0
$\begingroup$

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 ...

$\endgroup$
2
$\begingroup$

HINT:

$$\dfrac{\sum_{r=1}^n(r^2+n^2)^2}{n^5}=\dfrac1n\sum_{r=1}^n\left(1+\dfrac{r^2}{n^2}\right)^2$$

For the rest see The limit of a sum $\sum_{k=1}^n \frac{n}{n^2+k^2}$

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Future Math person Nov 25 '16 at 8:52
  • $\begingroup$ @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? $\endgroup$ – lab bhattacharjee Nov 25 '16 at 9:02
  • $\begingroup$ 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? $\endgroup$ – Future Math person Nov 25 '16 at 9:09
  • $\begingroup$ $\int_{0}^{1} (1+x^2)^2 dx$ sorry $\endgroup$ – Future Math person Nov 25 '16 at 9:15
  • $\begingroup$ @SubhashisChakraborty, Right. What's your query? $\endgroup$ – lab bhattacharjee Nov 25 '16 at 10:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.