Infinite series convergence value problem

Question

Is it possible to find the convergent value of the series: $\sum\limits_{x=0}^{\infty}(x+3) \cdot a^x $ where $a$ is a constant less than 1?

I thought about expanding:

$\sum\limits_{x=0}^{\infty}(x+3) \cdot a^x =\sum\limits_{x=0}^{\infty}x \cdot a^x + 3 \sum\limits_{x=0}^{\infty} a^x$

and the second term is just $\cfrac{3}{1-a}$, but I don't know what to do with the first term. Is there a general method to solve these types of problems? I that there are many types of convergent tests but not sure about methods to actually find convergent value. Thanks in advance.

See this and its "linked" section. – David Mitra Jul 23 '13 at 23:16 — David Mitra, Jul 23 '13 at 23:16
Please don't use $x$ as the index variable. – mrf Jul 23 '13 at 23:46 — mrf, Jul 23 '13 at 23:46

score 2 · Accepted Answer · answered Jul 23 '13 at 23:18

$$\sum_{x=0}^{\infty} a^x = \dfrac1{1-a}$$ $$\sum_{x=0}^{\infty} xa^x = \sum_{x=0}^{\infty} a (xa^{x-1}) = \sum_{x=0}^{\infty} a\dfrac{d}{da}\left(a^x\right) = a\dfrac{d}{da}\left(\sum_{x=0}^{\infty} a^x\right) = a \dfrac{d}{da}\left(\dfrac1{1-a}\right) = \dfrac{a}{(1-a)^2}$$

score 1 · Answer 2 · answered Jul 23 '13 at 23:19

1

Continuing your own thoughts(Hint): $$(\frac{1}{1-x})' = (\sum_{n=0}^{\infty}x^n)' = \sum_{n=1}^{\infty}nx^{n-1} = \sum_{n=0}^{\infty}(n+1)x^n$$

answered Jul 23 '13 at 23:19

Alex Botev

1,106

Infinite series convergence value problem

2 Answers2