Limits of infinite series

Question

How do I go about finding the answer to this? I'm not really sure on how to find the general sum formula.

Answer 1

You have that $$ n\left( {\frac{n} {{n^2 + n\pi }} + \cdots \frac{1} {{n^2 + n\pi }}} \right) < n\left( {\frac{1} {{n^2 + \pi }} + \cdots \frac{1} {{n^2 + n\pi }}} \right) < n\left( {\frac{1} {{n^2 + \pi }} + \cdots \frac{1} {{n^2 + \pi }}} \right) $$ Therefore $$ \frac{{n^2 }} {{n^2 + n\pi }} < n\left( {\frac{1} {{n^2 + \pi }} + \cdots \frac{1} {{n^2 + n\pi }}} \right) < \frac{{n^2 }} {{n^2 + \pi }} $$ Since $$ \mathop {\lim }\limits_{n \to \infty } \frac{{n^2 }} {{n^2 + n\pi }} = \mathop {\lim }\limits_{n \to \infty } \frac{{n^2 }} {{n^2 + \pi }} = 1 $$ by squeeze theorem you have that $$ \mathop {\lim }\limits_{n \to \infty } n\left( {\frac{1} {{n^2 + \pi }} + \cdots \frac{1} {{n^2 + n\pi }}} \right) = 1 $$