Monotonicity of $f(n)= \frac{1}{n} \sum_{i=1}^{n-1} \frac{1}{i}$ Define $f: \mathbb{N} \rightarrow \mathbb{R}$ as $f(n)= \frac{1}{n} \sum_{i=1}^{n-1} \frac{1}{i}$.
I was wondering how to tell if $f$ is a increasing or decreasing function?
Thanks and regards!
 A: A simpler proof would be to notice that $\displaystyle f(n) \gt \frac{1}{n}$ (for $n \gt 2$)
Thus $\displaystyle (n+1)f(n+1) - nf(n) = \frac{1}{n} \lt f(n)$ and so $f(n) \gt f(n+1)$.
A: The following 3 condidions are equivalent:
$$f(n)>f(n+1)$$
$$\frac{1+\frac12+\dots+\frac1{n-1}}n>\frac{1+\frac12+\dots+\frac1{n}}{n+1}$$
$$n+\frac{n+1}2+\dots+\frac{n+1}{n-1}+1>n+\frac n2+\dots+\frac n{n-1}+1$$
In the last inequality, the corresponding terms on the LHS are greater (or equal) as the corresponding terms on the RHS. (They both have the same number of terms.) At least one of these inequalities is strict. 
EDIT: (From the comments I see that this was not clear enough.)
There is the same number of terms, since I divided $n+1$ (obtained by multypling the first term in the second inequality) between $n$ and $1$ (the first and the last term in the LHS).
A: A formal proof would be
$$
\begin{align}
f(n+1) - f(n) &= \frac{1}{n+1} \sum_{i=1}^{n} \frac{1}{i} - \frac{1}{n} \sum_{i=1}^{n-1} \frac{1}{i}
\\
&= \frac{n}{n+1}\frac{1}{n} \left(\sum_{i=1}^{n-1} \frac{1}{i} + \frac{1}{n}\right) - \frac{1}{n} \sum_{i=1}^{n-1} \frac{1}{i}
\\
&= (\frac{n}{n+1} - 1)\frac{1}{n} \sum_{i=1}^{n-1} \frac{1}{i} + \frac{1}{n(n+1)}
\\
& = \frac{1}{n(n+1)} \left(1 - \sum_{i=1}^{n-1} \frac{1}{i}\right) < 0
\end{align}
$$
for $\forall n \geq 3$
A: Consider:
$$\begin{align*}
n(n+1)(f(n)-f(n+1))&=n(n+1)\left(\frac{1}{n}\sum_{i=1}^{n-1}\frac{1}{i}-\frac{1}{n+1}\sum_{i=1}^{n}\frac{1}{i}\right)\\
&=\sum_{i=1}^{n-1}\left(\frac{n+1}{i}-\frac{n}{i}\right) - 1\\
&=\sum_{i=1}^{n-1}\frac{1}{i} - 1\\
&\geq 0\end{align*}$$ for $n\geq2$.
In specific $f(n)-f(n+1)\geq0$ in general.
Edit: Little writing error.
