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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?

Eric Brown
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1 Answers1

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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.

user
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