# Thread: Convergence of a Series (proof)

1. ## Convergence of a Series (proof)

Hello. I am having some difficulty proving the following:

If $\displaystyle \sum_{n=1}^\infty a_n$ converges and $\displaystyle \{b_n\}$ is monotonic and bounded, then $\displaystyle \sum_{n=1}^\infty a_nb_n$ converges.

I tried using the Cauchy Criterion and partial summation to handle this, but I haven't had any luck. Your insight would be appreciated.

2. Originally Posted by roninpro
Hello. I am having some difficulty proving the following:

If $\displaystyle \sum_{n=1}^\infty a_n$ converges and $\displaystyle \{b_n\}$ is monotonic and bounded, then $\displaystyle \sum_{n=1}^\infty a_nb_n$ converges.

I tried using the Cauchy Criterion and partial summation to handle this, but I haven't had any luck. Your insight would be appreciated.

It's a result of Dirichlet's test: if $\displaystyle \sum\limits_{n=1}^\infty a_n$ is a bounded series (i.e., its partials sums sequence is bounded) and if $\displaystyle \{b_n\}_{n=1}^\infty$ is a descending monotone sequence that converges to zero, then $\displaystyle \sum\limits_{n=1}^\infty a_nb_n$ converges:

Since $\displaystyle \{b_n\}$ is monotone and bounded it converges to a finite limit, say L. Assume it is monotone ascending (if it is descending it is very simmlilar) , so we get $\displaystyle (L-b_n) \xrightarrow [n\to \infty] {} 0$ monotonically descending and $\displaystyle \sum\limits_{n=1}^\infty a_n$ is bounded because it is convergent, and thus by Dirichlet's test the series $\displaystyle \sum\limits_{n=1}^\infty a_n(L-b_n)$ converges, but then:

$\displaystyle \sum\limits_{n=1}^\infty a_nL-\sum\limits_{n=1}^\infty a_n(L-b_n)$ is the difference of two convergent series and thus it converges, and $\displaystyle \sum\limits_{n=1}^\infty a_nL-\sum\limits_{n=1}^\infty a_n(L-b_n)=\sum\limits_{n=1}^\infty a_nb_n$ by arithmetic of limts, and we're done.

Tonio