Putnam 2012 B4 generalization

 Let $c \gt 0$ be a real number and $a_n$ be a sequence such that $a_0 = 1$ and

$$a_{n+1} = a_n  + e^{-c a_n}$$

Show that

$$\lim_{n \to \infty} (ca_n - \log n) = \log c$$

($\log$ is $\log$ to base $e$)

The Putnam problem was with $c = 1$ and only asked for proof of existence of the limit.

Scroll down for a solution.

Consider $b_n = e^{a_n}$ then we get that $b_{0} = e$ and

$$b_{n+1} = b_n e^{1/(b_n)^c}$$

We can easily show that $a_n$ is unbounded (proof by contradiction) and so is $b_n$ and thus $\frac{1}{b_{n}^c} \to 0$.

The recurrence for $a_n$ gives us

$$b_{n+1}^c = b_{n}^c e^{c/(b_n)^c}$$

Expanding the $e^{\dots}$ part we get

$$b_{n+1}^c = b_{n}^c ( 1 + \frac{c}{b_{n}^c} + O\left(\frac{1}{b_{n}^{2c}}\right)) = b_{n}^c + c + O\left(\frac{1}{b_{n}^c}\right)$$

This telescopes to give us

$$b_{n}^c - b_{0}^c = nc + \sum_{k=0}^{n} O\left(\frac{1}{b_{k}^c}\right)$$

And so

$$\frac{b_{n}^c - b_{0}^c}{n} = c + \frac{\sum_{k=0}^{n} O\left(\frac{1}{b_{k}^c}\right)}{n}$$

Thus

$$\frac{b_{n}^c}{n} \to c$$

Taking logarithms gives the result.