Is the sequence $a_n=(1 + n^{-1/2})^\sqrt n$ bounded and monotone?

Question

It's clear that $a_n=(1 + n^{-1/2})^\sqrt n =(1+ \frac{1}{\sqrt n})^\sqrt n $, which seems to be pretty similar to, $ (1+ \frac{1}{n}) ^n$ but how can I continue to solve this problem?

Answer 1

$n\longmapsto \left(1+\frac{1}{n}\right)^n$ and $n\longmapsto \sqrt n$ are increasing. Therefore, $(a_n)$ is a composition of two increasing functions, and thus, it's also increasing.

Answer 2

Consider the function $$ f(x)=(1+1/x)^x$$

Note that $f(x)$ is monotone and bounded over $[1,\infty).$

Thus $$ a_n=(1+ \frac{1}{\sqrt n})^\sqrt n$$ is monotone and bounded.

Answer 3

I was going to cite this answer to show that $\left(1+\frac1n\right)^n$ was increasing. Then I realized that that answer only showed this increase for integer $n$, so composing with $\sqrt{n}$ would not really work. Thus, I feel this increase needs to be shown for real $n$.

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For $u\gt-1$, $$ \frac{\mathrm{d}}{\mathrm{d}u}\left(u-\log(1+u)\right) =\frac{u}{1+u}\\ $$ which means that for $-1\lt u\lt0$, $u-\log(1+u)$ is decreasing, and for $u\gt0$, $u-\log(1+u)$ is increasing. Thus, $u-\log(1+u)$ has a minimum at $u=0$. That is, for $u\gt-1$, $$ u-\log(1+u)\ge0 $$

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For $x\gt0$, we have $u=-\frac1{1+x}\gt-1$, and therefore, $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\,x\log\left(1+\frac1x\right) &=\overbrace{-\frac1{1+x}\ \ }^u-\overbrace{\log\left(1-\frac1{1+x}\right)}^{\log(1+u)}\\ &\ge0 \end{align} $$

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This means that for $x\gt0$, $\left(1+\frac1x\right)^x$ is increasing since its logarithm is increasing.

Now we can set $x=\sqrt{n}$ and get that the composition of increasing functions $$ \left(1+\frac1{\sqrt{n}}\right)^{\large\sqrt{n}} $$ is increasing.