Infinite summation of reciprocals with infinity in the denominator $$\lim_{n\to\infty} \sum_{i=0}^n \frac{1}{i+n} $$
A friend asked me to evaluate this as a fun exercise since he didn't know what it was. I tried and came up with an upper bound of 1 but I don't know where to go from there. We've just started integrals so that's where the idea comes from.
 A: Let $H_n=\sum_{i=1}^n\frac{1}{n}$ denote the partial sum of the harmonic series. Your limit equals
$$
\lim_{n\to\infty}H_{2n}-H_n=\log 2,
$$
since $H_n=\log n+C+o(1)$, and the constant term $C$ (whatever it is) cancels out.
A: Since it is about integrals you may quickly see that
$$\sum_{i=0}^n \frac{1}{i+n} = \sum_{i=0}^n \frac{1}{\frac in +1}\frac 1n \stackrel{n \to \infty}{\longrightarrow}\int_0^1\frac 1{x+1}dx= \log 2$$
A: $$S_n=\sum_{i=0}^n \frac{1}{i+n}=H_{2 n}-H_{n-1}$$ For large $p$, use the asymptotics
$$H_p=\gamma +\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12
   p^2}+\frac{1}{120 p^4}+O\left(\frac{1}{p^6}\right)$$
Apply it twice and continue with Taylor expansion to get
$$S_n=\log (2)+\frac{3}{4 n}+\frac{1}{16 n^2}-\frac{1}{128 n^4}+O\left(\frac{1}{n^6}\right)$$ which shows the limit and how it is approached.
Moreover, it is a good shortcut method for evaluation the partial sums. For example
$$S_{10}=\frac{178964263}{232792560}\approx 0.768771403$$ while the above truncated series gives
$$S_{10}\sim \log (2)+\frac{96799}{1280000}\approx 0.768771399$$
