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may someone show how to compute $\lim_{n \rightarrow \infty} \left(\left(\frac{9}{4} \right)^n+\left(1+\frac{1}{n} \right)^{n^2} \right)^{\frac{1}{n}}$?

According to W|A it's e, but I don't know even how to start...

Please help, thank you.


marked as duplicate by Dan Rust, Michael Albanese, AlexR, Brian Rushton, Jack D'Aurizio Jan 11 '14 at 14:59

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Clearly, $$ \left(1+\frac{1}{n}\right)^{\!n^2}< \left(\frac{9}{4}\right)^n+\left(1+\frac{1}{n}\right)^{\!n^2}<2\,\mathrm{e}^n, $$ and therefore $$ \left(1+\frac{1}{n}\right)^{\!n}< \left(\left(\frac{9}{4}\right)^n+\left(1+\frac{1}{n}\right)^{n^2}\right)^{1/n}\le 2^{1/n}\mathrm{e} $$ which implies that $$ \mathrm{e}=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{\!n}\le\lim_{n\to\infty}\left(\left(\frac{9}{4}\right)^n+\left(1+\frac{1}{n}\right)^{n^2}\right)^{1/n}\le \lim_{n\to\infty}2^{1/n}\mathrm{e}=\mathrm{e}. $$ Hence the limit of $\,\left(\left(\frac{9}{4}\right)^n+\left(1+\frac{1}{n}\right)^{n^2}\right)^{1/n},\,\,$ as $\,n\to\infty$, exists and it is equal to e.

  • $\begingroup$ Great! Thanks a lot. $\endgroup$ – Galc127 Jan 10 '14 at 9:03

By Taylor series we have

$$\left(1+\frac{1}{n} \right)^{n^2}\sim_\infty e^n $$

but $$\frac 9 4<e$$ hence $$\left(\frac{9}{4} \right)^n=_\infty o(e^n)$$ hence $$\left(\left(\frac{9}{4} \right)^n+\left(1+\frac{1}{n} \right)^{n^2} \right)^{\frac{1}{n}}\sim_\infty e$$

  • $\begingroup$ Great answer, thank you! $\endgroup$ – Galc127 Jan 10 '14 at 9:01
  • 2
    $\begingroup$ Yes it is great answer cause of using basics of Calculus appropriately. +1 $\endgroup$ – mrs Jan 10 '14 at 9:48

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