6

More precisely, suppose $f(k)$ is increasing, differentiable, $f(1) \ge 1$, and $f'(n)/f(n) \le d/n$ for some $d>0$.

Then, for any $x > 1$, as $n \to \infty$, $$\dfrac{f(n)}{x^n}\sum_{k=1}^n \dfrac{x^k}{f(k)} \to \dfrac{x}{x-1}. $$

This is the most reasonable generalization I could find of my answer to Computing $\lim_n \frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k}$.

That question is the case $x=2, f(n) = n$.

I have what I think is a proof, which uses the same technique as my answer referenced above. But it a little messy, and my hope is that someone can come up with a nicer proof.

I will post my proof in a few days if no answer is given.

marty cohen
  • 107,799

2 Answers2

1

From the conditions $f(n)\le f(n+1)$ and $\frac{f'(n)}{f(n)}\le d/n$ we have

$$1\le \frac{f(n+1)}{f(n)}\le \left(1+\frac1n\right)^d \tag 1$$

Note that the right-hand side of the inequality in $(1)$ can be obtained by integrating the inequality $\int_n^{n+1}\left(\log(f(x)\right)'\,dx\le \int_n^{n+1}d\frac{1}{x}\,dx$.

Application of the squeeze theorem reveals

$$\lim_{n\to \infty}\frac{f(n+1)}{f(n)}=1 \tag 2$$

Alongside $(1)$, it is straightforward to show from the condition $f'(n)/f(n)\le d/n$ that

$$\lim_{n\to \infty}\frac{x^n}{f(n)}=\infty \tag 3$$

Note that $\int_1^n \left(\log(f(x)\right)'\,dx\le \int_1^{n}d\frac{1}{x}\,dx$ yields $f(n)\le f(1)\,n^d$ and $\frac{x^n}{f(n)}\ge \frac{x^n}{f(1)\,n^d}\to \infty$.

Using $(2)$ we find

$$\begin{align} \lim_{n\to \infty}\left(\frac{\sum_{k=1}^{n+1}\frac{x^k}{f(k)}-\sum_{k=1}^{n}\frac{x^k}{f(k)}}{\frac{x^{n+1}}{f(n+1)}-\frac{x^n}{f(n)}}\right)&=\lim_{n\to \infty}\left(\frac{\frac{x^{n+1}}{f(n+1)}}{\frac{x^{n+1}}{f(n+1)}-\frac{x^{n}}{f(n)}}\right)\\\\ &=\lim_{n\to \infty}\left(\frac{x}{x-\frac{f(n+1)}{f(n)}}\right)\\\\ &=\frac{x}{x-1}\tag 4 \end{align}$$

Finally, equipped with $(2)$, the Stolz-Cesaro Theorem is applicable from which we obtain the coveted limit

$$\lim_{n\to \infty}\frac{\sum_{k=1}^n\frac{x^k}{f(k)}}{\frac{x^n}{f(n)}}=\lim_{n\to \infty}\left(\frac{\sum_{k=1}^{n+1}\frac{x^k}{f(k)}-\sum_{k=1}^{n}\frac{x^k}{f(k)}}{\frac{x^{n+1}}{f(n+1)}-\frac{x^n}{f(n)}}\right)=\frac{x}{x-1}$$

as conjectured!

Mark Viola
  • 179,405
0

As in zhw's answer in the linked question we can also here apply Stolz–Cesàro to get

$$\lim_{n\to\infty} \frac{f(n)}{x^n}\sum_{k=1}^n\frac{x^k}{f(k)} = \lim_{n\to\infty} \frac{x}{x - \frac{f(n+1)}{f(n)}} = \frac{x}{x-1}$$

since $\lim_{n\to\infty}\frac{f(n+1)}{f(n)} = 1$. For this final limit we can for example apply Grönwall's inequality to $f'(x) \leq \frac{d}{x}f(x)$ giving us $1\leq \frac{f(n+1)}{f(n)} \leq \left(\frac{n+1}{n}\right)^d$ and the limit follows by the squeeze theorem. This can also be deduced by the mean value theorem applied to $g(x) = \log f(x)$ which gives $1 \leq \frac{f(n+1)}{f(n)} \leq e^{\frac{d}{n}}$.

Winther
  • 24,478