Limit of a convergent series and a divergent sequence

[terhorst]

Homework Statement

Show that if:

$$lim_{k\to\infty}b_k\to+\infty$$,

$$\sum_{k=1}^\infty a_k$$ converges and,

$$\sum_{k=1}^\infty a_k b_k$$ converges, then

$$lim_{m\to\infty} b_m \sum_{k=m}^\infty a_k = 0$$

Homework Equations

The Attempt at a Solution

I only have an idea why this is true--$$\sum a_k$$ converges, so the tails become small, and $$\sum a_k b_k$$ converges, so $$a_k$$ shrinking dominates $$b_k$$ blowing up. I know that $$|b_k| < \epsilon / |a_k|$$ for any epsilon and large k, and I know that $$\left| \sum_{m=k}^\infty a_m \right| < \epsilon$$ for large k, but I can't find the way to combine these two statements.

 

[mighty2000]

Your idea is on the right track. Here is a more formal proof:

Since \sum a_k converges, we know that for any \epsilon > 0, there exists a positive integer N such that for all n > m > N, we have \left| \sum_{k=m}^n a_k \right| < \epsilon.

Now, since \lim_{k \to \infty} b_k = +\infty, we can choose a positive integer M such that for all k > M, we have b_k > \frac{1}{\epsilon}.

Combining these two statements, we have for all n > m > \max\{N, M\}, we have
\left| b_m \sum_{k=m}^n a_k \right| = |b_m| \left| \sum_{k=m}^n a_k \right| < |b_m| \epsilon < \frac{\epsilon}{\epsilon} = 1.

Since this holds for all n > m > \max\{N, M\}, we can take the limit as m \to \infty to get
\lim_{m \to \infty} b_m \sum_{k=m}^\infty a_k = 0.