# Limits of infinite series

How do I go about finding the answer to this? I'm not really sure on how to find the general sum formula. • Use the squeeze theorem with the bounds$$\frac1{n^2+n\pi}\le\frac1{n^2+k\pi}\le\frac1{n^2}$$ – Peter Foreman Sep 21 '19 at 16:20
• please tell me how you got those bounds :( – Tiffany Sep 21 '19 at 16:27
• The LHS is just the lowest term in the summation and the RHS is greater than the greatest term in the summation. – Peter Foreman Sep 21 '19 at 16:28

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 $$\mathop {\lim }\limits_{n \to \infty } n\left( {\frac{1} {{n^2 + \pi }} + \cdots \frac{1} {{n^2 + n\pi }}} \right) = 1$$