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!
 A: 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 $\gamma-\gamma+\ln(2)$.
A: Please Read: This is an incorrect answer. I will, however, leave it up to show a common mistake that can be made in this type of problem.
$$\lim_{n \to \infty}\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{n+n}$$
Remember that $$\lim_{n \to \infty}\frac{1}{n} = 0$$
From this theorem, and by applying limit properties, you would get
$$\lim_{n \to \infty}\frac{1}{n+1}+\lim_{n \to \infty}\frac{1}{n+2}+\lim_{n \to \infty}\frac{1}{n+3}+...+\lim_{n \to \infty}\frac{1}{n+n}$$
$$0+0+0+0+...$$
$$0$$
What's the mistake? $0+0+0...$ is an indeterminate form
A: If you know about harmonic numbers
$$S_n=\sum_{i=1}^n \frac 1 {n+i}=H_{2 n}-H_n$$ For large values of $p$, we have
$$H_p=\gamma +\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12
   p^2}+O\left(\frac{1}{p^4}\right)$$ Apply it twice and simply to get
$$S_n=\log (2)-\frac{1}{4 n}+\frac{1}{16 n^2}+O\left(\frac{1}{p^4}\right)$$ which shows the limit and how it is approached.
Moreover, this gives a quite good approximation of the partial sums. For example
$$S_{10}=\frac{155685007}{232792560}\approx 0.6687714$$ while the above truncated expansion would give
$$S_{10}\sim \log(2)-\frac{39}{1600}\approx 0.6687722$$
