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Show that $$\lim \frac{1}{n} \sum_{i=1}^n \frac{1}{i} =0 $$

I've proved that this sequence converges (it is bounded and decreasing). NOW, I need to find a sequence that is bigger than this one and goes to zero. Maybe something using geometric serie of 1/2

Thanks in advance!

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marked as duplicate by BCLC, Arnaud D., Strants, trancelocation, Martin Sleziak May 15 '18 at 13:14

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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We can approximate a finite sum with a definite integral (see here). We obtain that $$ \log(n+1)=\int_1^{n+1}x^{-1}\mathrm dx\le\sum_{i=1}^n\frac1i\le1+\int_1^nx^{-1}\mathrm dx=1+\log n. $$ Now we need to show that $$ \lim_{n\to\infty}\frac{\log n}n=0. $$ This can be done by using l'Hôpital's rule (a more general statement is proved here).

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  • $\begingroup$ What is the reason to downvote my answer? $\endgroup$ – Cm7F7Bb Feb 21 '16 at 8:55
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$$\sum_{i=1}^n \frac{1}{i} = \sum_{1\le i\le\sqrt{n}} \frac{1}{i} + \sum_{\sqrt{n}<i\le n} \frac{1}{i} \le \sum_{1\le i\le\sqrt{n}} 1 + \sum_{\sqrt{n}<i\le n} \frac{1}{\sqrt{n}} \le \sqrt{n} + \sqrt{n} = 2\sqrt{n}.$$

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  • $\begingroup$ i need some help. how do you establish the inequality: $$\sum_{\sqrt{n}<i \le n} \frac{1}{\sqrt{n}} \le \sqrt{n} $$ ? do you mind explaining to me? $\endgroup$ – karhas Sep 22 '15 at 17:36
  • $\begingroup$ @karhas This is the sum of a constant. Just multiply by the number of terms. $\endgroup$ – Erick Wong Sep 22 '15 at 19:03
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In the same direction as user121270 and coolydudey60 in their comments $$ \frac{1}{n} \sum_{i=1}^n \frac{1}{i} =\frac{H_n}{n}$$ and for large values of $n$ $$\frac{H_n}{n}=\frac{\gamma +\log \left(n\right)}{n}+\frac{1}{2 n^2}-\frac{1}{12 n^3}+O\left(\left(\frac{1}{n}\right)^4\right)$$ and then what V.C. proposed in his answer $\frac{1+\log(n)}{n}$ seems to be very good.

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Using Stolz-Cesaro theorem we get $$\lim\limits_{n\to\infty} \frac{\sum_{i=1}^n \frac1i}n = \lim\limits_{n\to\infty} \frac{\frac1{n+1}}{(n+1)-n} = \lim\limits_{n\to\infty} \frac1{n+1} = 0.$$

You can consider this as a special case of a more general fact that if $\lim\limits_{n\to\infty} a_n=L$, then also $$\lim\limits_{n\to\infty} \frac{a_1+\dots+a_n}n=L.$$ (If a sequence is convergent, then the arithmetic means of the first $n$ element converge to the same limit.)

See for example this question (and other questions shown there among linked questions): Prove convergence of the sequence $(z_1+z_2+\cdots + z_n)/n$ of Cesaro means

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I just write an alternate approach, which may be useful:

Consider $\frac{1}{n}\sum_{i=1}^n\frac{1}{i}=\ln(a_n)$ then: $$1\leq a_n=[e^{\frac{1}{1}}e^{\frac{1}{2}}...e^{\frac{1}{n}}]^{\frac{1}{n}}\leq \frac{e^{\frac{1}{1}}+e^{\frac{1}{2}}+...+e^{\frac{1}{n}}}{n}$$

Since $\lim_{n\rightarrow\infty}e^{\frac{1}{n}}=1$, the right-hand-side go to $1$ as $n$ approach $\infty$ (Cesaro mean), then $\lim_{n\rightarrow\infty} a_n=1$, which implies your limit is $\ln(1)=0$.

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  • $\begingroup$ The justifying of the $e^{1/n}$ sum using Cesàro means is overkill. The question is exactly equivalent to the Cesàro mean of $1/n$ in the first place... $\endgroup$ – Erick Wong Aug 26 '14 at 21:21
  • $\begingroup$ sure why not :)) $\endgroup$ – Leaning Aug 27 '14 at 7:31
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With Stolz-Cesaro Theorem:

\begin{align} \color{#66f}{\large\lim_{n\ \to\ \infty}{1 \over n}\sum_{i\ =\ 1}^{n}{1 \over i}} =\lim_{n\ \to\ \infty} {\sum_{i\ =\ 1}^{n + 1}1/i - \sum_{i\ =\ 1}^{n}1/i \over \pars{n + 1} - n} =\lim_{n\ \to\ \infty}{1/\pars{n + 1} \over 1}=\lim_{n\ \to\ \infty}{1 \over n + 1}= \color{#66f}{\large 0} \end{align}

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It's much easier to prove that a Harmonic series is $O(\log n)$ and the result follows immediately

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Another more general way is the following theorem: Let's $a_n\ge0$ and $\sum_{n=1}^{\infty}a_n\lt \infty$ than $$ \lim_{N\rightarrow \infty}\frac{\sum_{n=1}^{N}na_n}{N}=0. $$ In this you just need to put $a_n=\frac{1}{n^2}$.

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