# Expressing a sum as a definite integral

So i'm asked to find the limit by expressing this summation below it as a definite integral:

$\lim_{n\to\infty} n^{-5}[(1^{2}+n^{2})^{2}+(2^{2}+n^{2})^{2}+(3^{2}+n^{2})^{2}+...+((n-1)^{2}+n^{2})^{2}+(n^{2}+n^{2})^{2}]$

I'm not sure how I'm supposed to express this as a definite integral though. I don't see any way for me to convert it to the form of a reimann sum so I can apply the definition ...

$$\dfrac{\sum_{r=1}^n(r^2+n^2)^2}{n^5}=\dfrac1n\sum_{r=1}^n\left(1+\dfrac{r^2}{n^2}\right)^2$$
For the rest see The limit of a sum $\sum_{k=1}^n \frac{n}{n^2+k^2}$
• That's true but we have $...+((n-1)^{2}+n^{2})^{2}+(n^{2}+n^{2})^{2}$. Does the previous term have any role in this? – Future Math person Nov 25 '16 at 8:52
• @SubhashisChakraborty, $$(1^{2}+n^{2})^{2}+(2^{2}+n^{2})^{2}+(3^{2}+n^{2})^{2}+...+((n-1)^{2}+n^{2})^{2}+(n^{2}+n^{2})^{2}$$ $$=\sum_{r=1}^n(r^2+n^2)$$ right? – lab bhattacharjee Nov 25 '16 at 9:02
• Yes. That's definitely true. Then you just rewrote it like so in the above statement. That means integral becomes $\int_{0}^{1} x^2 dx$ i think since he bounds go from 0 to 1 and the terms inside the summation are effectively an x term... Is that correct? – Future Math person Nov 25 '16 at 9:09
• $\int_{0}^{1} (1+x^2)^2 dx$ sorry – Future Math person Nov 25 '16 at 9:15