Sequence

[Tolaso J Kos]

Examine the convergence the sequence:

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

[Grigorios Kostakos]

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

\(\color{gray}\bullet\) By Cauchy's root test we have that for \(|a|<1\) the sum \(\sum_{n=1}^{\infty}{|{n\,a^{n}}|}\) converges. So, the same happens to the product \(\prod_{n=1}^{\infty}(1+|n\,a^{n}|)\,.\) But \[\displaystyle\mathop{\prod}\limits_{n=1}^{\infty}|{1+n\,a^{n}|}\leqslant\mathop{\prod}\limits_{n=1}^{\infty}(1+|n\,a^{n}|)\] and the product \(\prod_{n=1}^{\infty}(1+n\,a^{n})\) converges absolutely.

\(\color{gray}\bullet\) For \(a\leqslant-1\) the \(\lim_{n\to+\infty}{n\,a^{n}}\) does not exists. So \(\sum_{n=1}^{\infty}{n\,a^{n}}\) diverges and, therefore, \(\prod_{n=1}^{\infty}(1+n\,a^{n})\) diverges.

\(\color{gray}\bullet\) For \(a\geqslant1\), \(\lim_{n\to+\infty}{n\,a^{n}}=+\infty\). So \(\sum_{n=1}^{\infty}{n\,a^{n}}\) diverges and, therefore, \(\prod_{n=1}^{\infty}(1+n\,a^{n})\) diverges -in fact tends to \(+\infty\).