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

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    $\begingroup$ Use the squeeze theorem with the bounds$$\frac1{n^2+n\pi}\le\frac1{n^2+k\pi}\le\frac1{n^2}$$ $\endgroup$ – Peter Foreman Sep 21 '19 at 16:20
  • $\begingroup$ please tell me how you got those bounds :( $\endgroup$ – Tiffany Sep 21 '19 at 16:27
  • $\begingroup$ The LHS is just the lowest term in the summation and the RHS is greater than the greatest term in the summation. $\endgroup$ – Peter Foreman Sep 21 '19 at 16:28
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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 $$ \mathop {\lim }\limits_{n \to \infty } n\left( {\frac{1} {{n^2 + \pi }} + \cdots \frac{1} {{n^2 + n\pi }}} \right) = 1 $$

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