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

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.

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

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.
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 $$\sum\limits_{n=1}^{N} a_n\cdot b_n=\sum\limits_{n=1}^{N}\frac{1}{(n+1)\ln(n+1)}\sim \ln\ln{N},\quad N\to\infty.$$
• I need to do this at the initial point $n=0$ – Diane Vanderwaif Jan 22 '14 at 20:49
• 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. – mkl314 Jan 22 '14 at 21:29