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I'm preparing for exam and I have to solve this problem.

Please, how do I prove that $\lim\limits_{n\to\infty}\frac{1}{n}\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)=0.$ Any Theorems to guide as well?

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marked as duplicate by Martin Sleziak, user99914, Namaste, Strants, rtybase May 18 '18 at 18:08

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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A quick way to show this is if you know that $(1+ 1/2 + ... + 1/n) \approx \log n$ then the limit is $\frac{\log n}{n} \rightarrow 0$.

But you don't actually need to know the asymtpotics of the sum to prove the limit is $0$, for example, observe that for any fixed $k \in \mathbb{N}$ $$ \lim_{n\rightarrow\infty} \frac{1}{n}(1+ 1/2 + ... + 1/n) = \lim_{n\rightarrow\infty}\frac{1}{n}(1+ ... 1/(k-1)) + \lim_{n\rightarrow\infty}\frac{1}{n}(1/k + ... 1/n) = 0 +\lim_{n\rightarrow\infty}\frac{n-k}{n}\frac{1}{n-k}(1/k + ... 1/n) = \lim_{n\rightarrow\infty}\frac{1}{n-k}(1/k + 1/(k+1) ... 1/n) \le \lim_{n\rightarrow\infty}\frac{1}{n-k}(1/k + 1/k ... 1/k) = \lim_{n\rightarrow\infty}\frac{1}{n-k}(\frac{n-k}{k}) = \frac{1}{k} $$ Thus the limit is $\le \frac{1}{k}$ for any integer $k$, and since it is clearly nonnegative, this implies it is $0$.

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This is the Cesaro limit of the sequence of reciprocals. As already the sequence itself converges, $\frac1n\to 0$, the Cesàro limit exists and has the same value.

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  • $\begingroup$ you mean $1+\frac{1}{2}+\cdots \frac{1}{n}\to some \quad finite \quad value$? $\endgroup$ – Frank Moses May 18 '18 at 6:49
  • $\begingroup$ That's not what he says... $\endgroup$ – user370967 May 18 '18 at 6:50
  • $\begingroup$ @FrankMoses See Cesàro summation. $\endgroup$ – user1551 May 18 '18 at 6:59
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Note that we can break up the inner sum into two parts, as follows:

\begin{align} 1+\frac12+\frac13+\cdots+\frac1n & = \left(1+\frac12+\frac13+\cdots+\frac{1}{\lfloor\sqrt{n}\rfloor}\right) + \left(\frac{1}{\lfloor\sqrt{n}\rfloor+1}+\frac{1}{\lfloor\sqrt{n}\rfloor+2}+\cdots+\frac1n\right) \\ & \leq (1+\sqrt{n})+\sqrt{n} \\ & = 1+2\sqrt{n} \end{align}

Therefore,

$$ \frac1n \left(1+\frac12+\frac13+\cdots+\frac1n\right) \leq \frac{1+2\sqrt{n}}{n} $$

can be made as small as desired, and the limit is $0$.

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Hint:

Except for the Cesaro-Cauchy theorem, can you prove the inequality: $$\frac{1+\frac{1}{2}+\dots+\frac{1}{n}}{n}\leq\frac{2}{\sqrt{n}},\ n\geq1?$$

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Let $n=2^k-1$ with $k\in\mathbb{N}$. Then we have:

$$\frac{1}{n}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+...+\frac{1}{n}\right)\leq\frac{1}{n}\left(1+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+...+\frac{1}{2^{k-1}}\right)\\ =\frac{1}{n}\left(1+1+1+...+1\right)\\ =\frac{k}{n}=\frac{k}{2^k-1}$$

Since $\lim_{n\rightarrow\infty}$ implies $\lim_{k\rightarrow\infty}$, and exponentials go to infinity much faster than any polynomial, we immediately see that

$$0\leq\lim_{n\rightarrow\infty}\frac{1}{n}\left(1+\frac{1}{2}+...+\frac{1}{n}\right)\leq\lim_{k\rightarrow\infty}\frac{k}{2^k-1}=0$$

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As an alternative to Stolz-Cesaro, by the harmonic series we have

$$\frac{1}{n}\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)\sim \frac{\log n}n\to 0$$

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Since $1/x$ is a decreasing function on $x\ge 1$, $\ln n=\int_1^n dx/x\in [\sum_{k=2}^{n+1}\frac{1}{k},\,\sum_{k=1}^n\frac{1}{k}]$. Asymptotically $\sum_{k=1}^n\frac{1}{k}\sim\ln n$, so your limit is $\lim_{n\to\infty}\frac{\ln n}{n}=\lim_{x\to\infty}xe^{-x}$. That this is $0$ follows from $\int_0^{\infty}xe^{-x}dx=1$ being finite, since $xe^{-x}$ decreases for $x>1$.

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