Given that $\sum b(n)$ is convergent, is $\sum a(n)$ convergent?

Question

Consider

• $\sum_\limits{n=1}^{\infty} b(n)$, such that for all $n$, $0<b(n) ≤1$
• $\sum_\limits{n=1}^{\infty} a(n)$ such that for all $n$, $-b(n)≤a(n)≤b(n)$.

Given that $\sum b(n)$ is convergent, is $\sum a(n)$ convergent? Prove it in detail.

Answer 1

Yes.

$$ \sum_{n=1}^\infty a(n) \leq \bigg|\sum_{n=1}^\infty a(n)\bigg|\leq\sum_{n=1}^\infty |a(n)|\leq\sum_{n=1}^\infty b(n) < \infty $$

The third inequality follows from $-b\leq a \leq b \iff |a|\leq b$.

Being a bit more pedantic - we use the fact that absolute convergence implies convergence. In other words, if $\sum_{n=1}^\infty |a(n)|$ converges, then $\sum_{n=1}^\infty a(n)$ converges. The former converges because every element of the series is bounded by the corresponding $b(n)$. (Being ever more pedantic, $\sum_{n=1}^\infty |a(n)|$ converges because the truncated series is a bounded and increasing monotone sequence, thus it converges.)

Answer 2

If $\sum a_n$ be diverge, then $\lim a_n=\ell\neq0$ then $$0<\ell=\lim |a_n|≤\lim b_n$$

Answer 3

The second condition implies $|a(n)|\le b(n)$ for every $n.$ Since $\sum b(n) < \infty,$ $\sum |a(n)|<\infty$ by the comparison test. Thus $\sum a(n)$ converges absolutely, hence converges.