4
$\begingroup$

Here's Prob. 8, Chap. 3 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

If $\sum a_n$ converges, and if $\left\{ b_n \right\}$ is monotonic and bounded, prove that $\sum a_n b_n$ converges.

My effort:

Let $b = \lim_{n \to \infty} b_n$.

Let's first suppose that $\left\{ b_n \right\}$ is monotonically decreasing. Then $b_n \geq b$ for all $n$. So we have $$ b_0 - b \geq b_1 - b \geq \cdots \geq 0.$$ Moreover the sequence of the partial sums of $\sum a_n$ is bounded and $\lim_{n \to \infty} \left( b_n - b \right) = 0$.

So by Theorem 3.42 in Rudin, we can conclude that $\sum a_n (b_n - b)$ converges. And since $a_n b_n = a_n (b_n - b) + a_n b$ and since $\sum a_n$ converges, therefore $\sum a_n b_n$ converges by Theorems 3.47 in Rudin.

Now let's suppose that $\left\{ b_n \right\}$ is monotonically increasing. Then $\left\{ -b_n \right\}$ is monotonically decreasing and so $\sum a_n \left(-b_n\right)$ converges. Therefore by Theorem 3.47 in Rudin $\sum a_n b_n$ converges as well.

Is my proof correct? If not, then where is it wanting?

$\endgroup$
3
  • $\begingroup$ I'm not sure what this theorem 3.42 is, but assuming you are using it correctly there's no mistake in your argument; also for the last argument I suppose convergence of $\sum a_n(-b_n)$ enough to say that $\sum a_nb_n$ converges as $\sum a_nb_n=-\sum a_n(-b_n)$ $\endgroup$
    – user160738
    Commented Dec 25, 2016 at 15:18
  • $\begingroup$ @user160738 Theorem 3.42 in Rudin reads as follows: Suppose the partial sums $A_n$ of $\sum a_n$ form a bounded sequence; $b_0 \geq b_1 \geq b_2 \geq \cdots$; and $\lim_{n \to \infty} b_n = 0$. Then $\sum a_n b_n$ converges. $\endgroup$ Commented Dec 25, 2016 at 17:08
  • $\begingroup$ @user160738 can you please have a look at my post again and answer the question(s)? $\endgroup$ Commented Dec 30, 2016 at 17:46

1 Answer 1

5
$\begingroup$

Set $A_n=a_1+\cdots+a_n$, then $$ \sum_{k=1}^na_kb_k=\sum_{k=1}^n (A_k-A_{k-1})b_k=\sum_{k=1}^nA_kb_k -\sum_{k=2}^nA_{k-1}b_k=\sum_{k=1}^nA_kb_k -\sum_{k=1}^{n-1}A_{k}b_{k+1}=\sum_{k=1}^{n-1}A_k(b_k-b_{k+1})+A_nb_n $$ Clearly, $A_nb_n$ converges as a product of two converging sequences, while $s_n=\sum_{k=1}^{n-1}A_k(b_k-b_{k+1})$, also converges, because it converges absolutely: $$ \sum_{k=1}^{n-1}|A_k| |b_k-b_{k+1}|\le \big(\sup A_n\big)\cdot |b_1-b|, $$ where $\,b=\lim b_n$.

$\endgroup$
3
  • $\begingroup$ (it's called the summation by parts) $\endgroup$
    – reuns
    Commented Dec 26, 2016 at 1:50
  • $\begingroup$ @YiorgosS.Smyrlis can you please have a look at my effort and comment on how good it is? That would be so kind of you! $\endgroup$ Commented Dec 30, 2016 at 17:45
  • $\begingroup$ @user1952009 can you please comment on my effort? $\endgroup$ Commented Dec 30, 2016 at 17:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .