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So I know that showing $(1+\frac{1}{n})^n$ is an increasing sequence has probably appeared on this site about 100 times, but my professor said he thinks the induction step is most easily seen if you expand this out and compare the summation term by term. I tried this and I did not what see why the inequality was obvious. Can somebody help me out here? Thanks!

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  • $\begingroup$ Use the binomial theorem. Just expand for $n$ and $n+1$. $\endgroup$ – Avinash N Oct 24 '18 at 20:11
  • $\begingroup$ Right, and after I did that the reason for the inequality still wasn't obvious to me $\endgroup$ – Math is hard Oct 24 '18 at 20:12
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To show that $\left(1+\frac1n\right)^{n}$ is increasing, all we need to show is that $\frac{\left(1+\frac1{n+1}\right)^{n+1}}{\left(1+\frac1n\right)^{n}}\ge1 $. Proceeding, we have

$$\begin{align} \frac{\left(1+\frac1{n+1}\right)^{n+1}}{\left(1+\frac1n\right)^{n}}&=\frac{n+1}{n}\left(\frac{n(n+2)}{(n+1)^2}\right)^{n+1}\\\\ &=\frac{n+1}{n}\left(1-\frac{1}{(n+1)^2}\right)^{n+1}\\\\ (\text{Using Bernoulli's Inequality})&\ge \frac{n+1}{n}\left(1-\frac{1}{n+1}\right)\\\\ &=1 \end{align}$$

And we are done!

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  • $\begingroup$ this is the best proof i've seen so far thanks! $\endgroup$ – Math is hard Oct 24 '18 at 20:34
  • $\begingroup$ Thank you Michael. Much appreciated. And feel free to accept the answer. $\endgroup$ – Mark Viola Oct 25 '18 at 0:07
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The $k$-th term in the binomial expansion looks like

$$ \frac{1}{n^k} \binom{n}{k} = \frac{n(n-1)\cdots(n-k+1)}{n^k k!} = \frac{1}{k!}\left( 1\right)\left( 1 - \frac1n\right)\left( 1 - \frac2n\right)\cdots\left( 1-\frac{k - 1}n\right). $$

(We've distributed a $1/n$ into each factor of the numerator.) Now compare factor by factor when we increase $n$ to $n + 1$.

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  • $\begingroup$ and in addition to this, the expansion of the $n+1$ will have one additional term in the summation, but the idea is that we can throw that out and still get the inquality to hold by comparing term by term what you have posted. Right? $\endgroup$ – Math is hard Oct 24 '18 at 20:17
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    $\begingroup$ @MichaelVaughan Yes, the additional term is positive so it fights right into the inequality. $\endgroup$ – Trevor Gunn Oct 24 '18 at 20:20

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