1
$\begingroup$

The limit is a Riemann sum

$$\lim_{n\rightarrow\infty}\sum_{i=1}^n\left(\frac{i^4}{n^5}+\frac{i}{n^2} \right)$$ $\delta x=\frac{1}{n}$, so I distribute it to the terms to get

$$\lim_{n\rightarrow\infty}\sum_{i=1}^n\left(\frac{i^4}{n^6}+\frac{i}{n^3} \right)$$ Now that they have similar denominators I multiply $\frac{i}{n^3}\cdot(n^3)$ to get $$\lim_{n\rightarrow\infty}\sum_{i=1}^n\left(\frac{i^4}{n^6}+\frac{2i}{n^6} \right)$$

Combining the terms I get $$\lim_{n\rightarrow\infty}\sum_{i=1}^n\left(\frac{2i^4}{n^6}\right)$$ Knowing that $i^2=\frac{n(n+1)(2n+1)}{6}$ I split up the $i^4$ into$$\lim_{n\rightarrow\infty}\frac{2}{n^6}\sum_{i=1}^n\left(\frac{n(n+1)(2n+1)}{6}\right)+ \left(\frac{n(n+1)(2n+1)}{6}\right)$$ Am I on the right track?

$\endgroup$
1
$\begingroup$

We have that by Riemann sum

$$\lim_{n\rightarrow\infty}\sum_{i=1}^n\left(\frac{i^4}{n^5}+\frac{i}{n^2} \right)=\lim_{n\rightarrow\infty}\frac1n\sum_{i=1}^n\left(\frac{i^4}{n^4}+\frac{i}{n} \right)=\int_0^1(x^4+x) dx$$

As an alternative refer to Faulhaber's formula.

$\endgroup$

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.