Solutions to $\sum{a_n} = \prod{(a_n+1)}$

Question

$$\sum_n^\infty{a_n} = \prod_n^\infty{(a_n+1)}$$

Can you give a nontrivial example of a real sequence which satisfies this equation? By "trivial" I mean sequences such as $-1,1,0,0,0\dots$ which result in the series and infinite product both being zero.

Answer 1

Rewrite: $$a_1+\sum_{n=2}^\infty a_n=(a_1+1)\prod_{n=2}^\infty (a_n+1)$$ Just set $$a_1=\frac{\prod_{n=2}^\infty (a_n+1)-\sum_{n=2}^\infty a_n}{1-\prod_{n=2}^\infty (a_n+1)}$$

This means you have an immensely infinite number of solutions. Almost any sequence for which the product exists is ok, if you just prefix it with a first term that satisfies the above condition.

The sequence $\{a_1,1,0,0,0,\ldots\}$ is just one example, but any other is just fine.

Answer 2

Let $$f(x) = \prod_{n=1}^\infty \left(1+\frac{x(-1)^n}{n^2}\right) - \sum_{n=1}^\infty \frac{x(-1)^n}{n^2}$$

then $f$ is continious with $f(0) = 1$ and $f(-3) < - 1$. By the intermediate value theorem there exist a $x\in(-3,0)$ s.t. $f(x) =0$. Numerically we find $x\approx −2.02467$.