I'm working in this exercise:  Find $$\lim_{n \to \infty} \frac{1}{n+1}+\frac{1}{n+2}+\cdots +\frac{1}{2n}.$$   I'm thinking about partitions and seeing this like an integral but I don't know which limits of integration should I use, maybe $0$ to $1$ since it's kinda close.  Any help is appreciated.
My foreign school book is not clearest at this point, what is really needed with this? What does it mean that "assign definite integral for $\lim_{n\rightarrow \infty}\sum_{k=1}^{n} \frac{1}{n+k}$"? I am not sure, whether I should assign it like this $\lim_{n\rightarrow \infty} \int_{1}^{n} \frac{1}{n+x} dx$ and noticing that it is continuous then trying to find borders so that $F(a) - F(b)$?  (the book messes up all kind of Leibniz stuff at this point, a bit messy -- and just stating $\int f(x)\,dx= F(a)-F(b)$, but not even paying attention to different borders. As far as I know, it is important to specify whether borders depend on the integration factor)
I'm running the following script on Ubuntu 14.04:  #!/bin/bash  apt-get purge -y nginx apt-get install -y nginx  date   When I run it like cat /tmp/script | bash, apt-get starts installing, then "date" is printed (not the actual date, but the command name), then the remaining apt-get output is printed.  If however I run the script like /tmp/script, it works as expected: printing the date after apt-get finished.  Why does this happen and how can I force bash to work when being piped to the same way it does when invoked directly? Evaluate $\prod \frac {2k - 1} {2k}$ I came across to this: evaluate $$\prod_{k = 1}^{n} \frac {2k - 1} {2k}.$$  I brought it to a closed-form expression in asymptotics but I suspect my asymptotics is bad and I also want to see different solutions, hence the question.  My original solution: Write   $$\begin {eqnarray} \prod_{k = 1}^{n} \frac {2k - 1} {2k} &amp; = &amp; \exp \left (\sum_{k = 1}^{n} \log \left (1 - \frac {1} {2k}\right) \right) \nonumber \\ &amp; = &amp; \exp \left (- \sum_{k = 1}^{n} \sum_{m = 1}^{\infty} \frac {1} {(2k)^m m}\right). \end {eqnarray}$$  Let $$S = \sum_{m = 1}^{\infty} \frac {1} {(2k)^m m} \tag {1}.$$ Multiply both sides of $(1)$ by $2k$ to get $2k S &lt; S + 1$, and by $4k$ to get $4kS &gt; S + 2$. Hence, $$\frac {2} {4k - 1} &lt; S &lt; \frac {1} {2k - 1}.$$ Then, $$\frac {1} {2} \left( \log (4n - 1) - \log 3 \right) - 2 + O \left(\frac {1} {n^2}\right) &lt; \sum_{k = 1}^{n} \sum_{m = 1}^{\infty} \frac {1} {(2k)^m m} &lt; \frac {1} {2} \log (2n - 1) - 1 + O \left(\frac {1} {n^2}\right).$$ Hence, $$\sqrt {\frac {c} {2n - 1}} &lt; \prod_{k = 1}^{n} \frac {2k - 1} {2k} &lt; \sqrt {\frac {3 c^2} {4n - 1}},$$ where $c$ is a constant. Any better approach? Which command controls the open file limits? 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