^n}) - \frac{n^2}{2^{(n+1)}}$$

$$\frac{S}{2} = \sum_{i=1}^n\frac{2i-1}{2^i}- \frac{n^2}{2^{(n+1)}}$$

$$\frac{S_\infty}{2} = \sum_{i=1}^\infty\frac{2i-1}{2^i}- \lim_{n\to\infty}\frac{n^2}{ 2^{(n+1)}}$$

$$\frac{S_\infty}{2} = \sum_{i=1}^\infty\frac{2i-1}{2^i}$$

$$\frac{S_\infty}{2}-\frac{S_\infty}{4} = \sum_{i=1}^\infty\frac{2i-1}{2^i} - \sum_{i=1}^\infty\frac{2i-1}{2^{(i+1)}}$$

$$\frac{S_\infty}{4} = (\frac{1}{2} + \frac{3}{4} +\frac{5}{8} + ...) - (\frac{1}{4} + \frac{3}{8} +\frac{5}{16} + ...)$$

$$\frac{S_\infty}{4} = \frac{1}{2} + (\frac{1}{2} +\frac{1}{4} + \frac{1}{8} +...)$$

$$S_\infty = 2 + 2(1 +\frac{1}{2} + \frac{1}{4} +...) = 2 + 4= 6$$

Answer 3

To give a rigorous answer, we first need to find the radius of convergence of the series $$\displaystyle\sum_{n=0}^{\infty}\dfrac{n^2}{2^n}=\displaystyle\sum_{n=0}^{\infty}n^2\left(\dfrac{1}{2}\right)^n$$

To do so, let's apply the Cauchy-Hadamard theorem. Consider the general term $a_n=n^2$. Then,

$$\displaystyle\lim_{n\to\infty}\sqrt[n]{|n^2|}=1$$

The radius of convergence will be $\rho= 1^{-1}=1$. Thus, we will be able to compute the infinite sum, as $\dfrac{1}{2}\in(-1,1)$.

Recall the sum of the geometric series $\displaystyle\sum_{n=0}^{\infty}r^n=\dfrac{1}{1-r}$ when $|r|<1$. Now consider a function $f(x)=\displaystyle\sum_{n=0}^{\infty}x^n$. On the interval $(-1,1)$ the series converges quasi-uniformly, so we can differentiate twice as follows:

$$f'(x)=\left(\displaystyle\sum_{n=0}^{\infty}x^n\right)'=\left(\dfrac{1}{1-r}\right)'\Leftrightarrow f'(x)=\displaystyle\sum_{n=0}^{\infty}nx^{n-1}=\dfrac{1}{(1-x)^2}$$

Multiplying now by $x$, $$g(x)=\displaystyle\sum_{n=0}^{\infty}nx^{n}=\dfrac{x}{(1-x)^2}$$ Differentiating $g(x)$,

$$g'(x)=\left(\displaystyle\sum_{n=0}^{\infty}nx^{n}\right)'=\left(\dfrac{x}{(1-x)^2}\right)'\Leftrightarrow g'(x)=\displaystyle\sum_{n=0}^{\infty}n^2x^{n-1}=\dfrac{x+1}{(1-x)^3}$$

Multiplying again by $x$ both sides we get $$\displaystyle\sum_{n=0}^{\infty}n^2x^{n}=\dfrac{x(x+1)}{(1-x)^3}$$

Substituting $x$ by $\dfrac{1}{2}$,

$$\displaystyle\sum_{n=0}^{\infty}n^2\left(\dfrac{1}{2}\right)^{n}=\dfrac{\frac{1}{2}\left(\frac{1}{2}+1\right)}{\left(1-\frac{1}{2}\right)^3}=6$$

Answer 4

for the infinity sum we get $$\sum_{i=0}^n\frac{i^2}{2^i}=2^{-n}(-6+3\cdot2^{n+1}-4n-n^2)$$ can you finish this?