Sum $\sum_{k=0}^\infty \frac{k+1}{3^k}$

Question

$$\sum_{k=0}^\infty \frac{k+1}{3^k}$$

Steps:

$$\sum_{k=0}^\infty \frac{k}{3^k} + \sum_{k=1}^\infty \frac{1}{3^k}$$

$$Sn = \frac {a1(1-r^n)}{1-r}$$

$$r = \frac{second term}{first term}$$ $$a1 = \frac {1}{3^0}$$ $$a2 = \frac {1}{3} + \frac {1}{3^1}$$

$a1 = 0+1 = 1$
a2 = $\frac{2}{3}$

r = 2/3

$$S∞ = \frac {a1(1-r^∞)}{1-r}$$ $$S∞ = \frac {1(1- (2/3)^∞)}{1-(2/3)}$$ $$S∞ = \frac {1(1- 0)}{1-(2/3)}$$ $$S∞ = \frac {1}{1/3}$$ $$S∞ = 3$$

Here I get the sum being equal to three when in fact it is apparently 9/4. I am finding very limited information on how to solve infinite sum of series.

Answer 1

hint:

$$\sum_{k=0}^n(k+1)x^k=\biggl(\sum_{k=0}^nx^{k+1}\biggr)'=\biggl(\sum_{k=1}^{n+1}x^{k}\biggr)'.$$

Some details:

As a result,$$\sum_{k=0}^\infty(k+1)x^k=\biggl(\sum_{k=1}^{\infty}x^{k}\biggr)'.$$ Now it is well-known that $$\sum_{k=1}^{\infty}x^{k}=\frac x{1-x}=-1+\frac1{1-x}, \enspace\text{so }\sum_{k=0}^\infty(k+1)x^k=\\frac1{(1-x)^2}.$$

Answer 2

$$\begin{align} \sum_{k=0}^\infty \frac{k+1}{3^k} &=\frac{3}{2}\bigg(1-\frac{1}{3}\bigg)\sum_{k=0}^\infty \frac{k+1}{3^k}\\ &=\frac{3}{2}\sum_{k=0}^\infty \bigg(\frac{k+1}{3^k}-\frac{k+1}{3^{k+1}}\bigg)\\ &=\frac{3}{2}\bigg(\frac{1}{3^0}-\frac{1}{3^1}+\frac{2}{3^1}-\frac{2}{3^{2}}+\frac{3}{3^2}-\frac{3}{3^3}+\frac{4}{3^3}-\frac{4}{3^4}+...\bigg)\\ &=\frac{3}{2}\bigg(\frac{1}{3^0}+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...\bigg)\\ &=\frac{3}{2}\bigg(\frac{1}{1-\frac{1}{3}}\bigg)\\ &=\frac{3}{2}\cdot \frac{3}{2} \\ &=\frac{9}{4} \\ \end{align}$$

Answer 3

HINT

In general, $$ S(x) = \sum_{k=0}^\infty x^k = \frac{1}{1-x}, \quad \forall |x| < 1. $$ Therefore, $$ S'(x) = \frac{d}{dx}\sum_{k=0}^\infty x^k = \sum_{k=0}^\infty \frac{d}{dx}x^k = \sum_{k=1}^\infty k x^{k-1} = \frac{1}{x} \sum_{k=1}^\infty k x^k $$ but transforming the other side, $$ S'(x) = \frac{d}{dx} \frac{1}{1-x} = \frac{1}{(1-x)^2}. $$ Therefore, $$ \frac{x}{(1-x)^2} = xS'(x) = \sum_{k=1}^\infty k x^k, \quad \forall |x| <1. $$