Our professor gives us the following ungraded exercise for our analytic number theory class:
Let $ E $ be a set with one element. Suppose $ (b_n) $ is a sequence with $ |b_n| \leq \lambda < 1 $, and let $ a_n = 1 + b_n $.
1) Find $ (b_n) $ so that $ \sum b_n $ converges, but $ \prod a_n $ diverges.
2) Find $ (b_n) $ so that $ \prod a_n $ converges, but $ \sum b_n $ diverges.
I am not sure how to do this problem. Any help is appreciated.
Best Answer
Suppose $$ b_{2n-1}=\frac1{\sqrt{n+1}}\quad\text{and}\quad b_{2n}=-\frac1{\sqrt{n+1}} $$ Then, Dirichlet's Convergence Test guarantees convergence of the series. In fact, $$ \sum_{k=1}^\infty b_k=0 $$ However, $$ \left(1+b_{2n-1}\right)\left(1+b_{2n}\right)=1-\frac1{n+1} $$ Therefore, $$ \prod_{k=1}^{2n}\left(1+b_k\right)=\frac1{n+1} $$
$$ a_{2n-1}=\frac{\sqrt{n}+1}{\sqrt{n}}\quad\text{and}\quad a_{2n}=\frac{\sqrt{n}}{\sqrt{n}+1} $$ Then, Dirichlet's Convergence Test guarantees convergence of the log of the series. In fact, $$ \prod_{k=1}^\infty a_k=1 $$ However, $$ \begin{align} \left(a_{2k-1}-1\right)+\left(a_{2k}-1\right) &=\frac1{\sqrt{n}}-\frac1{\sqrt{n}+1}\\ &=\frac1{n+\sqrt{n}} \end{align} $$ Therefore, $$ \begin{align} \sum_{k=1}^{2n}\left(a_k-1\right) &=\sum_{k=1}^n\frac1{k+\sqrt{k}}\\ &\ge\int_1^{n+1}\frac{\mathrm{d}x}{x+\sqrt{x}}\\ &=2\log\left(\frac{\sqrt{n+1}+1}2\right) \end{align} $$