# Problem involving series

• May 9th 2012, 02:15 PM
spoon737
Problem involving series
The following problem is in a section on convergence tests for series (comparison, ratio, root, integral, p-series, etc.) I've been stuck on it for about a day, so any help would be much appreciated.

Suppose that $\Sigma a_{n}$ is a convergent series with positive terms. Let $\{S_{n}\}$ be the sequence of partial sums for $\Sigma a_{n}$. Let $S = lim \ S_{n}$ and let $\rho_{k} = \frac{a_{k+1}}{a_{k}}$. Suppose there is a number $N$ for which $\rho_{k} \ge \rho_{k+1}$ if $k \ge N$ and $\rho_{N} < 1$. Show that $S - S_{N} \le \frac{a_{N+1}}{1 - \rho_{N}}$.

So far all I've been able to do is show that $S - S_{N} = a_{N+1}(1 + \rho_{N+1} + \rho_{N+1}\rho_{N+2} + \rho_{N+1}\rho_{N+2}\rho_{N+3} + ...)$ which means I need to show that $1 + \rho_{N+1} + \rho_{N+1}\rho_{N+2} + \rho_{N+1}\rho_{N+2}\rho_{N+3} + ... \le \frac{1}{1 - \rho_{N}}$, but I'm not sure where to take it from here, or if this is even a good approach.
• May 9th 2012, 06:44 PM
wsldam
Re: Problem involving series
So you know $0 < \rho_N < 1$. What can you say about $\Sigma (\rho_N)^n$? Does it converge? Where to? You also know that $\rho_N \geq \rho_{N+1}$. So then what can you say about the sum that you found?
• May 9th 2012, 07:24 PM
spoon737
Re: Problem involving series
Ah, I see. Term for term, my series is less than or equal to a geometric series converging to $\frac{1}{1 - \rho_{N}}$. Thanks a billion!