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$$\lim_{N \to \infty} \frac{1}{N} \sum_{n=1}^{N} \frac{1}{n}$$

I'm working on a proof that, if I could show the previous expression evaluates to zero could be helpful, but I'm really doubtful about it. It must have been talked about before on this site I just haven't been able to find it.

I know that the sum of the harmonic series diverges and the limit of $\lim_{N \to \infty} \frac{1}{N}=0$, so it comes down to a zero times infinity situation.

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    $\begingroup$ Use the Stolz's theorem. The answer is $0$. $\endgroup$ Jan 21, 2020 at 0:12
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    $\begingroup$ In general, if $a_n\to L$ then $\frac{a_1+a_2+...+a_n}{n}\to L$. Your sequence is the sequence of means of $\frac{1}{n}$, so it converges to $0$. $\endgroup$
    – Mark
    Jan 21, 2020 at 0:15
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    $\begingroup$ One other approach is via the harmonic series asymptotics $\sum_{k=1}^n \frac{1}{k} = \log(n) + \gamma + \mathcal{O}(\frac{1}{n})$. $\endgroup$
    – Riley
    Jan 21, 2020 at 0:28
  • $\begingroup$ @MichaelRozenberg I am not seeing quite how to apply the Stolz-Cesàro theorem. It looks like we need an expression of this form: $\frac{a_{n+1}-a_n{b_{n+1}-b_n}$, and prove first that this limit exists. Is this correct? $\endgroup$ Jan 21, 2020 at 14:18
  • $\begingroup$ Yes, of course. Also, we need that $n$ would be increasing, which happens. $\endgroup$ Jan 21, 2020 at 14:27

4 Answers 4

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if you want to do this directly and without integration, consider

for an upper bound:
$\frac{1}{N} \sum_{n=1}^{N} \frac{1}{n} = \frac{1}{N}\cdot\big( \sum_{n=1}^{N} 1\cdot\frac{1}{n}\big)\leq \frac{1}{N}\cdot \big(N\big)^\frac{1}{2} \big(\sum_{n=1}^{N} \frac{1}{n^2}\big)^\frac{1}{2} = \frac{1}{\sqrt{N}}\cdot \big(\sum_{n=1}^{N} \frac{1}{n^2}\big)^\frac{1}{2} \leq \frac{C}{\sqrt{N}}$
for some constant $C$, where the first inequality is Cauchy-Schwarz.

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  • $\begingroup$ Wow this one really speaks to me. I knew when you brought that $1$ out you were up to something sneaky. $\endgroup$ Jan 21, 2020 at 3:38
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Hint:

You have a Riemann sum: $$\sum_{n=2}^N\frac 1n<\int_1^N\frac{\mathrm dt}t=\ln(N),\quad\text{ so}\quad\displaystyle\frac{1}{N} \sum_{n=1}^{N} \frac{1}{n}\le\frac1N+\frac{\ln(N)}{N}.$$

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HINT:

Find a simple upper bound of the integral of $\int_1^N \frac1x\,dx$ to show that $\sum_{n=1}^N\frac1n\le \log(N)+1$.

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  • $\begingroup$ I see, I guess the thing I am missing is the connection between the series and the integral. Do we just say that $\sum_{n=1}^N \frac{1}{n} < \int_1^N \frac{1}{x} dx$ as a general rule because the integral is smoother so it fills in the area between the steps left by the summation? $\endgroup$ Jan 21, 2020 at 0:49
  • $\begingroup$ No, we compare $1/x$ on an interval $[n,n+1]$ to its maximum or minimum there. Since $1/x$ is decreasing, this is easy. But this shows an upper bound for $\sum_{n=1}^N\frac{1}{n}$ is $\int_1^{N+1}\frac{1}{x}\;dx$. This still is plenty to solve the OP. $\endgroup$
    – GEdgar
    Jan 21, 2020 at 1:39
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    $\begingroup$ @gedgar $\log(N+1)$ is not an upper bound for $\sum_{n=1}^N \frac1n$. However, for a monotonically decreasing function $f$, we have the estimate (i.e., upper bound) $$\sum_{n=1}^N f(n)\le f(1) +\int_1^N f(x)\,dx$$ $\endgroup$
    – Mark Viola
    Jan 21, 2020 at 4:37
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If you are aware of harmonic numbers $$\sum_{n=1}^{N} \frac{1}{n}=H_N$$ For large $N$, we have $$H_N=\gamma +\log \left({N}\right)+\frac{1}{2N}+O\left(\frac{1}{N^2}\right)$$ $$\frac{1}{N} \sum_{n=1}^{N} \frac{1}{n}=\frac{H_N}{N}=\frac \gamma N+\frac {\log \left({N}\right)} N+\frac{1}{2N^2}+O\left(\frac{1}{N^3}\right)\sim \frac \gamma N$$ Just by curiosity, compute for $N=10$; the exact value is $\frac{7381}{25200}\approx 0.292897$ while the truncated expansion would give $\frac{\gamma }{10}+\frac{\log (10)}{10}+\frac{1}{200}\approx 0.292980$.

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