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!

• 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 ${k\over n}$. – Christopher Carl Heckman May 10 '16 at 0:03
• Just a heads up: I think you accidentally changed the sign in the denominator from the second-last summation to the last. – alphacapture May 10 '16 at 0:05
• @alphacapture : Great eyes! Now he doesn't even have to do an improper integral ... – Christopher Carl Heckman May 10 '16 at 0:05
• One more thing after my first hint: $\displaystyle {k\over n}$ will equal $x$ in the function. – Christopher Carl Heckman May 10 '16 at 0:14
• 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. – alphacapture May 10 '16 at 0:21