Convergence and Limit of Infinite Series with Exponential Terms

[willworkforfood]

I am having trouble evaluating the sum $$\sum_{i=1}^{\infty}\frac{i}{4^i}$$ by hand.

My TI-89 is giving me an answer of 4/9 or 0.44 repeating, but I am uncertain how to go about solving this by hand and proving the calculator's result. To my knowledge, no identity or easy quick fix like the Integral test exists. I tried solving for the sum S by messing around with the first 5 or 6 terms, but it was all to no avail. Can anyone here point me in the direction towards evaluating this by hand in terms of a method(not asking anyone to do the problem itself for me)?

 

[Galileo]

Do you know the geometric series?

$$\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$$
if |x|<1.

You can make this look like your sum by differentiation.

 

[amcavoy]

I will provide a general case, you can figure out the rest:

$$\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$$

Now taking the derivative of both sides gives the following:

$$\sum_{n=0}^{\infty}nx^{n-1}=\frac{1}{(1-x)^2}$$

Multiply both sides by x:

$$\sum_{n=0}^{\infty}nx^n=\frac{x}{(1-x)^2}$$

You can apply this to your problem by changing the lower indices and changing x to (1/4).

I hope this helps,

Alex

Edit: It looks like someone got to it just before I submitted this

 

[willworkforfood]

Thanks for all the help :D

 

[bbailey]

I am trying to find the sum of the infinite series i=1 to infinity, of (e^n)/(3^(n-1)). I know it converges, but I am struggling trying to find the limit. can someone help me.