Find the following limit: $$\lim_{n \to \infty} \sum_{k=1}^{n} \frac{k^4}{k^5+n^5}$$ I had an idea of using upper Riemann sum for function $x^4$ on interval $[0,1]$ but I don't know how to deal with $k^5$ in denominator which ruins my approach. Kindly asking for some help.
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$\begingroup$ Maybe expand terms and see if anything cancels? Also please add some context for the problem. $\endgroup$– Тyma GaidashSep 4, 2021 at 2:08
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4$\begingroup$ Surely you can put this in the form $\frac{1}{n}\sum_{k=1}^n f(k/n)$. Divide numerator and denominator by $n^5$. $\endgroup$– RRLSep 4, 2021 at 2:10
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4$\begingroup$ $\sum^n_{k=1}\frac{n^4}{n^5+k^5}=\frac{1}{n}\sum^n_{k=1}\frac{\big(\tfrac{k}{n}\big)^4}{1+\big(\tfrac{k}{n}\big)^5}\xrightarrow{n\rightarrow\infty}\int^1_0\frac{x^4}{1+x^5}\,dx$ $\endgroup$– MittensSep 4, 2021 at 2:16
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1 Answer
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Using 'Definite integrals as limit of a sum':
The above limit is equivalent to $$\int_0^1{\frac{x^4}{x^5+1}}dx.$$
Substituting $z=x^5$ gives us $$\int_0^1{\frac{1}{5(z+1)}}dz$$ which is equal to $\displaystyle\frac{\log{2}}{5}$.
