Show that the inequality holds $\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}\ge\frac{7}{12}$ We have to show that:
$\displaystyle\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}\ge\frac{7}{12}$
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 $n=2k+1$
 A: 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 $\ln 2>\frac{3}{5} > \frac{7}{12}$.
A: 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 $n = 2k+1$
$$\frac{1}{n} + \ldots + \frac{1}{2n} = \frac{1}{2k+1} + \frac{1}{2k+2} + \ldots + \frac{1}{3k} \stackrel{\downarrow}{+} \frac{1}{3k+1} + \ldots + \frac{1}{4k+2} \geq\\
\geq \overbrace{\frac{1}{3k} + \frac{1}{3k} + \ldots + \frac{1}{3k}}^{k \text{times}} + \overbrace{\frac{1}{4k} + \frac{1}{4k} + \ldots + \frac{1}{4k}}^{k \text{times}} + \frac{1}{4k+1} + \frac{1}{4k+2} \geq \\ \geq \frac{k}{3k} + \frac{k}{4k} + \geq \frac{1}{3} + \frac{1}{4} = \frac{7}{12}
$$
A: 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 $n=2k+1$?
A: (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
A: Hint 
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. 
A: 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 $$ e^{7/12} \leq 2 < 2+\frac{1}{n}.$$
A: 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
$$
S_n \ge \frac{6 \left(1+\frac1{n}\right)}{9-\frac1{n^2}}
$$
A: 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.
