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?
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1the Limit is also $e$ – Dr. Sonnhard Graubner Mar 12 '18 at 15:19
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To be totally correct, one really needs to show that $\left(1+\frac1x\right)^x$ is increasing for real $x\gt0$. – robjohn Mar 12 '18 at 16:36
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$n\longmapsto \left(1+\frac{1}{n}\right)^n$ and $n\longmapsto \sqrt n$ are increasing. Therefore, $(a_n)$ is a composition of two increasing functions, and thus, it's also increasing.
Surb
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Surb.Please elaborate . (1+1/n)^n is increasing, is that so obvious?Thanks, Peter – Peter Szilas Mar 12 '18 at 15:46
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@PeterSzilas : I let you manage this ;-) First, it's not hard and very famous ! secondly, if you search a little bit, I'm sure you can find such a result on MSE... The idea is to use binomial theorem... – Surb Mar 12 '18 at 15:56
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@Surb: this leaves me uncomfortable since most of the proofs using the binomial theorem, or Bernoulli's inequality for that matter, work only for integer $n$. Composing with $\sqrt{n}$ takes us out of that realm. – robjohn Mar 12 '18 at 16:25
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@robjohn: for $(1+\frac{1}{n})^n$ using binomial theorem is great (of course I don't use it on $a_n$). For the composition function, I use the general fact that if $f:(X,\leq_1)\to (Y,\leq _2)$ and $g:(Y,\leq_2)\to (Z,\leq_3)$ are increasing, then $g\circ f:(X,\leq_1)\to (Z,\leq_3)$ is increasing... – Surb Mar 12 '18 at 16:43
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@Surb: but what version of the Binomial Theorem are you using to prove that $\left(1+\frac1n\right)^n$ is increasing? Does it only show that $\left(1+\frac1n\right)^n\le\left(1+\frac1{n+1}\right)^{n+1}$? If so, then that is not really good enough since $\sqrt{n+1}-\sqrt{n}=\frac1{\sqrt{n+1}+\sqrt{n}}\lt1$. – robjohn Mar 12 '18 at 16:56
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@robjohn : yes it's to show $(1+\frac{1}{n})^n\leq (1+\frac{1}{n+1})^{n+1}$. I don't understand your counter example, sorry. – Surb Mar 12 '18 at 17:01
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@Surb: This approach shows that for $n\in\mathbb{Z}^+$, $\left(1+\frac1n\right)^n$ is increasing. However, it doesn't show what happens between integers and to compose with $\sqrt{n}$, which are not all in $\mathbb{Z}$, we should show that $\left(1+\frac1x\right)^x$ is increasing for real $x\gt0$. It is only your comment regarding the Binomial Theorem that I am questioning, not your answer, which does not refer to the Binomial Theorem. – robjohn Mar 12 '18 at 18:21
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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 $n$.
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 $u\gt-1$, $$ u-\log(1+u)\ge0 $$
For $x\gt0$, we have $u=-\frac1{1+x}\gt-1$, and therefore, $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\,x\log\left(1+\frac1x\right) &=\overbrace{-\frac1{1+x}\ \ }^u-\overbrace{\log\left(1-\frac1{1+x}\right)}^{\log(1+u)}\\ &\ge0 \end{align} $$
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.
robjohn
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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.
Mohammad Riazi-Kermani
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