Product of two sequences of functions [uniform convergence]

[timon]

Homework Statement

This is a homework question for a introductory course in analysis. given that
a) the partial sums of $f_n$ are uniformly bounded,

b) $g_1 \geq g_2 \geq ... \geq 0,$

c) $g_n \rightarrow 0$ uniformly,

prove that $\sum_{n=1}^{\infty} f_n g_n$ converges uniformly (the whole adventure takes place on some interval E in R).

Homework Equations

Suppose $x$ and $y$ are two sequences. Then,

$\sum_{j=m+1}^{n} x_jy_j = s_ny_{n+1} - s_my_{m+1} + \sum_{j=m+1}^{n} s_j(y_j - y_{j+1}).$

This is called partial summation, and is given as a hint with the exercise.

The Attempt at a Solution

Inspired by the Cauchy-criterion for uniform convergence of series of functions, I did the following.

$| \sum_{j=m+1}^{n} f_n g_n | = | (\sum_{i=1}^{n}) f_i g_{n+1} - (\sum_{i=1}^{m} f_i) g_{m+1} + \sum_{j=m+1}^{n} (\sum_{i=1}^{j} f_i) (g_j - g_{j_1} ) |$

$\leq |g_{n+1} \sum_{i=1}^{n} f_i| + |g_{m+1} \sum_1^m f_i | + | \sum_{j=m+1}^{n} (\sum_{i=1}^{j} f_i) (g_j - g_{j_1} ) |$
(the last step owing to the subadditivity of the modulus).
The first two terms can be made small since the partial sums of $f$ are bounded and g goes to zero, leaving the third term. I'm having trouble doing anything interesting with that though. Am I on the right track?

 

[micromass]

You can bound $\sum{f_j}$ by L. This leaves you with

$$L\sum{g_j-g_{j+1}}$$

But look at this sum carefully. Isn't that a telescoping sum??

 

[timon]

thanks a lot! I tried to do the same thing but couldn't get $g_i - g_{i+1}$ to converge. I feel somewhat silly now!