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So I have to prove:

For each natural number greater than or equal to 3, $$(1+\frac1n)^n<n$$

My work: Basis step: $n=3$ $$\left(1+\frac13\right)^3<3$$ $$\left(\frac43\right)^3<3$$ $$\left(\frac{64}{27}\right)<3$$ which is true.

Now the inductive step, assume $P(k)=\left(1+\frac1k\right)^k<k$ to be true and prove $P(k+1)=\left(1+\frac1{k+1}\right)^{k+1}<k+1$.

This is where I am stuck because usually you add or multiply by $k+1$ or some similar term.

Lord_Farin
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user111702
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4 Answers4

6

Hint: $$ \left( 1+\frac{1}{k+1} \right)^{k+1} = \left( 1 + \frac{1}{k+1}\right) \left( 1 + \frac{1}{k+1}\right)^{k} < \left( 1 + \frac{1}{k+1}\right) \left( 1 + \frac{1}{k}\right)^{k}\\ < \left( 1 + \frac{1}{k+1}\right)k $$ where the last inequality comes from your induction hypothesis.

Tom
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  • Alright I understand how you broke down the $$\left(1+\frac1{k+1}\right)^{k+1}=\left(1+\frac1{k+1}\right)\left(1+\frac1{k+1}\right)^k$$ but I do not see how you got the inequality from the hypothesis – user111702 Nov 26 '13 at 21:42
  • The first inequality is simply because $\frac{1}{k+1} < \frac{1}{k}$. The second inequality was by the induction hypothesis that $$\left( 1 + \frac{1}{k} \right)^k < k$$ – Tom Nov 26 '13 at 21:44
  • So from the hypothesis you multiply both sides by a factor of $$\left(1+\frac1{k+1}\right)$$ and get $$\left(1+\frac1{k+1}\right)\left(1+\frac1k\right)^k<\left(1+\frac1{k+1}\right)k$$ right? – user111702 Nov 26 '13 at 21:50
  • That's one way to see it. I just considered the following $$ \left( 1+\frac{1}{k+1} \right) \underbrace{ \left(1+\frac{1}{k} \right)^k }_{<k \text{ by hypothesis}} < \left(1 + \frac{1}{k+1} \right)k$$ – Tom Nov 26 '13 at 21:51
  • Alright so then how do you know that $$\left(1+\frac1{k+1}\right)\left(1+\frac1{k+1}\right)^k<\left(1+\frac1{k+1}\right)\left(1+\frac1{k}\right)^k$$ – user111702 Nov 26 '13 at 21:56
  • That's just because $\frac{1}{k+1} < \frac{1}{k}$. – Tom Nov 26 '13 at 21:58
  • So then that proves $$\left(1+\frac1{k+1}\right)^k<k$$ but how does that then lead to $$\left(1+\frac1{k+1}\right)^{k+1}<k+1$$ – user111702 Nov 26 '13 at 22:01
  • Multiply the $k$ through in $\left(1+\frac{1}{k+1}\right)k$ and you should see why you get $< k+1$. – Tom Nov 26 '13 at 22:02
  • I got $$k+\frac{k}{k+1}$$ $$\frac{k(k+1)}{k+1}+\frac{k}{k+1}$$ $$\frac{k^2+k+k}{k+1}$$ $$\frac{k^2+2k}{k+1}$$ $$\frac{k(k+2)}{k+1}$$ – user111702 Nov 26 '13 at 22:10
  • Do you see that $k + \frac{k}{k+1} < k + 1$? This comes down to just believing that $\frac{k}{k+1} < 1$ (the numerator is smaller than the denominator). – Tom Nov 26 '13 at 22:16
  • Oh I see or alternatively: $$k+1-\frac1{k+1}<k+1$$ – user111702 Nov 26 '13 at 22:19
4

$\left(1+\frac 1 n\right)^n=1+1+\binom n 2\frac 1 {n^2}+\binom n 3 \frac 1 {n^3}+\dotsb+\frac 1 {n^n}$

But $\binom n k \frac 1 {n^k}=\frac {n(n-1)\dotsm(n-k+1)}{k!n^k}<\frac 1 {k!}$.

So the expression we're interested in is less than $$1+1+\frac 1 {2!}+\frac 1{3!}+\dotsb+\frac 1 {n!}<1+1+\frac 1 2 +\frac 1 4 +\frac 1 8+\dotsb=3.$$

dfeuer
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1

Two approaches

Bernoulli-like Inductive Step $$ \begin{align} \frac{\left(1+\frac1{n+1}\right)^{n+1}}{\left(1+\frac1n\right)^n} &=\overbrace{\left(\frac{n(n+2)}{(n+1)^2}\right)^{n+1}}^{\lt1}\frac{n+1}n\\ &\lt\frac{n+1}n \end{align} $$ This shows that if $\left(1+\frac1n\right)^n\lt n$, then $\left(1+\frac1{n+1}\right)^{n+1}\lt n+1$

Non-Inductive Bernoulli Approach

Bernoulli's Inequality says that for $n\ge2$, $$ \begin{align} \left(1-\frac1{n+1}\right)^n &=\left(\left(1-\frac1{n+1}\right)^{n/2}\right)^2\\ &\ge\left(1-\frac{n/2}{n+1}\right)^2\\[3pt] &=\left(\frac{n/2+1}{n+1}\right)^2 \end{align} $$ which is the reciprocal of $$ \begin{align} \left(1+\frac1n\right)^n &\le\left(\frac{n+1}{n/2+1}\right)^2\\[6pt] &\lt2^2 \end{align} $$ Thus, $\left(1+\frac1n\right)^n\lt n$ if $n\ge4$.

We just need to verify the inequality for $n=3$: $$ \left(1+\frac13\right)^3=\frac{64}{27}\lt3 $$

robjohn
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1

This is equivalent to proving $$P(n)\colon\qquad (n+1)^n <n^{n+1}$$ for $n\ge3$.

Inductive step: We assume that $P(k-1)$ holds, i.e., that $$k^{k-1}<(k-1)^k.$$ We will prove $P(k)$ by contradiction.

Assume that $P(k)$ does not hold, i.e., $$(k+1)^k \ge k^{k+1}.$$

By multiplying the inequalities $$ \begin{align*} (k-1)^k &> k^{k-1}\\ (k+1)^k &\ge k^{k+1} \end{align*} $$ we get $$(k^2-1)^k \ge k^{2k},$$ i.e., $(k^2-1)^k \ge (k^2)^k$, which is a contradiction.

This can be used to show that the sequence $\sqrt[n]{n}$ is eventually decreasing. I guess there are a few posts about this question, but I was not able to find some such post quickly.

Martin Sleziak
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