A little bit earlier I posted this question about the equation using the $\gamma$ Euler-Mascheroni constant. There I kinda understood what was going on with help of fellow MSE colleagues, so I decided to test it on the following limit.
$$\lim_{n\to\infty}\left(\frac{1}{2n+1}+\cdots+\frac{1}{9n}\right)$$
The mentioned equation is as follows:
$1+\frac{1}{2}+\cdots+\frac{1}{n}=\gamma+\mathcal{E}_n+\ln(n)\space,\ \mathcal{E}_n\longrightarrow 0\space\ when\space\ n\longrightarrow\infty$
By the same procedure as in the mentioned question,
$$1+\frac{1}{2}+\cdots+\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n}+\frac{1}{2n+1}+\cdots+\frac{1}{9n}-\left(1+\frac{1}{2}+\cdots+\frac{1}{2n}\right)= \\ =/H_n=\gamma+\mathcal{E}_n+\ln(n)/= \\ =\gamma+\mathcal{E}_{9n}+\ln(9n)-(\gamma+\mathcal{E}_{2n}+\ln(2n))=\mathcal{E}_{9n}-\mathcal{E}_{2n}+\ln\left(\frac{9}{2}\right)$$
$$\text{And thereby,}$$
$$\lim_{n \to \infty}\left(\frac{1}{2n+1}+\cdots+\frac{1}{9n}\right)=\lim_{n \to \infty}\underbrace{\mathcal{E}_{9n}}_{\rightarrow 0}-\underbrace{\mathcal{E}_{2n}}_{\rightarrow 0}+\ln\left(\frac{9}{2}\right)=\ln\left(\frac{9}{2}\right)$$
My question, is everything alright with this ?
I find the method very nice and furthermore interesting since I haven't been seeing it a lot around here.
Thanks
Best Answer
Yes, you are right.
Another way:
$$\lim_{n\rightarrow+\infty}\sum_{k=2n+1}^{9n}\frac{1}{k}=\lim_{n\rightarrow+\infty}\frac{1}{n}\sum_{k=1}^{7n}\frac{1}{2+\frac{k}{n}}=\int_0^7\frac{1}{2+x}dx=\ln4.5.$$