Help with evaluating this Riemann-Sum.

Question

How to find :$\displaystyle{\lim_{n \to \infty}}$${\dfrac{1}{n^3}}$$\sum_{r=1}^n {\dfrac{r^4}{(2r-1)(2r+1)}}$

My approach: I tried to solve this by trying to convert it into a definite integral.

I rewrote the expression as :${\dfrac{1}{4}}$$\displaystyle{\lim_{n \to \infty}}\sum_{r=1}^n $${\dfrac{r^3}{n^3}}$ $\Big($${\dfrac{1}{(2r+1)}}$$+$${\dfrac{1}{(2r-1)}}$$\Big)$

Now what to do after this to convert it in to a definite integral?

Answer 1

$\dfrac{r^4}{4r^2-1}= \dfrac{1}{16}\cdot \dfrac{(16r^4-1)+1}{4r^2-1}= \dfrac{1}{16}\left(4r^2+1+\dfrac{1}{(2r-1)(2r+1)}\right)$

$\sum\limits_{r=1}^n \frac1{16} = \frac{n(n+1)}{32}$ $\sum\limits_{r=1}^n \frac{r^2}4 = \frac{n(n+1)(2n+1)}{24}$ $\sum\limits_{r=1}^n \frac1{16\cdot(4r^2-1)} = \frac1{16}-\frac{n+1}{32n+16}$

From here, we simply sum these three results, and take their sum to infinity.

E.g.

$$\sum\limits_{n=1}^\infty \frac1{n^3}\cdot\bigg(\frac1{16}-\frac{n+1}{32n+16}+\frac{n(n+1)(2n+1)}{24}+\frac{n(n+1)}{32}\bigg)$$

Answer 2

hint: $\dfrac{r^4}{4r^2-1}= \dfrac{1}{16}\cdot \dfrac{(16r^4-1)+1}{4r^2-1}= \dfrac{1}{16}\left(4r^2+1+\dfrac{1}{(2r-1)(2r+1)}\right)$

Answer 3

Your approach can also work, but I think you forgot about a great opportunity to simplify the problem.

Instead of $\sum\limits_{r=1}^{n} \frac{r^4}{4r^2-1}$, why don't you compute $\sum\limits_{r=1}^{n} \frac{r^4}{4r^2}$?

If you can show that the difference is $o(n^3)$, then you win.