What is $\sum_{n=1}^{\infty}a_1 a_2...a_n$?

Solution 1:

Let's denote $l(x)=\ln\big((e^x-1)/x\big)$, with $l(0)=0$ understood as a limit. Clearly, $\{a_n\}$ is decreasing, hence $\lim\limits_{n\to\infty}a_n$ exists and is equal to $0$, the only root of $a=l(a)$. This, together with $l(x)\sim x/2$ for small $x$, implies the convergence of the sum; let it be $f(x)$, where $x=a_1$. The convergence is uniform w.r.t. $x$ in any finite range (since $l(x)$ is increasing), hence $f(x)$ is continuous. From the definition, we have $f(x)=x\big[1+f\big(l(x)\big)\big]$; substituting $f(x)=e^x-1+g(x)$, we obtain $g(x)=xg\big(l(x)\big)$. By induction, $$g(a_1)=a_1 g(a_2)=\ldots=a_1 a_2\cdots a_ng(a_{n+1}),$$ and since the RHS tends to $0$ when $n\to\infty$, we have $g(x)=0$ identically.

Finally, the answer is $e^{a_1}-1$.