Product of two sequences of functions [uniform convergence]

timon
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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?
 
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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??
 
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!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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