Evaluate $\lim_{n\to \infty} \sum_{r=1}^n {\frac{r^4}{4r^2-1}}$ $$\lim_{n\to \infty} \sum_{r=1}^n {\frac{r^4}{4r^2-1}}$$
I am a class 12th student and this question is part of a previously given assignment.My solution is as follows...
$$\lim_{n\to \infty} \sum_{r=1}^n {\frac{r^4}{4r^2-1}}$$
$$=\frac{1}{16}\lim_{n\to \infty}
 \sum_{r=1}^n\left({4r^2+1}+ \frac{1}{4r^2-1}\right)$$
$$=\lim_{n\to \infty}\frac{n(n+1)(2n+1)}{24} \;+\;\lim_{n\to \infty}{(n/16)}+\;\;\lim_{n\to \infty}\sum_{r=1}^n\frac{1}{16(4r^2-1)}$$
what should i do after that? , 
$\mathbf {i\,think\,answer\,is \, \infty} $ 
 A: You could write
$$\frac{r^4}{4 r^2-1}=\frac{1}{16}+\frac{r^2}{4}+\frac{1}{32}\left(\frac{1}{2 r-1}-\frac{1}{2 r+1}\right)$$ making, as you almost did,
$$\sum_{r=1}^n {\frac{r^4}{4r^2-1}}=\frac{n}{16}+\frac{ n (n+1) (2 n+1)}{24} +\frac{1}{32}\sum_{r=1}^n\left(\frac{1}{2 r-1}-\frac{1}{2 r+1}\right)$$ and notice a telescoping sum
$$\sum_{r=1}^n\left(\frac{1}{2 r-1}-\frac{1}{2 r+1}\right)=1-\frac{1}{2 n+1}$$
Adding all terms and simplifying, this would give
$$S_n=\sum_{r=1}^n {\frac{r^4}{4r^2-1}}=\frac{n^4+2 n^3+2 n^2+n}{6 (2 n+1)}$$
A: Just notice that $$\frac{n(n+1)(2n+1)}{24}+\frac{n}{16}+\sum_{r=1}^n\frac{1}{16(4r^2-1)}\ge\frac{n}{16}\ \ \ \forall n\in\mathbb{N},$$ so $$\lim_{n\to\infty}\left[\frac{n(n+1)(2n+1)}{24}+\frac{n}{16}+\sum_{r=1}^n\frac{1}{16(4r^2-1)}\right]\ge\lim_{n\to +\infty}\frac{n}{16}=+\infty$$
A: You can immediately tell the sum diverges, because as n goes to infinity, so does the terms of the sum. Think about it, what should happen as n goes to infinity ? If you are adding an infinite amount of terms, an infinite amount of them must go to cero, as n goes to infinity. Otherwise, your sum blows up!
