For which $a \in \mathbb{R^+}$ does the sequence $$\gamma_n=(1+a)(1+2a^2)\cdots(1+na^n)$$ converge? Give a brief explanation.
My attempt:
Is easy to see that
- $a=0 \implies \gamma_n \equiv1$;
- $a \ge 1 \implies$ $\gamma_n$ diverges.
Convergence: if $0<a<1$, $\gamma_n$ converges $\iff$ $\sum_{k=0}^{\infty}\ln(1+ka^k)$ converges.
$$\sum_{k=0}^\infty \ln(1+ka^k) \le \sum_{k=0}^\infty ka^k=\frac{a}{(1-a)^2}<+\infty$$ so $\gamma_n(a)$ converges in $[0,1) \space \blacksquare$
Is the solution correct? If yes, Is there a briefer explanation?