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I'm having problems trying to compute the limit of the sum

$\lim_{n\to\infty}\sum_{k=1}^{n}(\frac{k}{n})^{4}(\frac{2k-1}{n^{2}})$

As you can see you can't take $\xi_{k}$ as the extreme of some partition, for example $k/n$. because we obtain $\frac{1}{n}\sum_{k=1}^{n}(\frac{k}{n})^{4}(\frac{2k-1}{n})=\frac{1}{n}\sum_{k=1}^{n}\frac{2k^{5}-k^{4}}{n^{5}}=\frac{1}{n}\sum_{k=1}^{n}(2(\frac{k}{n})^{5}-\frac{1}{n}(\frac{k}{n})^{4})$ but what do I do with the second $\frac{1}{n}$?

I'm thinking on taking another values for $\xi_{k}$ and another interval distinct from $[0,1]$ but I don't get which one could be useful.

Another idea I had is that as we can take inside the sum $(\frac{k}{n})^{4}\frac{2k-1}{n}=2(\frac{k}{n})^{4}(\frac{\frac{k-1}{n}+\frac{k}{n}}{2})$ with $x_{k-1}=\frac{k-1}{n}$ and $x_{k}=\frac{k}{n}$ the second parentheses would be the mid point of $[x_{k-1},x_{k}]$ but in the first parentheses I have one extrema $x_{k}=\frac{k}{n}$ so this would let me to take two distinct points in each subinterval

Gary
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Iesus Dave Sanz
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    $$ 2 \times \frac{1}{n}\sum\limits_{k = 1}^n {\left( {\frac{k}{n}} \right)^5 } - \frac{1}{n} \times \left( {\frac{1}{n}\sum\limits_{k = 1}^n {\left( {\frac{k}{n}} \right)^4 } } \right) \to 2\int_0^1 {x^5 dx} - 0 \times \int_0^1 {x^4 dx} = 2\int_0^1 {x^5 dx} $$ – Gary Oct 28 '21 at 01:39
  • @Gary You're right. Thank you – Iesus Dave Sanz Oct 28 '21 at 01:40

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