My best guess for what the series should be:
[tex]\displaystyle\sum_{n=1}^\infty\frac{n^2x^n}{8\cdot16\cdot24\cdot\cdots\cdot8n}[/tex]
The denominator is the product of consecutive multiples of 8, and actually contains a factorial:
[tex]\displaystyle\prod_{k=1}^n8k=8^n\prod_{k=1}^nk=8^nn![/tex]
So the series is
[tex]\displaystyle\sum_{n=1}^\infty\frac{n^2}{n!}\left(\frac x8\right)^n[/tex]
From here, you can apply any of the applicable convergence tests. My personal go-to is the ratio test:
[tex]\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2}{(n+1)!}\left(\frac x8\right)^{n+1}}{\frac{n^2}{n!}\left(\frac x8\right)^n}\right|=\frac{|x|}8\lim_{n\to\infty}\frac{n+1}{n^2}=0[/tex]
0 < 1, so the series converges for all [tex]x[/tex].
By using the ratio convergence test, we will see that the sum converges for all values of x.
We assume that the series is:
[tex]\sum_{n = 1}^{\infty} (\frac{x}{8})^n*\frac{n^2}{n!}[/tex]
To take the convergence test, we can use the ratio method, this means that we take the quotient between the n-th term and the (n + 1)-th term in the limit where n tends to infinity, we will get:
[tex]r = \lim_{n \to \infty} |\frac{\frac{(n + 1)^2}{(n + 1)!} (x/8)^{n + 1}}{\frac{(n)^2}{(n)!} (x/8)^{n }} |\\\\r = \lim_{n \to \infty} | \frac{1}{n}* (x/8)| =0[/tex]
This means that, as n increases, the terms of the sequence are smaller and smaller. Thus, the sum converges. And you can see that it is independent of the value of x, so we conclude that the sum converges for all values of x.
If you want to learn more about sums, you can read:
https://brainly.com/question/6561461
