Prove that $\sum_{k=1}^n\frac{1}{k}-\ln(n)$ converge.

Question

I have to prove that the sequence $(x_n)_{n\geq 1}$ define by $$x_n=\sum_{k=1}^n\frac{1}{k}-\ln(n)$$ converge. I have shown that $(x_n)$ is decreasing but I fail to show that it's undervalued. I tried to show that it's undervalued by $0$, but with no success.

Answer 1

A simple trick is to exploit a telescopic product:

$$ \prod_{j=1}^{n}\left(1+\frac{1}{j}\right)=n+1\tag{2} $$ in order to have:

$$\begin{eqnarray*} H_n-\log(n) &=& H_n-\log(n+1)+\log\left(1+\frac{1}{n}\right)\\ &=& \sum_{j=1}^{n}\left(\frac{1}{j}-\log\left(1+\frac{1}{j}\right)\right)+O\left(\frac{1}{n}\right)\tag{2}\end{eqnarray*} $$ then the convexity of the function $f(x)=\frac{1}{x}$ over $\mathbb{R}^+$ gives: $$\frac{1}{j}-\log\left(1+\frac{1}{j}\right)=\frac{1}{j}-\int_{j}^{j+1}\frac{dx}{x}=\int_{0}^{1}\frac{x}{j(x+j)}\,dx\leq\frac{1}{2j^2}\tag{3}$$ hence $H_n-\log(n)$ converges to some constant less than $\frac{\pi^2}{12}$. By exploiting the inequality $\frac{1}{j}-\log\left(1+\frac{1}{j}\right)\geq \frac{1}{2j(j+1)}$ we also have that $\lim_{n\to +\infty}\left(H_n-\log n\right)\geq\frac{1}{2}.$

Answer 2

Not sure what you are allowed to use, so here are 2 ways:

1) $\log n = \int_{1}^{n} \frac{dx}{x}$, so you have a sum and corresponding integral, you can either use integral test or Euler-Maclaurin formula. 2) Use the asymptotic expansion of Harmonic series, you have $ \gamma + O(\frac{1}{n})$ left, which converges.

Answer 3

Hint

As you said $$x_n=\sum_{k=1}^n\frac{1}{k}-\ln(n)$$ is decreasing.

Exactly the same way you can prove that $$y_n=\sum_{k=1}^n\frac{1}{k}-\ln(n+1)$$ is increasing and $$y_n \leq x_n$$

These imply that $y_1$ is a lower bound for $x_n$ (and that $x_1$ is an upper bound for $y_n$).

Answer 4

To prove it elementarily, we may make use of the Leibniz's rule, which asserts that a sufficient condition for an alternating series $\sum_{n}a_{n}$ to converge is simply that $a_{n}$ is decreasing and $\to 0$.

Noting that for all $n \geq 1$ we have $$ \sum_{k=1}^{n}\frac{1}{k} - \log n = \big( 1 - \int_{x=1}^{2}\frac{1}{x}\big) + \big( \frac{1}{2} - \int_{x=2}^{3}\frac{1}{x}\big) + \cdots + \big( \frac{1}{n-1} - \int_{x=n-1}^{n}\frac{1}{x}\big) + \frac{1}{n}, $$ and that both $\frac{1}{k}, \int_{k-1}^{k}\frac{1}{x}$ are decreasing and $\to 0$, we have by Leibniz's rule the convergence of $\sum_{k=1}^{n}\frac{1}{k} - \log n$.