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We have to show that:


To be honest I don't have idea how to deal with it. I only suspect there will be need to consider two cases for $n=2k$ and $n=2k+1$

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Is there a particular condition on n (e.g. natural number, etc)? –  Bridgeburners Jul 31 '14 at 13:58
yes n is natural number –  Gregor Jul 31 '14 at 13:59
See also: math.stackexchange.com/questions/883615/… –  Martin Sleziak Jul 31 '14 at 15:49

8 Answers 8

up vote 4 down vote accepted

For $n = 2k$

$$\frac{1}{n} + \ldots + \frac{1}{2n} = \frac{1}{2k} + \frac{1}{2k+1} + \ldots + \frac{1}{3k} \stackrel{\downarrow}{+} \frac{1}{3k+1} + \ldots + \frac{1}{4k} \geq\\ \geq \overbrace{\frac{1}{3k} + \frac{1}{3k} + \ldots + \frac{1}{3k}}^{k+1 \text{times}} + \overbrace{\frac{1}{4k} + \frac{1}{4k} + \ldots + \frac{1}{4k}}^{k \text{times}} = \frac{k+1}{3k} + \frac{k}{4k} \geq \frac{1}{3} + \frac{1}{4} = \frac{7}{12} $$

For $n = 2k+1$

$$\frac{1}{n} + \ldots + \frac{1}{2n} = \frac{1}{2k+1} + \frac{1}{2k+2} + \ldots + \frac{1}{3k} \stackrel{\downarrow}{+} \frac{1}{3k+1} + \ldots + \frac{1}{4k+2} \geq\\ \geq \overbrace{\frac{1}{3k} + \frac{1}{3k} + \ldots + \frac{1}{3k}}^{k \text{times}} + \overbrace{\frac{1}{4k} + \frac{1}{4k} + \ldots + \frac{1}{4k}}^{k \text{times}} + \frac{1}{4k+1} + \frac{1}{4k+2} \geq \\ \geq \frac{k}{3k} + \frac{k}{4k} + \geq \frac{1}{3} + \frac{1}{4} = \frac{7}{12} $$

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Hint By the monotonicity of $x \mapsto \frac{1}{x}$, we have

$$\frac{1}{n} + \dots + \frac{1}{2n} = \sum_{k=n}^{2n} \frac{1}{k} \geq \int_n^{2n} \frac{1}{x} \, dx.$$

A direct calculation of the latter integral yields an even sharper bound:

$$\frac{1}{n} + \dots + \frac{1}{2n} \geq \ln 2 > \frac{7}{12}.$$

For the last inequality note that it follows from $e \leq 3$ that

$$e^{3/5} \leq 3^{3/5} <2;$$

hence $\ln 2>\frac{3}{5} > \frac{7}{12}$.

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+1 but now one must show $\ln 2 \ge \frac{7}{12}$ –  I. J. Kennedy Jul 31 '14 at 14:18
$\ln 2 \geq 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} = \frac{7}{12}$ by truncating alternating series expansion of $\ln(1+x)$ at $x=1$ at even number of term so that we always get positive contributions to the series (cf the proof of alternating series test). –  dioid Jul 31 '14 at 14:31
@I.J.Kennedy True enough. But it is not difficult to see that $$e^{3/5} \leq 3^{3/5} <2$$ and this implies $$\ln 2 > \frac{3}{5} \geq \frac{7}{12}.$$ –  saz Jul 31 '14 at 14:35

You are right: consider the case $n=2k$ and $n=2k+1$. The case $n=2k$ is easier and is as follows.

$$\frac1{2k}+\cdots+\frac{1}{3k}\geq (k+1)\frac{1}{3k}\ge\frac 13$$ $$\frac1{3k+1}+\cdots+\frac{1}{4k}\geq k\frac{1}{4k}=\frac1{4}.$$ So $$\frac1n+\frac1{n+1}+\cdots+\frac1{2n}\geq \frac13+\frac14=\frac7{12}.$$

Can you deal with the case $n=2k+1$?

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Try induction with LHS $> \dfrac{2n}{3n+1}$. Then show for $n>2$ this is larger than RHS.

Note: your approach of splitting into two cases should also work, if you pair terms symmetrically distant from middle, and bound them with the sum of extreme terms.

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(sorry I don't have enough reputation to comment and time to check that my answer is OK, but it could help you while a better answer comes)

Perhaps you could use this theorem, and observe that:

$\forall a>0,\ \displaystyle \int_a^{2a} \dfrac{1}{x}\mathrm{d}x = \ln(2)$

EDIT: saz proposed the same demonstration but properly justified, which I find much more elegant than fiddling with the elements of the sum! :D

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The sum is: $$S=\sum_{k=1}^N\dfrac{1}{k+N}=\Psi(2N+1)-\Psi(N+1)$$ Because $S$ doesn't have extrema, we calculate two limits: $$\lim_{N\to1}S=0.5$$ and $$\lim_{N\to\infty}S=\ln(2)$$ This answers to your question.

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Let $k$ be a positive integer and let $x$ be such that $x \in [k-1,k]$. You have $$ n+k-1 \leq x +n $$ and $$ \int_{k-1}^k \frac{1}{x+n} dx \leq \int_{k-1}^k \frac{1}{n+k-1} dx =\frac{1}{n+k-1} $$ thus, summing from $k=1$ to $n+1$, $n\geq1$, we get $$ \int_{0}^{n+1} \frac{1}{x+n} dx \leq \sum_{k=1}^{n+1}\frac{1}{n+k-1}=\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n} $$ or $$ \ln \left(2+\frac{1}{n}\right) \leq \frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n} $$ But $$ e^{7/12} \leq 2 < 2+\frac{1}{n}.$$

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one approach is to pair off the integers in the harmonic sum as $n+k$ and $2n-k$. so if $n$ is even we have: $$ S_n = \sum_{k=0}^{\frac{n}2-1}\left(\frac1{n+k} + \frac1{2n-k} \right) + \frac2{3n} $$ now $$\frac1{n+k} + \frac1{2n-k} = \frac{3n}{(n+k)(2n-k)} \ge \frac{4n}{3n^2} = \frac4{3n} $$ and there are $\frac{n}2$ such terms, so we have: $$ S_n \ge \frac23 + \frac2{3n} $$ when $n$ is odd we arrive (i think) at the numerically similar estimate $$ S_n \ge \frac{6 \left(1+\frac1{n}\right)}{9-\frac1{n^2}} $$

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