Putnam 2016 Part 3

Random number: 77

8th January 2017

So I think this is a solution for the first B problem:

B1. Let $x_0,x_1,x_2,...$ be the sequence such that $x_0 = 1$ and for $n \geq 0$, $$x_{n+1} = \text{ln} (e^{x_n} - x_n)$$ (as usual, the function ln is the natural logarithm). Show that the infinite series $$x_0 + x_1 + x_2 + ...$$ converges and find its sum.

So here's my spiffy solution. I don't want to go full hardcore on all the little real-analysis-ey details, but it should stand up to the test.

It is easy to show that $x_n$ is a decreasing sequence (by the monotonic increasing of the log function) and that it is bounded below by zero. Hence, there exists some $\delta \geq 0$ such that $\lim_{n \to \infty} x_n = \delta$. Suppose that this limit is greater than 0. Then: $$\forall \epsilon > 0, \exists N \in \mathbb{N}, \forall n > N : \delta \leq x_n \leq \delta + \epsilon$$ We know that $x_n \leq 1$, so we have for large enough $n$: $$\text{ln}(e^{\delta}-x_n) \leq \text{ln}(e^{x_n} - x_n) \leq \text{ln}(e^{\delta + \epsilon} - x_n)$$ But the middle part is equal to $x_{n+1}$, and the $x_n$ can be simplified with our first limit inequality, resulting in: $$x_{n+1} \leq \text{ln}(e^{\delta + \epsilon} - \delta)$$ But $\text{ln}(e^{\delta + \epsilon} - \delta)$ is a function in $\epsilon$ that is continuous at $\epsilon = 0$, hence we can find a small enough $\epsilon$ such that we have $$x_{n+1} \leq \text{ln}(e^{\delta} - \delta) < \delta$$ which proves that $\delta$ is not the limit of the strictly decreasing sequence $x_n$ if it is greater than 0, hence $\lim_{x \to \infty} x_n = 0$

We can then notice that, if we define $S_n$ as the partial sum from $x_0$ to $x_n$, that the relationship $$x_{n+1} = \text{ln}(e - S_n)$$ holds (proving it by easy induction). We then have the relationship that $$\lim_{n \to \infty} x_{n+1} = \lim_{n \to \infty} \text{ln} (e - S_n) = 0$$ Now due to the fact that ln is a strictly increasing function, and that $S_n$ is also strictly increasing, it is clear that $S_n$ must converge to $(e - 1)$.

I'm still not done with part A though; like I'm really sucking with the Group Theory question; it just feels like something I'll facepalm over if I see the answer.