Curious property of monotonic functions If $f:\mathbb{R}\to\mathbb{R}$ is continuous and monotonically increasing on the interval $[1,\infty]$ with $f'(x)\leq\frac{1}{x}$ on the interval $[1,\infty]$ then is it true that:
$$\lim_{n \rightarrow \infty}  \frac{1}{nf(n)}\sum_{k=1}^n  f(k)=1$$
This is by no means a theorem also, its just a guess I made after experimenting with sums of the natural logarithm. It makes intuitive sense to me, because any monotonic function $f(x)$ with a derivative that isn't monotonic results in that function increasing as $x$ increases, but the rate at which it increases is decreasing, therefore in a sense its sort of approaching a constant value, ie 'increaseing at a very slow rate', making all terms slightly less then $f(x)$ , ie $f(x-1), f(x-2),\ldots$ etc, all very close in value to $f(x)$. Meaning the summands towards the end of the summation should all be very close in value, while the smaller ones like $f(1),f(2),\ldots$ etc are neglible. So athough clearly $\sum_{k=1}^n  f(k)<nf(n)$, 
 I still find it reasonable that
 $\lim_{n \rightarrow \infty}  \frac{1}{nf(n)}\sum_{k=1}^n  f(k)=1$
A disproof/proof of the theorem would be nice, but in addition some background intuition would also be greatly appreciated.
 A: If $f$ is non-decreasing,
$$
\sum_{k=1}^nf(k)\le nf(n)
$$
Since $0\le f'(x)\le1/x$, we have for $k\le n$
$$
f(n)-f(k)\le \log(n)-\log(k)
$$
Therefore, for $n\ge1$,
$$
\begin{align}
\sum_{k=1}^nf(k)
&\ge\sum_{k=1}^n{\large(}f(n)-\log(n)+\log(k){\large)}\\
&=nf(n)-n\log(n)+\sum_{k=1}^n\log(k)\\
&\ge nf(n)-n\log(n)+\int_1^n\log(x)\,\mathrm{d}x\\[6pt]
&=nf(n)-n\log(n)+n\log(n)-n+1\\[12pt]
&=nf(n)-n+1
\end{align}
$$
So we have for $n\ge1$,
$$
nf(n)-n+1\le\sum_{k=1}^nf(k)\le nf(n)
$$
Thus, if $\lim\limits_{n\to\infty}f(n)=\infty$, we have
$$
\sum_{k=1}^nf(k)\sim nf(n)
$$
However, if $\lim\limits_{n\to\infty}f(n)=L$, then for any $\epsilon>0$, there is an $N$ so that if $n\gt N$, then
$$
L-\epsilon\le f(n)\le L
$$
Then, for $n\gt N$,
$$
\begin{align}
\sum_{k=1}^nf(k)
&\ge\sum_{k=1}^Nf(k)+\sum_{k=N+1}^nL-\epsilon\\
&=\sum_{k=1}^Nf(k)+(n-N)(L-\epsilon)\\
&\ge\sum_{k=1}^Nf(k)+n(f(n)-\epsilon)-N(L-\epsilon)\\
&=n(f(n)-\epsilon)+\left(\sum_{k=1}^Nf(k)-N(L-\epsilon)\right)
\end{align}
$$
Since $\epsilon$ was arbitrary, we have, if $\lim\limits_{n\to\infty}f(n)=L$,
$$
\sum_{k=1}^nf(k)\sim nf(n)
$$
Faster growth:
Suppose that $f'(x)=1/x^\alpha$, where $\alpha<1$; e.g. $f(x)=\frac1{1-\alpha}x^{1-\alpha}$. Then the Euler-Maclaurin Sum Formula says
$$
\sum_{k=1}^nf(n)=\frac1{(1-\alpha)(2-\alpha)}n^{2-\alpha}+O\left(n^{1-\alpha}\right)
$$
which means that
$$
\frac1{nf(n)}\sum_{k=1}^nf(n)=\frac1{2-\alpha}+O\left(1/n\right)
$$
Therefore,
$$
\lim_{n\to\infty}\frac1{nf(n)}\sum_{k=1}^nf(n)=\frac1{2-\alpha}<1
$$
A: Your conjecture is true for monotonic $f:[1,\infty)\rightarrow[a,\infty)$ for any $a>0$ and in this case the limit equals $1$.  To prove this, I will show that $$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{\left(f(n)-f(k)\right)}{f(n)}=0.\ \ \ \ \ \ \ \ \ \ \ (1)$$ There are two cases, either $f$  is unbounded and $\lim_{n\rightarrow\infty}f(n)=\infty,$ or it is bounded and $\lim_{n\rightarrow\infty}f(n)=c.$ (Recall that bounded monotonic sequences converge) 
Case 1: The derivative condition tells us that for an integer $A$ 
$$f(n)-f(A)\leq\sum_{k=A}^{n}\frac{1}{k}\leq\log\left(\frac{n}{A}\right)+C$$ and so $$\sum_{k=1}^{n}f(n)-f(k)\leq\sum_{k=1}^{n}\left(\log\left(\frac{n}{k}\right)+C\right)=O(n).$$ Thus, $$\frac{1}{nf(n)}\sum_{k=1}^{n}\left(f(n)-f(k)\right)=O\left(\frac{1}{f(n)}\right),$$ and since $f(n)\rightarrow\infty,$ the limit is $0$. 
Case 2: 
When $f$ has a positive limit, equation $(1)$ is equivalent to $$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^{n}\left(f(n)-c\right)=0.$$ To prove this, let $\epsilon>0.$ Since $f$ has a limit, there exists $N(\epsilon)$ such that for all $n>N,$ $|f(n)-c|\leq\epsilon.$ This implies that $$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^{n}|f(n)-c|\leq\epsilon.$$ Since this holds for every $\epsilon>0,$ it follows that the limit equals $0$.
Remark: Note that the conjecture is not true for $f:[1,\infty)\rightarrow[0,\infty)$ as $f(x)=\frac{1}{x}$ provides a counter example.
