Prove that for any positive integer $n>1$ $$ \frac{1}{n+1} + \frac{1}{n+2} + \frac{1}{n+3} \ldots + \frac{1}{2n} > \frac{13}{24} $$ I can prove that the series is greater than $\frac{12}{24}$ however i can't prove that it is greater than $\frac{13}{24}$.
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You might find this MathJax tutorial helpful in future :) – Shaun May 25 '14 at 12:10
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This is trivial, as others have pointed out, because the sum is increasing as a function of $n$. A less trivial variant is this question, which asks for a proof of the upper bound $\frac{25}{36}$. But don't look at joriki's answer! My answer is much better, as joriki himself would agree :-) – TonyK May 25 '14 at 12:30
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The LHS $x_n=\frac1{n+1}+\frac1{n+2}+\cdots+\frac1{2n}$ is such that $$ x_{n+1}-x_n=\frac1{2n+2}+\frac1{2n+1}-\frac1{n+1}=\frac1{2(n+1)(2n+1)}\gt0, $$ for every $n\geqslant1$, hence $x_n\geqslant x_2=\frac7{12}$ for every $n\geqslant2$.
Note that $x_1=\frac12$ and that $\frac7{12}=\frac12+\frac1{12}\gt\frac12+\frac1{24}=\frac{13}{24}$.
Unrelated but possibly worthwhile of notice is the fact that the increasing sequence $(x_n)$ is bounded since $\lim\limits_{n\to\infty}x_n=\log2\approx0.69$.
Did
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A less efficient method compared to Did's is the following
For each $k \in \{1,\ldots,n\}, \displaystyle\int_k^{k+1} \frac{1}{n+x} dx \leq \frac{1}{n+k}$
Summing these inequalities yields $\displaystyle \int_1^{n+1} \frac{1}{n+x} dx \leq \sum_{k=1}^n \frac{1}{n+k}$
Hence $\displaystyle \ln\frac{2n+1}{n+1} \leq \sum_{k=1}^n \frac{1}{n+k}$
For $n\geq 3, \displaystyle \frac{13}{24} < \ln\frac{2n+1}{n+1} $
This, however fails for $n=2$, but Did covered that case.
Gabriel Romon
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