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In some note I see the writer claims that the following inequality is obvious using an induction argument:

\begin{equation} \left(\left(\frac{1}{n}\right)^n+\left(\frac{2}{n}\right)^n+\cdots+\left(\frac{n}{n}\right)^n\right)^{\frac{1}{n}}<1+\frac{1}{n} \end{equation} But it's not obvious to me. Could you please help me to understand the claimed obvious solution?

John Wayland Bales
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Karaj
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    Raise both sides to the $n$ and clear denominators. We need to show $$\sum_{k=1}^n{k^n}\le(n+1)^n.$$ I haven't been able to do this yet, but it looks a lot more amenable to induction. On second thought, you can do this by comparison with $\int_1^{n+1}{x^n}dx$ can't you? No induction required. – saulspatz Jun 23 '18 at 22:47

2 Answers2

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Since my comment turned out to be an answer, I'll post it as such.

Raise both side to the $n$ and clear denominators. We must show $$\sum_{k=1}^n{k^n}\le n^n\left(1+\frac1n\right)^n=(n+1)^n$$ However, $$\sum_{k=1}^n{k^n}\le\int_1^{n+1}{x^n\mathrm{dx}}=\frac{x^{n+1}}{n+1}\Big|_1^{n+1}<\frac{(n+1)^{n+1}}{n+1}=(n+1)^n$$

saulspatz
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  • The integral goes to n, not n+1. Therefore the result is $n^{n+1}/(n+1)$. And, actually, $n^{n+1} > (n+1)^n$ for $n \ge 3$. – marty cohen Jun 23 '18 at 23:54
  • Looking again, if the integral goes to n+1, then the first inequality is correct. So your proof does go through. Upvoted. – marty cohen Jun 24 '18 at 00:00
  • @martycohen Just a typo. I'll fix it. – saulspatz Jun 24 '18 at 00:10
  • Your proof will then be correct. You might take a look at mine, which gives a slightly more accurate result. My technique gives sum $< n^n+n^{n+1}/(n+1) = n^n(1+n/(n+1))$. – marty cohen Jun 24 '18 at 00:16
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I can't prove it by induction, but I can prove it.

We want to show that $\Big(\sum_{k=1}^n(k/n)^n\Big)^{1/n‎} < 1+‎\dfrac{1}{n}‎ $.

If $f(x) = x^n$, since $f'(x) > 0$ for $n \ge 2$ and $0 < x < 1$, $(k/n)^n \lt n\int_{k/n}^{(k+1)/n} x^n dx \lt ((k+1)/n)^n $.

Therefore

$\begin{array}\\ \sum_{k=1}^{n-1}(k/n)^n &=\sum_{k=0}^{n-1}(k/n)^n\\ &\lt \sum_{k=0}^{n-1}n\int_{k/n}^{(k+1)/n} x^n dx\\ &= n\sum_{k=0}^{n-1}\int_{k/n}^{(k+1)/n} x^n dx\\ &= n\int_{0}^{1} x^n dx\\ &= \dfrac{n}{n+1}\\ \text{so}\\ \sum_{k=1}^{n}(k/n)^k &\lt 1+\frac{n}{n+1}\\ \end{array} $

Taking the $n$-th root $(\sum_{k=1}^{n}(k/n)^n)^{1/n} \lt (1+\dfrac{n}{n+1})^{1/n} $.

By Bernoulli's inequality, $(1+x)^n \ge 1+nx$ or $(1+x/n)^n \ge 1+x$.

Taking the $n$-th root, $(1+x)^{1/n} \le 1+x/n$.

Putting $x=n/(n+1)$, $(1+n/(n+1))^{1/n} \le 1+(n/(n+1))/n =1+1/(n+1) $.

Therefore $\Big(\sum_{k=1}^n(k/n)^n\Big)^{1/n‎} < 1+‎\dfrac{1}{n+1}‎ $ which is slightly better that what was asked.

We can accurately estimate the sum by noting that $(1-k/n)^n \approx e^{-k} $ for small $k$, so that

$\begin{array}\\ \sum_{k=1}^n(k/n)^n &=\sum_{k=0}^{n-1}((n-k)/n)^n\\ &=\sum_{k=0}^{n-1}(1-k/n)^n\\ &\approx\sum_{k=0}^{n-1}e^{-k}\\ &\approx \dfrac{1}{1-1/e}\\ &= \dfrac{e}{e-1}\\ &=1+ \dfrac{1}{e-1}\\ &\approx 1.582\\ \end{array} $

Therefore $(\sum_{k=1}^n(k/n)^n)^{1/n} \approx (1+ \dfrac{1}{e-1})^{1/n} \approx 1+ \dfrac{1}{n(e-1)} $.

marty cohen
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