Find the limit of $\displaystyle\lim_{n\to\infty}\frac 1 {n^5}(1^4+2^4…+n^4)$ using definite integrals.
It's equal to: $\displaystyle\lim_{n\to\infty} \sum^n_{i=1}\frac 1 i$ but now I'm not sure how to turn it to an integral.
$\Delta x_i=\frac 1 n, f(x_i)=1$ so the integral would be: $\displaystyle\int 1dx$ ? How can I find the bounds?
Best Answer
Hint: rewrite the sum as: $$\frac1n\sum_{k=1}^n\Bigl(\frac kn\Bigr)^4.$$ This is an upper Riemann sum for the function $x^4$ on the interval $[0,1]$.