Is there any smart trick to find the following limit? $$ \lim_{n \rightarrow \infty} \left(\frac{1}{\sqrt{2n^4+1}} + \frac{2}{\sqrt{2n^4+2}} + \cdots + \frac{n}{\sqrt{2n^4+n}}\right). $$



$$\lim_{n\to +\infty}\frac{1}{\sqrt{2n^4}}\sum_{k=1}^{n}k = \lim_{n\to +\infty}\frac{n(n+1)}{2\sqrt{2}\,n^2}=\color{red}{\frac{1}{2\sqrt{2}}}.$$ We get the same limit if we replace the initial $\sqrt{2n^4}$ with $\sqrt{2n^4+n}$.
It follows that our actual limit also equals $\frac{1}{2\sqrt{2}}$.

  • $\begingroup$ +1. Your shortcut avoids the $Hurwitz\ Zeta\ Function$. Clever. $\endgroup$ – Felix Marin Oct 17 '16 at 1:33

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