$\sum_{k=1}^n 1/k - \log n$

Question

I got this question : $$a_n = \sum_{k=1}^n \frac 1k - \log n$$ I proved that $\lim a_n $ exist. Now I have to prove: $$ 0<a_n-\lim a_n\le \frac 1n $$ for every $n \in \mathbb N$.

I tried induction,without any luck in the induction step. I think there is another way , maybe, with integrals.

Thanks for helping.

Answer 1

The given inequality is equivalent with $$1+\cdots +\frac{1}{n-1}-\ln n < \gamma$$ It now follows by remarking that $$b_n=1+\cdots +\frac{1}{n}-\ln (n+1)$$ (which has the same limit $\gamma$ as $a_n$) is increasing. This being easy to see from the inequality $$\ln (1+\frac{1}{n})\leq \frac{1}{n}$$

Answer 2

$\displaystyle (\lim a_n) -a_n = \sum_{k=n}^\infty a_{k+1}-a_k = \sum_{k=n}^\infty \frac{1}{k+1} - \ln\left(1-\frac{1}{k+1} \right)$

Since $\forall x> 0, x-\frac{x^2}2\leq \ln(1+x)< x$,

$$-\frac 12 \sum_{k=n}^\infty \frac{1}{(k+1)^2} \leq (\lim a_n) -a_n < 0$$

A standard integral estimate yields $\displaystyle \frac 12 \sum_{k=n}^\infty \frac{1}{(k+1)^2}\leq \frac 12 \int_{n-1}^\infty \frac 1{(t+1)^2} dt = \frac 1{2n}$, hence the result $$0<a_n-\lim a_n\le \frac 1n$$