Putnam 2016 B1 Short (and incorrect) "answer"

Question

I took the Putnam exam for the first time on December, and I would like to discuss my (incorrect) answer for B1, which states:

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

Below is what I did. Essentially, I defined a generating function and manipulated it. Unfortunately, I got the answer as $e$ instead of the correct answer $e-1$. I am pretty sure that I got the answer wrong because I didn't check whether my function was convergent or not.

After raising each side of the given equation and rearranging, we get $$-x_n=e^{x_{n+1}} -e^{x_n}.$$ Let $b_n = e^{x_n}$. Then $-x_n = b_{n+1} - b_n$. Let $f(x) = b_0 + b_1x+b_2x^2+...$ . Then $f(x)-xf(x)$ is $(1-x)f(x) = b_0 - x_0 - x_1 - x_2 - ...$

$b_0$ can be calculated to be equal to $e$. Let $x=1$. Then $0 = e -x_0-x_1-x_2-x_3...$

Moving everything except $e$ to the left side of the equality gives us the desired answer. (I thought!!!)

If anyone can prove that my function is divergent, that would be great.

Answer 1

Because $x_n\to 0$, $b_n\to 1$, so $b_0+b_1+b_2+\cdots\to +\infty$. Plugging in $x=1$ you get $0\cdot \infty$, so what you really need is a limit, $\lim\limits_{x\nearrow 1}(1-x)f(x)=1+\lim\limits_{x\nearrow 1}(1-x)\sum\limits_{n=0}^\infty(b_n-1)x^n=1$, which gives you the $1$ you're missing.

The last equation is true because $b_n-1\approx x_n$ and $\lim\limits_{n\to\infty}\frac{x_{n+1}}{x_n}=0$ so the series has radius of convergence greater than $1$. The previous equation was manipulating by adding and subtracting $1$ using the geometric series for $\dfrac{1}{1-x}$.

Answer 2

You can also use the fact that given $x_n=b_n-b_{n+1}$ the partial sums telescope $$\sum_{n=1}^kx_n=b_1-b_{k+1}$$ Since $x_n\to 0$ then $b_n\to 1$ and then you have that $$\lim _{k\to \infty} \sum_{n=1}^kx_n=b_1-1=e-1$$