Convergence of a particular infinite sum

[krissycokl]

Homework Statement

Let $b_n$ be a bounded sequence of nonnegative numbers. Let r be a number such that $0 \leq r < 1$.
Define $s_n = b_1*r + b_2*r^2 + ... + b_n*r^n$, for all natural numbers n.
Prove that ${s_n}$ converges.

Homework Equations

Sum of first n terms of geometric series = $sum_n = (a_1)(1-r^{n+1})/(1-r)$

The Attempt at a Solution

Clearly, ${s_n}$ is monotonically increasing.
Since ${b_n}$ is bounded, $|b_n| \leq M$, for all natural numbers n.

I want to use the fact that if ${s_n}$ is both monotonically increasing and is bounded, then it must converge. The part of the problem that has stumped me for the past 45 minutes is how to show that ${s_n}$ is bounded.

The only material we've covered regarding infinite series thus far is for purely geometric series, which doesn't fit this problem precisely--but I included the formula anyway.

Help would be greatly appreciated! I have an exam on Monday and getting so completely stymied by a simple problem is not doing wonders for my confidence.

 

[Dick]

sn=b1*r+b2*r^2+...bn*r^n<=M*r+M*r^2+...+M*r^n, right? That's just a hint. Does it help?

 

[krissycokl]

Aghhhh, so simple! Something about the way I do proofs is just <i>wrong</i>, I always get caught in roadblocks of thinking and miss simple detours like that.

$s_n \leq M*r + M*r^2 + ... + M*r^n$
Then $s_n \leq M(r-r^{n+1})/(1-r)$
Then $s_n \leq Mr/(1-r) = M'$
Then $|s_n| \leq M'$ for all natural numbers n, so $s_n$ is bounded.
Thus, since $s_n$ is bounded and monotonically increasing, we have $s_n$ converges.

Or, I think that's right, anyway.

Thanks a bunch!

 

[Dick]

Aghhhh, so simple! Something about the way I do proofs is just <i>wrong</i>, I always get caught in roadblocks of thinking and miss simple detours like that.

$s_n \leq M*r + M*r^2 + ... + M*r^n$
Then $s_n \leq M(r-r^{n+1})/(1-r)$
Then $s_n \leq Mr/(1-r) = M'$
Then $|s_n| \leq M'$ for all natural numbers n, so $s_n$ is bounded.
Thus, since $s_n$ is bounded and monotonically increasing, we have $s_n$ converges.

Or, I think that's right, anyway.

Thanks a bunch!

It's exactly right. Good take on the hint.