How to show that $\lim \frac{1}{n} \sum_{i=1}^n \frac{1}{i}=0 $? [duplicate]

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!

Solution 1:

$$\sum_{i=1}^n \frac{1}{i} = \sum_{1\le i\le\sqrt{n}} \frac{1}{i} + \sum_{\sqrt{n}<i\le n} \frac{1}{i} \le \sum_{1\le i\le\sqrt{n}} 1 + \sum_{\sqrt{n}<i\le n} \frac{1}{\sqrt{n}} \le \sqrt{n} + \sqrt{n} = 2\sqrt{n}.$$

Solution 2:

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).

Solution 3:

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.

Solution 4:

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

Solution 5:

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 $\ln(1)=0$.