Limit of series with infinite product

In our Calc 1 class this is one of the proposed series: $\displaystyle\sum_{n\geqslant 1}\frac1{n}\prod_{k=1}^n \left(1-\frac{\pi}{k}\right)$. We have only been taught basic criteria, so no integration, just comparison test, quotient test, etc... Any way to tell this series diverges? (Because that is my intuition but I am unable to prove it)

Solution 1:

I think it's convergent: For $n \ge 4$ consider $$ \prod_{k=4}^n\left(1-\frac{\pi}{k} \right) = \prod_{k=4}^n\frac{k-\pi}{k} \le \prod_{k=4}^n\frac{k-3}{k} = \frac{(n-3)!}{4\cdot 5 \cdot \dots \cdot n} $$ $$ =6\frac{(n-3)!}{n!}=\frac{6}{(n-2)(n-1)n}. $$ Thus, for $n \ge 4$ $$ \frac{1}{n}\prod_{k=1}^n\left|1-\frac{\pi}{k} \right| \le \left(\prod_{k=1}^3\left|1-\frac{\pi}{k} \right|\right)\frac{6}{(n-2)(n-1)n^2} $$ and the right hand side in this inequality gives a convergent majorant.

Solution 2:

We know that $ \left(\forall x\in\mathbb{R}\right), 1+x\leq\operatorname{e}^{x} $, thus, $ \forall n\geq 3 $ : $$ \left\vert\prod_{k=2}^{n}{\left(1-\frac{\pi}{k}\right)}\right\vert =\prod_{k=2}^{n}{\left(1-\frac{\pi}{k}\right)}\leq\prod_{k=2}^{n}{\operatorname{e}^{-\frac{\pi}{k}}}=\operatorname{e}^{-\pi\left( H_{n}-1\right)}\leq\operatorname{e}^{-\pi\ln{n}+\pi}=\frac{\operatorname{e}^{\pi}}{n^{\pi}} $$

Hence : $ \left(\forall n\geq 3\right),\ \left\vert\prod\limits_{k=1}^{n}{\left(1-\frac{\pi}{k}\right)}\right\vert\leq\frac{\operatorname{e}^{\pi}\left(\pi - 1\right)}{n^{\pi}} $.

Since $\sum\limits_{n\geq 3}{\frac{\operatorname{e}^{\pi}\left(\pi -1\right)}{n^{\pi +1}}} $ converges, then $ \sum\limits_{n\geq 3}{\frac{1}{n}\left\vert\prod\limits_{k=1}^{n}{\left(1-\frac{\pi}{k}\right)}\right\vert} $ converges, which means our series converges.