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.

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 $$\sum\limits_{n=1}^{N} a_n\cdot b_n=\sum\limits_{n=1}^{N}\frac{1}{(n+1)\ln(n+1)}\sim \ln\ln{N},\quad N\to\infty.$$