Convergece of a sequence

Examine the convergence of the sequence:

$$\gamma_n=\left ( 1+a \right )\left ( 1+2a^2 \right )\left ( 1+3a^3 \right )\cdots\left ( 1+na^n \right)$$

for the different values of $a \in \mathbb{R}$.

Solution



$$\gamma_{n}=\displaystyle\mathop{\prod}\limits_{k=1}^{n}(1+k\,a^{k})\,,\quad n\in\mathbb{N}$$ We use that for a sequence of positive real numbers $b_n, \; n\in\mathbb{N}$,  the product $\prod \limits_{n=1}^{\infty}(1+b_n)$ converges if-f the sum $\sum \limits_{n=1}^{\infty} b_n$   converges.

$\color{gray}\blacksquare$  By Cauchy's root test we have that for $|a|<1$ the sum $\sum \limits_{n=1}^{\infty}|n\,a^{n}|$ converges. So, the same happens to the product $\prod_{n=1}^{\infty}(1+|n\,a^{n}|)$.  But

$$\mathop{\prod}\limits_{n=1}^{\infty}|{1+n\,a^{n}|}\leqslant\mathop{\prod}\limits_{n=1}^{\infty}(1+|n\,a^{n}|)$$ and the product $\prod \limits_{n=1}^{\infty}(1+n\,a^{n})$ converges absolutely.

$\color{gray}\blacksquare$ For $a<-1$ the $\displaystyle \lim_{n\to+\infty}{n\,a^{n}}$ does not exist. So $\sum \limits_{n=1}^{\infty}{n\,a^{n}}$diverges and, therefore, $\prod \limits_{n=1}^{\infty}(1+n\,a^{n})$ diverges. If $a=-1$ the sequence , obviously, converges as constant.

$\color{gray}\blacksquare$ For $a\geqslant 1$ , $ \displaystyle \lim_{n\to+\infty}{n\,a^{n}}=+\infty$. So $\sum \limits_{n=1}^{\infty}{n\,a^{n}}$ diverges and, therefore, $\prod \limits_{n=1}^{\infty}(1+n\,a^{n}) $ diverges -in fact tends to $+\infty$.