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So here it is: $$\lim_{n \to \infty} \sqrt[n]{e^n+(1+\frac{1}{n})^{n^2}}~~~(n \in \mathbb{N})$$

I tried to use the Squeeze theorem like so: $e \leq \sqrt[n]{e^n+(1+\frac{1}{n})^{n^2}} \leq e \sqrt[n]{2}$

And if that is true, the limit is simply $e$.

Only that, for it to be true, I first need to prove that $(1+\frac{1}{n})^{n^2} \leq e^n$. So how do I prove it?

Alternatively, how else can I calculate the limit? I prefer to hear about the way I tried, though.

Thanks

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  • $\begingroup$ It's known that $(1 + \frac{1}{n})^n < e$. What you want should follow easily from that. $\endgroup$ – Arthur Dec 12 '14 at 14:18
  • $\begingroup$ Yes, it is true that $(1+1/n)^n$ is increasing. $\endgroup$ – André Nicolas Dec 12 '14 at 14:18
  • $\begingroup$ math.stackexchange.com/questions/295584/… $\endgroup$ – Guy Fsone Nov 10 '17 at 16:32
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    $\begingroup$ @GuyFsone You're flooding the review queue with questionable claims of duplicity. Knock it off. $\endgroup$ – amWhy Nov 11 '17 at 0:19
  • $\begingroup$ taking another post an rise to power n and take the nth root . id this not a duplicate? $\endgroup$ – Guy Fsone Nov 11 '17 at 6:59
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Hint: $$ (1+u)^b = \exp(b\log (1+u)) \le \exp(bu) $$ when $b>0$, because $\log(1+u)\le u$.

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Do you already have the theorem that $(1+\frac{1}{n})^n \le e$ for any natural number $n$? If so, then you do indeed have the inequality you want, and the rest of your solution is correct.

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You want $(1+\frac{1}{n})^n < e$, i.e. $e^{1/n} > 1 + \frac{1}{n}$. This is true because

$$e^{1/n} = 1 +\frac{1}{n} +\frac{1}{2n^2} + \cdots > 1 +\frac{1}{n}$$

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