How to show $a_n = (1 + \frac{2}{n} )^n$ is bounded above?

Asked Apr 23 '23 at 09:57

Active Apr 23 '23 at 09:58

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Can someone help me about this question? I have proved $a_n = (1 + \frac{2}{n})^n$ is increasing, but I have been confused by how to show it is bounded above.

I have show that for $1 \leq k \leq n$

${n \choose k} \frac{2^k}{n^k} = \frac{2^k}{k!} \frac{n(n-1)···(n-k)}{n^k} \leq \frac{2^k}{k!}$

Then, I was confusing on how can I show $a_n$ is bounded above a number.

Need someone's help, thanks.

edited Apr 23 '23 at 09:58

asked Apr 23 '23 at 09:57

murtabak

The inequality you are wanting to show follows from the fact that $n(n-1)\cdots (n-k)\leq n^k$ as $n\geq n-1\geq n-2 \geq \cdots \geq n-k$. Thus $n(n-1)\cdots (n-k)/n^k\leq 1$. – JayP Apr 23 '23 at 10:07
Looks like you could use the binomial theorem to show that $a_n$ is bounded based on the inequality you have already proven. – Surb Apr 23 '23 at 10:08
Related to https://math.stackexchange.com/questions/1185660/prove-1-frac1nn-is-bounded-above – JayP Apr 23 '23 at 10:09
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Does this answer your question? Proof, without logs/exp that (1+x/n)^n converges – Anne Bauval Apr 23 '23 at 10:31
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It is equivalent to prove that $\log(a_n)$ is bounded from above. – Jean Marie Apr 23 '23 at 10:39
Wouldn’t be simpler to start with $\sqrt{a_n}=\left(1+\frac 1{\frac 2n}\right)^\frac n2\to e$? – WindSoul Apr 23 '23 at 11:15
@WindSoul Sir, so my goal is to show if $\sqrt(a_n)$ is bounded above, then $a_n$ is also bounded above right? – murtabak Apr 24 '23 at 05:37
@JeanMarie alright Sir, I will check it, thanks sir. – murtabak Apr 24 '23 at 05:39

How to show $a_n = (1 + \frac{2}{n} )^n$ is bounded above?

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