$1+1/2+...+1/n=\log n + c + o(1)$ as $n\to\infty$

Question

I'm trying to show the claim, without knowing integrals.

I start with:

$$\log \left(\frac{n+1}{n}\right)=\frac{1}{n}+\mathcal{o}\left(\frac{1}{n}\right)=\gamma(n)\frac{1}{n}$$

$$\log \left(\frac{n+1}{n}\right)=\log (n+1)-\log (n)=\sum_{k=1}^n\log \left(\frac{k+1}{k}\right)-\log (n)=\sum_{k=1}^n\gamma(k)\frac{1}{k}-\log(n)$$

where $\lim_{n\to\infty}\gamma(n)=1$. Equating the right sides, and writing $\gamma(n)$ as $1+\alpha(n)$, where $\lim_{n\to\infty}\alpha(n)=0$, we have:

$$\gamma(n)\frac{1}{n}=\sum_{k=1}^n\frac{1}{k}+\sum_{k=1}^n\frac{\alpha(k)}{k}-\log(n)$$

$$\sum_{k=1}^n\frac{1}{k}=\log(n)-\sum_{k=1}^n\frac{\alpha(k)}{k}+\gamma(n)\frac{1}{n}$$

Now it remains to show that $\sum_{k=1}^n\frac{\alpha(k)}{k}$ converges to complete the proof, but I can't. Any suggestions?

Thanks in advance.

Answer 1

We have that

$$\log \left(\frac{n+1}{n}\right)=\frac{1}{n}+\mathcal{O}\left(\frac{1}{n^2}\right)$$

then

$$\log \left(\frac{n+1}{n}\right)=\log (n+1)-\log (n)=\sum_{k=1}^n\log \left(\frac{k+1}{k}\right)-\log (n)=$$$$=\sum_{k=1}^n\frac{1}{k}+\sum_{k=1}^n\mathcal{O}\left(\frac{1}{k^2}\right)-\log(n)$$

Answer 2

This is a lengthy comment. One way to show that $S_n=-\log n +\sum_{j=1}^n \frac {1}{j}$ converges WITH integrals is

(i). $S_{n+1}-S_n=\frac {1}{n+1}+\log (1-\frac {1}{n+1})<0$ because $$|\log (1-\frac {1}{n+1})|=\int_{1-\frac {1}{n+1}}^1 \frac {1}{x}dx>$$ $$>\int_{1-\frac {1}{n+1}}^1 1\cdot dx=\frac {1}{n+1}.$$ (ii). Let $T_n=\sum_{j=1}^n \frac {1}{j}.$ Then $$S_{n+1}=\frac {1}{n+1}+T_n -\sum_{j=1}^n \int_j^{j+1}\frac {1}{x}dx >$$ $$>\frac {1}{n+1}+T_n-\sum_{j=1}^n\int_j^{j+1}\frac {1}{j}dx=$$ $$=\frac {1}{n+1}+T_n-\sum_{j=1}^n\frac {1}{j}=$$ $$=\frac {1}{n+1}+T_n-T_n>0.$$ So the sequence $(S_n)_{n\in \Bbb N}$ is decreasing but is bounded below by $0.$

The limit is called Euler's Constant and is commonly denoted by $\gamma $ (gamma). A faster convergence is obtained with $S'_n=T_n-\log (n+\frac {1}{2}).$