Complication in summation How do I deal with this summation?
$$\sum_{j=n}^{2n-1} \frac{1}{n+j}$$
Do I just substitute all j's with n? That seems to be too easy.
*edit: This is a part of a larger problem, which is to find the limit of the summation as n tends to infinity.

If $u_n$ denotes the sum and you are asked if sequence $(u_n)$ converges when $n \to \infty$, the answer is yes; Moreover, the limit is $\ln(3/2)$. 
Here is why: we can write
$$u_n=\sum_{k=0}^{k=n-1}\frac{1}{2n+k}=\frac{1}{n}\sum_{k=0}^{k=n-1}\frac{1}{2+\dfrac{k}{n}}$$ which tends, as a Riemann sum, to 
$$\int_0^1 \dfrac{1}{2+x}dx=\ln(\dfrac32).$$
