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The Question: Find $$\lim_{n \rightarrow \infty } \frac{1}{\sqrt{nn}} + \frac{1}{\sqrt{n(n+1)}}+ \frac{1}{\sqrt{n(n+2)}}+\cdots+ \frac{1}{\sqrt{n(n+n)}} .$$

The attempt:

I rewrote this as a series: $\displaystyle\sum_{k=0}^{n} \frac{1}{\sqrt{n(n+k)}}.$ Then I tried to represent this sum as a Riemann Sum. The width of the partitions is represented as the regular partitions:

$$\displaystyle\sum_{k=0}^{n} \frac{1}{\sqrt{n(n+k)}} = \sum_{k=0}^{n} \frac{\sqrt{n}}{n\sqrt{(n+k)}} = \sum_{k=0}^{n}\left(\frac{1}{n}\right)\sqrt{\frac{n}{n-k}}.$$

I am not sure where to go from here. Do you think I am on the right track?

Please I wants hints. Try not and solve the problem completely.

Thank you very much!

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  • $\begingroup$ You have ${1\over n}$, which is $\Delta x$. Now you need to take the rest of the term $\displaystyle\left(\sqrt{n\over n+k}\right)$ and write it as just a function of ${k\over n}$. $\endgroup$ – Christopher Carl Heckman May 10 '16 at 0:03
  • $\begingroup$ Just a heads up: I think you accidentally changed the sign in the denominator from the second-last summation to the last. $\endgroup$ – alphacapture May 10 '16 at 0:05
  • $\begingroup$ @alphacapture : Great eyes! Now he doesn't even have to do an improper integral ... $\endgroup$ – Christopher Carl Heckman May 10 '16 at 0:05
  • $\begingroup$ One more thing after my first hint: $\displaystyle {k\over n}$ will equal $x$ in the function. $\endgroup$ – Christopher Carl Heckman May 10 '16 at 0:14
  • $\begingroup$ Perhaps this will help: try writing out $\int_{0}^{1}{f(x)dx}$ in terms of the limit of a sum as the number of terms goes to infinity, and try to match that expression with yours. $\endgroup$ – alphacapture May 10 '16 at 0:21
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Now \begin{align*} \lim_{n\to \infty} \sum_{k=0}^{n} \frac{1}{n} \frac{1}{\sqrt{1+\frac{k}{n}}} &= \int_{0}^{1} \frac{dx}{\sqrt{1+x}} \\ &= \left[ 2\sqrt{1+x} \right]_{0}^{1} \\ &= 2(\sqrt{2}-1) \end{align*}

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