I have a series from n=1 to infinity of

n/(4^n).

Can this convert into a geometric series? and so how?

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- Sep 20th 2010, 02:27 PMguyonfire89sum 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 PMPlato
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 PMguyonfire89
Can someone explain how x=1/4?

- Oct 4th 2010, 03:00 PMTheCoffeeMachine
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.