1
$\begingroup$

Prove or disprove: if $\sum\limits_{n=0}^\infty a_n$ converges and $\lim\limits_{n→\infty}⁡b_n =0$, then $\sum\limits_{n=0}^\infty a_n b_n$ converges.

I tried this example $$a_n= b_n = \frac {(-1)^n} {\sqrt {n+1}},$$ this will show that $$\sum_{n=0}^\infty a_n b_n = \sum_{n=0}^\infty \frac {1} { {n+1}},$$ which is the Harmonic series, diverges.

However, I tried every convergent test but I couldn't show that $$\sum_{n=0}^\infty \frac {(-1)^n} {\sqrt {n+1}}$$ converges.

$\endgroup$
  • $\begingroup$ Hint: for the last question, use the alternating series convergence test. $\endgroup$ – Clement C. Jan 22 '14 at 20:03
  • $\begingroup$ It converges by Leibniz' test. $\endgroup$ – J.R. Jan 22 '14 at 20:03
  • $\begingroup$ I haven't learn any of these test yet $\endgroup$ – Diane Vanderwaif Jan 22 '14 at 20:09
  • $\begingroup$ But you still can read the Wikipedia article, though? (or do you mean you are not allowed to use anything not seen in class?) $\endgroup$ – Clement C. Jan 22 '14 at 20:11
  • $\begingroup$ I'm not allowed to do so :D $\endgroup$ – Diane Vanderwaif Jan 22 '14 at 20:11
1
$\begingroup$

Convergence of $$ \sum_{n=0}^\infty\frac{(-1)^n}{\sqrt{n+1}}. $$ Let $s_n=\sum_{k=0}^n\frac{(-1)^k}{\sqrt{k+1}}$. Then $s_{2n-1}$, $n\in\mathbb N$, is increasing, while $s_{2n}$, $n\in\mathbb N$, is decreasing and $$ s_1\le s_3\le\cdots\le s_{2n-1}\le s_{2n+1}\le s_{2n}\le s_{2n-2}\le\cdots s_2\le s_0. $$ Hence both these subsequence are convergent, as monotonic and bounded, and their limits coincide, as $$ s_{2n}-s_{2n-1}=\frac{1}{\sqrt{2n+1}}\longrightarrow 0. $$ Thus $\{s_n\}$ converges.

This is the alternating series test: If $a_n>0$, $a_n$ decreasing and $a_n\to 0$, then $\sum_{n=1}^\infty (-1)^n a_n$ converges.

$\endgroup$
0
$\begingroup$

Take $a_n=\frac{(-1)^n}{n+1}$ for $n\geqslant 0$, and let $b_n=\frac{(-1)^n}{\ln(n+1)}$ for $n\geqslant 1$. The series $\sum_{n=0}^{\infty} a_n$ converges to $\ln{2}$ while the series $\sum_{n=1}^{\infty} a_n\cdot b_n$ diverges since $$\sum\limits_{n=1}^{N} a_n\cdot b_n=\sum\limits_{n=1}^{N}\frac{1}{(n+1)\ln(n+1)}\sim \ln\ln{N},\quad N\to\infty.$$

$\endgroup$
  • $\begingroup$ I need to do this at the initial point $n=0$ $\endgroup$ – Diane Vanderwaif Jan 22 '14 at 20:49
  • $\begingroup$ Choosing $b_0=1$, instead of $b_0=0$, could change the sum of a convergent series but has no effect on a divergent series. $\endgroup$ – mkl314 Jan 22 '14 at 21:29

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.