Convergence of $\prod (1+ta_n)$ implies convergence of $\sum a_n$ and $\sum a_n^2$ 
Let $a_n$ be a sequence of real numbers and assume that $\prod _n(1+ta_n)$ converges for two non-zero values of $t$, say $t_1, t_2\in \mathbb R\setminus \{0, -1/a_1, \ldots, -1/a_i, \ldots \}$.
Prove that $\sum a_n$ and $\sum a_n^2$ converges.

I'm absolutely stuck on this problem. I can't use any of the usual convergence criteria because there's no positivity assumption on the $a_n$.
I'd appreciate some hints.
 A: By "the product converges", it is meant that it converges in $\mathbb{R}\setminus \{0\}$, otherwise the assertion is false, consider $a_n = \frac{1}{n+2}$ and $t_1 = -1,\, t_2 = -2$. Or it may be allowed that $1 + t a_n = 0$ for finitely many $n$, and the product formed by the nonzero terms converges in $\mathbb{R}\setminus \{0\}$. With the latter viewpoint - which is for example the standard interpretation in complex analysis, where infinite products of holomorphic functions play a considerable role - we don't need to exclude the values of $-1/a_n$.
A necessary condition for convergence of the product when $t \neq 0$ is that $a_n \to 0$, so by dropping finitely many terms from the product - which affects neither convergence of the product nor of the sums - we may assume that $\lvert t_k a_n\rvert \leqslant \varepsilon$ for all $n$ and $k \in \{1,2\}$, given any $\varepsilon \in (0,1)$ that we choose.
The convergence of the product $\prod (1 + t a_n)$ is then equivalent to the convergence of the series
$$\sum_{n = 1}^{\infty} \log (1 + t a_n).\tag{1}$$
By our assumptions, the series
$$\sum_{n = 1}^{\infty} \bigl[ t_1 \log (1 + t_2 a_n) - t_2 \log (1 + t_1 a_n)\bigr]\tag{2}$$
converges. Also, for $\lvert x\rvert \leqslant \varepsilon$, we have
$$\sum_{m = 3}^{\infty} \frac{\lvert x\rvert^m}{m} \leqslant \frac{\lvert x\rvert^3}{3}\sum_{m = 0}^{\infty} \lvert x\rvert^m = \frac{\lvert x\rvert^3}{3(1 - \lvert x\rvert)} \leqslant \frac{\varepsilon}{3(1 - \varepsilon)} x^2$$
and therefore
$$x - \biggl(\frac{1}{2} + \frac{\varepsilon}{3(1-\varepsilon)}\biggr) x^2 \leqslant \log (1 + x) \leqslant x - \biggl(\frac{1}{2} - \frac{\varepsilon}{3(1-\varepsilon)}\biggr)x^2.\tag{3}$$
Choosing $\varepsilon$ appropriately, $(2)$ and $(3)$ yield the convergence of $\sum a_n^2$, then $(1)$ and $(3)$ yield the convergence of $\sum a_n$.
