3
$\begingroup$

I was wondering if my proof for the following problem is correct:

Problem:

If $\displaystyle \sum_{n} a_n$ converges, and if $\{b_n\}$ is monotonic and bounded, prove that $\displaystyle \sum_{n} a_n b_n$ converges.

Proof:

Assume $\displaystyle \sum_{n} a_n$ converges and $\{b_n\}$ is monotonic and bounded. Then the partial sums of $a_n$ form a bounded sequence; that is, $\displaystyle A_k = \sum_{n=1}^{k} a_n$ is a bounded sequence.

Theorem 3.14 states that:

Suppose $\{s_n\}$ is monotonic. Then $\{s_n\}$ converges if and only if it is bounded.

Thus, by Theorem 3.14, $\{b_n\}$ converges. Let $\displaystyle L = \lim_{n \rightarrow \infty} b_n$. Consider the sequence $\{b_n - L\}$. Then $\{b_n - L\}$ is also a monotonic and bounded sequence, so by Theorem 3.14, $\displaystyle \lim_{n \rightarrow \infty} b_n - L$ converges and $\displaystyle \lim_{n \rightarrow \infty} b_n - L = 0$.

Theorem 3.42 states that:

Suppose: (a) the partial sums $A_n$ of $\displaystyle \sum_{n} a_n$ form a bounded sequence; (b) $b_0 \geq b_1 \geq b_2 \geq \ldots$ (c) $\displaystyle \lim_{n \rightarrow \infty} b_n = 0$ Then $\displaystyle \sum_{n} a_n b_n$ converges.

Thus, by Theorem 3.42, $\displaystyle \sum_{n} a_n (b_n - L)$ converges, so $\displaystyle \sum_{n} a_n b_n - \sum_{n} a_n L$ converges. In particular, $\displaystyle \sum_{n} a_n b_n$ converges. $\blacksquare$

Thank you.

$\endgroup$
2
$\begingroup$

Just one comment: In your (almost) last line, you split up the infinite sum without knowing that each of the pieces exist. You should conclude that $\sum a_{n}b_{n}$ converges as follows:

$$ \begin{aligned} \sum a_{n}b_{n}&=\sum [a_{n}(b_{n}-L)+La_{n}]=\sum a_{n}(b_{n}-L)+L\sum a_{n}<\infty \end{aligned} $$

since $\sum a_{n}<\infty$.

$\endgroup$
  • $\begingroup$ Thank you, I didn't know that when you split up infinite sums, you need to justify why each sum exists. $\endgroup$ – Frederic Chopin Nov 12 '17 at 22:15
  • $\begingroup$ Think $\sum(1/n-1/n)=0\not=\sum (1/n)-\sum (1/n)$. $\endgroup$ – ervx Nov 12 '17 at 22:16
  • 1
    $\begingroup$ Also, I just realized something. I didn't prove that $\{b_n - L\}$ is a monotonically decreasing sequence, as it should be in Theorem 3.42. All is good if $\{b_n\}$ is monotonically decreasing, but what happens if $\{b_n\}$ is monotonically increasing? Then it is true that $\{b_n - L\}$ is a monotonically decreasing sequence. $\endgroup$ – Frederic Chopin Nov 12 '17 at 22:17
  • 1
    $\begingroup$ You can consider $L-b_{n}$ instead, in this case. $\endgroup$ – ervx Nov 12 '17 at 22:18
  • $\begingroup$ Oh yes! Good point for both $\endgroup$ – Frederic Chopin Nov 12 '17 at 22:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.