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$$\lim_{n\to\infty} \sum_{i=0}^n \frac{1}{i+n} $$ 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.

Abhijeet Saha
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3 Answers3

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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.

pre-kidney
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    The given result is not strong enough to derive the value. You need $H_n=\ln{n}+C+o(1)$. – Aphelli Jan 18 '20 at 16:52
  • Oh, you are right: in general $O(1)$ terms might not be of the form $C+o(1)$, for example something like $(-1)^n$, and this could ruin the cancellation. I will fix my answer. – pre-kidney Jan 18 '20 at 16:58
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Since it is about integrals you may quickly see that

$$\sum_{i=0}^n \frac{1}{i+n} = \sum_{i=0}^n \frac{1}{\frac in +1}\frac 1n \stackrel{n \to \infty}{\longrightarrow}\int_0^1\frac 1{x+1}dx= \log 2$$

trancelocation
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  • Very slick indeed although we've only done the Riemann sum definition of an integral; we haven't done the fundamental theorem of calculus yet so we only use power series but in evaluating our sums and take the limit for the integral but I guess this requires some actual integration, thanks for the nice solution. – Abhijeet Saha Jan 18 '20 at 06:01
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$$S_n=\sum_{i=0}^n \frac{1}{i+n}=H_{2 n}-H_{n-1}$$ 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 $$S_{10}\sim \log (2)+\frac{96799}{1280000}\approx 0.768771399$$

Claude Leibovici
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