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This is from the book Problems in Mathematical Analysis I by Kaczor and Nowak:

Show that, for $n\in \mathbb{N}$, $$\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n}>\frac{2}{3}$$ The solution in the back of the book says to apply the Arithmetic Harmonic mean inequality, but when I try to do so I get that $$\frac{\sum_{i=0}^n (n+i)}{n+1}\ge \frac{1}{\sum_{i=0}^n\frac{1}{n+i}}$$ and the left hand side is equal to $$\frac{n(n+1)+\frac{n(n+1)}{2}}{n+1}=\frac{3n}{2}$$ so it looks like the best I can do is $$\sum_{i=0}^n\frac{1}{n+i}\ge \frac{2}{3n}$$ what am I missing here?

user140776
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4 Answers4

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As pointed by r9m, Jensen's inequality is enough, since $\frac{1}{x}$ is a convex function on $\mathbb{R}^+$, hence:

$$\sum_{k=n}^{2n}\frac{1}{k}\geq \frac{1}{n}+n\left(\frac{1}{n}\sum_{k=n+1}^{2n} k\right)^{-1}=\frac{1}{n}+\frac{2n}{3n+1}\geq\frac{2}{3}.$$

As an alternative, $$ \frac{1}{k}\geq \log(k+1)-\log k+\frac{1}{2}\left(\frac{1}{k}-\frac{1}{k+1}\right) $$ implies: $$ \sum_{k=n}^{2n}\frac{1}{k}\geq \log\left(\frac{2n+1}{n}\right)+\frac{n+1}{2n(2n+1)}.$$

Jack D'Aurizio
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  • (+1) I didn't notice until after I posted that you used Jensen. It turns out that you don't need to pull the $\frac1n$ out of the sum before applying Jensen. – robjohn Jul 09 '15 at 23:18
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Since $f(x)=\frac{1}{x}$ is convex, Jensen's inequality gives $$ \begin{align} \frac1{n+1}\sum_{k=n}^{2n}\frac1k &\ge\left(\frac1{n+1}\sum_{k=n}^{2n}k\right)^{-1}\\ &=\frac{2(n+1)}{(4n^2+2n)-(n^2-n)}\\ &=\frac{2(n+1)}{3n(n+1)} \end{align} $$ Therefore, $$ \bbox[5px,border:2px solid #C0A000]{\sum_{k=n}^{2n}\frac1k \ge\frac23 \frac{n+1}{n}} $$

Vincenzo Oliva
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robjohn
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$$\frac{\frac1n+\frac1{n+1}+\ldots+\frac1{2n}}{n+1}\ge\frac{n+1}{n+(n+1)+\ldots+(2n)}=\frac{n+1}{n(n+1)+\frac{n(n+1)}2}\implies$$

$$\frac1n+\frac1{n+1}+\ldots+\frac1{2n}\ge\frac{(n+1)^2}{\frac32n(n+1)}=\frac23\frac{(n+1)^2}{n^2+n}\ge\frac23$$

Timbuc
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$$\sum_{k=n}^{2n}{1\over k} \ge \int_n^{2n}{dt\over t} = \ln(2).$$

ncmathsadist
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    How do you know $\ln(2) > \frac{2}{3}$? – Ian Jul 09 '15 at 22:20
  • You take the exponential – ParaH2 Jul 09 '15 at 22:22
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    How do you know $2>e^{2/3}$? (To me it seems that $\ln(2)>\frac{2}{3}$ would be easier to prove, actually, since in principle you could take $\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}$ out to a whole lot of terms. But it turns out that it takes about 40 terms to do that...) – Ian Jul 09 '15 at 22:22
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    $e<2.8 \implies e^2<2.8^2<8 \implies e^{2/3}<2$ – imj Jul 09 '15 at 22:27
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    @imj That's a nice argument, without too much horrible arithmetic. But then how do you prove $e<2.8$? About the best I can do off the top of my head is $2<e<3$. – Ian Jul 09 '15 at 22:31
  • Do a Riemann sum on $\int_1^2 dt/t$ with $n = 5$ subintervals. Pure carpentry. – ncmathsadist Jul 09 '15 at 22:37
  • @Ian $\sum_{k=0}^n \frac{3-4k^2}{(2k+1)!}$ decreases to $e$ (according to wiki), and it appears to be quite fast, as $n=2$ yields $2.725$ ($n=5$ gives 7 correct digits). No idea on how to prove that series work though :p – imj Jul 09 '15 at 23:23
  • @Ian Evaluating gives $$\int_0^1 \frac{x^2 (1 - x)^2}{1 + x^2} dx = \log 2 - \frac{2}{3},$$ but the integrand is nonnegative (and continuous and somewhere positive), so the integral is positive and hence rearranging gives $$\log 2 > \frac{2}{3}$$ as desired. – Travis Willse Jul 10 '15 at 01:20
  • In fact, using the naive inequality $\frac{x^2 (1 - x)^2}{1 + x^2} \leq x^2 (1 - x)^2$ and integrating gives that $\log 2 - \frac{2}{3} < \frac{1}{30},$ giving the bounds $$\frac{2}{3} < \log 2 < \frac{7}{10},$$ which are reasonably tight for the relatively little calculation involved. – Travis Willse Jul 10 '15 at 01:26