# sum of infinite series (clarification)

• Sep 20th 2010, 02:27 PM
guyonfire89
sum of infinite series (clarification)
I have a series from n=1 to infinity of

n/(4^n).

Can this convert into a geometric series? and so how?
• Sep 20th 2010, 02:43 PM
Plato
Quote:

Originally Posted by guyonfire89
I have a series from n=1 to infinity of
n/(4^n).
Can this convert into a geometric series? and so how?

Yes it can by differentiation.
Start with $\displaystyle \displaystyle f(x) = \sum\limits_{n = 0}^\infty {x^n } = \frac{1}{{1 - x}},~|x|<1$.

Differentiate to show $\displaystyle \displaystyle f'(x) = \sum\limits_{n = 1}^\infty {nx^n } = \frac{x}{{(1 - x)^2}}$

Then let $\displaystyle x=\frac{1}{4}$
• Oct 3rd 2010, 08:51 PM
guyonfire89
Can someone explain how x=1/4?
• Oct 4th 2010, 03:00 PM
TheCoffeeMachine
Quote:

Originally Posted by guyonfire89
Can someone explain how x=1/4?

You let $\displaystyle x = \frac{1}{4}$ and the LHS, that's the sum $\displaystyle \sum\limits_{n = 1}^\infty {nx^n }$ becomes $\displaystyle \sum\limits_{n = 1}^\infty {\frac{n}{4^n} }$ which is your original sum.
Do the same, i.e. let $\displaystyle x = \frac{1}{4}$ for the RHS, that's $\displaystyle \dfrac{x}{{(1 - x)^2}}$ and you will find what it evaluates to.