# Evaluate limit with sum

Evaluate $\displaystyle \lim_{n \to +\infty}\sum_{k=n+1}^{2n}\frac{1}{k}$. What are the ways of counting such things? My last topic in school was Riemann integral, can I use it here?

-
yes, you can use it here :-) – user20266 Apr 22 '12 at 12:50

You sure can!

$$\sum_{k=n+1}^{2n} \frac{1}{k} = \frac{1}{n} \sum_{k=n+1}^{2n} \frac{1}{(k/n)} .$$

The second version of writing the sum makes it clearer that it is the Riemann sum of $f(x) = 1/x$ obtained by dividing $[1,2]$ into $n$ pieces and setting up rectangles over those intervals. As such, your limit is $\int^2_1 \frac{1}{x} dx = \log 2.$

-

Yes, you can use it.

The technique is $$\int_{k+1}^{k+2}\frac{1}{x}d x \leq \frac{1}{k+1}\leq \int_{k}^{k+1}\frac{1}{x}d x.$$

-

The sum equals (the sum of 1/k from k=1 to 2n) minus (the sum of 1/k from k=n+2 to 2n). Then take the limit.

-