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In the interval [0,1] I have to find the limit of a Riemann sum

$$\lim _{n\to \infty }\sum _{i=1}^n\left(\frac{i^4}{n^5}+\frac{i}{n^2}\right)$$

so far I have this $$\lim _{n\to \infty }\sum _{i=1}^n\:\frac{i}{n}\left(\left(\frac{i}{n}\right)^3+1\right)$$ and tried to make it look like (a+ delta(X)i) but since a is 0 I feel kind of lost.

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Take $1/n $ common outside. So your sum changes to $\lim _{n\to \infty }\sum _{i=1}^n\left(\frac{i^4}{n^4}+\frac{i}{n}\right)(\frac 1 n)$ So it can be written as: $\int_0^1 (x^4 + x) dx $. Actually you have to bring in i/n and 1/n form where i/n changes to x and 1/n changes to dx. The limits of the integral change to L/n where L is the upper( or lower) limit

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Hint

Without Riemann $$\sum _{i=1}^n\left(\frac{i^4}{n^5}+\frac{i}{n^2}\right)=\frac 1 {n^5}\sum _{i=1} i^4+\frac 1 {n^2}\sum _{i=1} i$$ and use Faulhaber formulae.

Otherwise

$$\sum _{i=1}^\infty\frac{i}{n}\left(\left(\frac{i}{n}\right)^3+1\right)=\int_0^1x(x^3+1)\,dx$$

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  • $\begingroup$ you meant $i^{\color{red}{1}}$ in the second sum and sums are upto $n$ $\endgroup$ – farruhota Mar 24 at 6:56
  • $\begingroup$ @farruhota. Thanks for pointing the typo ! $\endgroup$ – Claude Leibovici Mar 24 at 6:59

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