So, I have recently encountered a problem which has something to do with the limit $$\lim_{n\to \infty} \frac{1+ \frac12 + \frac13 + … + \frac1n}{\log n}$$
I know intuitively that it should be $1$ as the sum $1+ \frac12 + \frac13 + … + \frac1n$ can be looked as the area under the curve $y=\frac1x$ from $x=1$ to $x=n$ which is $\int_{1}^n\frac 1x dx = \log n$
But I cannot prove it rigorously. So,any help would be appreciated. Thanks!
Best Answer
The way to do it is by using the Squeeze theorem. Simply notice that we have the bound $$\log(n+1)=\int_1^{n+1}\frac1xdx\leq H_n\leq1+\int_1^{n+1}\frac1xdx=1+\log(n+1),$$ where $H_n=1+1/2+\dots+1/n$, and then $$1\leq \frac{H_{n}}{\log(n+1)}\leq1+\frac1{\log (n+1)}.$$ Adding $\frac{1/(n+1)}{\log(n+1)}$ throughout, we get $$ 1+\frac{1}{(n+1)\log(n+1)}\leq\frac{H_{n+1}}{\log(n+1)}\leq1+\frac{1}{(n+1)\log(n+1)}+\frac1{\log(n+1)}. $$ Since the denominators are unbounded, taking the limit $n\to\infty$ we see that $H_{n+1}/\log(n+1)$ is bounded by two expressions which both tend to $1$, so that it must tend to $1$ too.