The sequence is the solution of the difference equation...

$\displaystyle a_{n+1}= \alpha_{n}\ a_{n}$ (1)

... that can be written as...

$\displaystyle \Delta_{n}= a_{n+1}-a_{n}= (\alpha_{n}-1)\ a_{n}= \beta_{n}\ a_{n}$ (2)

As illustrated in...

http://www.mathhelpforum.com/math-he...-i-188482.html
... the solution of (2) is...

$\displaystyle a_{n}= a_{1}+\sum_{k=1}^{n-1} \Delta_{k}$ (3)

In Your case is $\displaystyle 0 \le \alpha_{n} \le 1+\frac{1}{n^{2}}$ and the solution is more clear if we examine two possible alternatives...

a) if we suppose that $\displaystyle 1 \le \alpha_{n} \le 1+\frac{1}{n^{2}}$, then applying (3) we obtain...

$\displaystyle \lim_{n \rightarrow \infty} |a_{n}| \le |a_{1}|\ \sum_{k=1}^{\infty} \frac{1}{k^{2}}$ (4)

... and the series in (4) converges...

b) if we suppose that $\displaystyle 0<\alpha_{n} \le h <1$, then is...

$\displaystyle \lim_{n \rightarrow \infty} a_{n} =0$ (5)