Question

Need help with infinite sum of a power series

Answer 1

\[\sum_{n=0}^{\infty}(n+1)(1-p)^{n}\]
where 0<p<1

Answer 2

Hmm.... Let's try a trick here.
\[(n+1)(1-p)^n = - \frac{d}{dp} (1-p)^{n+1}\]
So
\[ \sum_{n=0}^\infty (n+1)(1-p)^n = -\sum_{n=0}^\infty \frac{d}{dp}(1-p)^{n+1}= -\frac{d}{dp}(1-p)\sum_{n=0}^\infty (1-p)^n\]
since 0 < p < 1, 1-p < 1, so
\[ \sum_{n=0}^\infty (1-p)^n = \frac{1}{1-(1-p)} = \frac{1}{p}\]
so finally, our sum becomes
\[-\frac{d}{dp}(1-p)\frac{1}{p} = \frac{d}{dp}1-\frac{1}{p} = \frac{1}{p^2}\]

Answer 3

Is there a name for that trick? it's a new one on me.

Answer 4

I'm not sure, actually. Maybe. I suppose it would be a cousin of the "differentiation under the integral sign" technique, or "integration by parametric differentiation" if you want big words. Are you familiar with that?

Answer 5

I've heard of differentiation under the integral sign but never learned it :/

Answer 6

I think I can extrapolate the idea though.

Answer 7

The integral form is
\[ \frac{d\ }{dt} \int_a^b f(x,t) \ dx = \int_a^b \frac{\partial f}{\partial t}(x,t) \ dx \]

Both with the series form and the integral form, you need to be careful changing the order the differentiation and summation; there are things that can go spectacularly wrong if the function in the integration or the series in the summation don't meet some standards of 'well-behaved'.

Answer 8

thanks everyone...I think i get it now