Question

Find the value of $\sum_{k=1}^{\infty} \frac{k^2}{2^k}$

Answer 1

6

Answer 2

Let the sum be $S$. This is an Arithmetico-Geometric Progression (AGP) sum.

We know the sum of an infinite geometric series for $|r|<1$:

$$\sum_{k=0}^{\infty} r^k = \frac{1}{1-r}$$

Differentiate with respect to $r$:

$$\sum_{k=1}^{\infty} k r^{k-1} = \frac{1}{(1-r)^2}$$

Multiply by $r$:

$$\sum_{k=1}^{\infty} k r^k = \frac{r}{(1-r)^2} \quad (*)$$

Differentiate $(*)$ with respect to $r$:

$$\sum_{k=1}^{\infty} k^2 r^{k-1} = \frac{d}{dr}\left(\frac{r}{(1-r)^2}\right)$$

Using the quotient rule $\frac{d}{dr}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$:

$$\frac{d}{dr}\left(\frac{r}{(1-r)^2}\right) = \frac{1 \cdot (1-r)^2 - r \cdot 2(1-r)(-1)}{((1-r)^2)^2}$$ $$= \frac{(1-r)^2 + 2r(1-r)}{(1-r)^4} = \frac{(1-r)(1-r+2r)}{(1-r)^4}$$ $$= \frac{1+r}{(1-r)^3}$$

So, we have:

$$\sum_{k=1}^{\infty} k^2 r^{k-1} = \frac{1+r}{(1-r)^3}$$

To get $\sum_{k=1}^{\infty} k^2 r^k$, multiply by $r$:

$$\sum_{k=1}^{\infty} k^2 r^k = \frac{r(1+r)}{(1-r)^3}$$

In this problem, $r = \frac{1}{2}$. Substitute $r=\frac{1}{2}$ into the formula:

$$S = \frac{\frac{1}{2}\left(1+\frac{1}{2}\right)}{\left(1-\frac{1}{2}\right)^3} = \frac{\frac{1}{2} \cdot \frac{3}{2}}{\left(\frac{1}{2}\right)^3}$$ $$S = \frac{\frac{3}{4}}{\frac{1}{8}} = \frac{3}{4} \times 8 = 6$$