Convergence of the sequence $ a_{n}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}-\int_{1}^{n} \frac{1}{x} d x$

Question

Let $$ a_{n}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}-\int_{1}^{n} \frac{1}{x} d x$$ for all $n \in \mathbb{N} .$ Show that $\left(a_{n}\right)$ converges.

Actually $ a_{n}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$ is the Reimann upper sum of $f(x)=\frac{1}{x}$ on the interval $[1,n]$ relative to a partition such that each subinterval of unit length, But I couldn't prove the convergence.

Can we use the fact $\ln (n+1) \leq 1+\frac{1}{2}+\ldots+\frac{1}{n} \leq 1+\ln n$

Answer 1

Rearrange the integral as a telescopic sum and use the fact that $\ln (1+x) = x+\mathcal O(x^2)$: $$\begin{split} \left (\sum_{k=1}^n \frac 1 k\right) -\int_1^n\frac{dx}x &= 1 + \left (\sum_{k=2}^n \frac 1 k\right)-\ln n\\ &= 1 + \sum_{k=2}^n \left(\frac 1 k -\ln k +\ln (k-1) \right)\\ &= 1 + \sum_{k=2}^n \left(\frac 1 k +\ln \left(1 - \frac 1 k \right) \right)\\ &= 1 + \sum_{k=2}^n \mathcal O \left(\frac 1 {k^2} \right)\\ \end{split}$$ The latter sum converges.