How to show the sequence $\frac{H_{n+1}}{n+1}$ decreases, where $H_n$ is partial sum of harmonic series.

Question

Few days back I asked this question If $a_n=b_n = \frac{(-1)^n}{n+1}$, how to deduce that the Cauchy product series converges? . I understand the proof. But I am struggling to prove the sequence is decreasing.

I tried to estimate $a_{n+1}/a_n$ and $a_n -a_{n+1}$. But I couldn't get anything. Can anyone explain how to prove using only real analysis approach.

Answer 1

$$ \frac{{H_{n + 1} }}{{n + 1}} - \frac{{H_n }}{n} = \frac{{nH_{n + 1} - (n + 1)H_n }}{{(n + 1)n}} = \frac{{\frac{n}{{n + 1}} + nH_n - (n + 1)H_n }}{{(n + 1)n}} = \frac{{\frac{n}{{n + 1}} - H_n }}{{(n + 1)n}} < 0 $$ since $$ \frac{n}{{n + 1}} < 1 \le H_n . $$

Answer 2

Alternative: $$ \frac{H_{n+1}}{n+1} = \int_{0}^{1} x^n(-\log(1-x))\,dx $$ by integration by parts. $-\log(1-x)$ is positive over $(0,1)$ and for a fixed $x\in(0,1)$ the term $x^n$ decreases as $n$ increases, so does the LHS.

This can be used to prove that $\log\left(\frac{H_{n+1}}{n+1}\right)$ is convex, too.