<s>[INST] Calculating limit approaches inifnity having infinite terms On my sample calculus mid-term exam, there is a weird question that asks us to calculate the limit that has an infinite term:
$\displaystyle\lim_{n \to \infty} {\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}}$
I am not sure what kind of technique should be used here.
Giving that this is mid-term of the first-year university calculus exam, we have only learned L'hopital's Rule and other basic techniques. But I don't see any technique that can fit to solve this problem 
Thanks!
 [/INST] Note that 
$$\displaystyle\lim_{n \to \infty} {\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}}=\displaystyle\lim_{n \to \infty} \frac{1}{n}{\sum_{k=1}^n\frac{1}{1+\frac{k}{n}}}$$
This is a Riemann Sum for $f(x)=\frac{1}{1+x}$ over the interval $[0,1]$, with the satndard partition $x_i=\frac{k}{n}$ for $1 \leq k \leq n$ and the right hand points of the interval as intermediate points.
Therefore
$$\displaystyle\lim_{n \to \infty} {\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}}=\int_0^1 \frac{1}{1+x}dx =\ln(1+x)|_0^1=\ln(2)$$
P.S. One can reach the came conclusion by using the well known identity
$$\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}+...-\frac{1}{2n}$$ and the standard definition of the Euler -Mascheroni constant, but this approach is typically beyond calculus.
Just for fun, here is how you get the limit with the E-M constant
$$\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}+...-\frac{1}{2n}  \\
=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2n}-2 (\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2n})\\
=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2n}- (\frac{1}{1}+\frac{1}{2}+...+\frac{1}{n})\\
=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2n}- \ln(2n)- (\frac{1}{1}+\frac{1}{2}+...+\frac{1}{n} -\ln(n))+\ln(2) 
$$
This converges to $\gamma-\gamma+\ln(2)$.
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