Solution check using the Euler-Mascheroni constant $\lim_{n\to\infty}\left(\frac{1}{2n+1}+\cdots+\frac{1}{9n}\right)$

Question

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

Answer 1

You are correct.

Next, as an exercise, prove that, using the $H_n$ approximation, if $a, b, c, d$ are integers with $0 < a < c$ then $\lim_{n \to \infty} \sum_{k=an+b}^{cn+d} \dfrac1{k} =\ln(\frac{c}{a}) $.

Then try with $a, b, c, d$ reals instead of integers.

Answer 2

Yes, you are right.

Another way:

$$\lim_{n\rightarrow+\infty}\sum_{k=2n+1}^{9n}\frac{1}{k}=\lim_{n\rightarrow+\infty}\frac{1}{n}\sum_{k=1}^{7n}\frac{1}{2+\frac{k}{n}}=\int_0^7\frac{1}{2+x}dx=\ln4.5.$$

Answer 3

Asm marty cihen suggested, harmonic numbers are interesting for the limit and more than.

You problem is $$S_n=\sum_{i=1}^{7n}\frac 1{2n+i}=H_{9 n}-H_{2 n}$$

For large $n$, remember that $$H_p=\left(\gamma +\log \left({p}\right)\right)+\frac{1}{2 p}-\frac{1}{12 p^2}+O\left(\frac{1}{p^3}\right)$$ If you just use the term in brackets, the limit is immediate.

If you want to go further, apply twice the formula to get $$S_n=(\log (9)-\log (2))-\frac{7}{36 n}+\frac{77}{3888 n^2}+O\left(\frac{1}{n^3}\right)$$

Try it for $n=5$. The exact value is $$S_5=\frac{13808926545210682009}{9419588158802421600}\approx 1.4659799 $$ while the trucated series gives $$S_5 \sim \log \left(\frac{9}{2}\right)-\frac{3703}{97200}\approx 1.4659807$$