Proving the convergence and find the limit of $S_n = \displaystyle \sum_{k=1}^{n} \displaystyle\Bigg(\frac{y + zx^{k+1}}{y + zx^k}\Bigg)^2 $

Question

Suppose $0 \leq x < 1$, and $y,z \in \mathbb{R}$ and let

$S_1 = \displaystyle\Bigg(\frac{y + zx^2}{y + zx}\Bigg)^2$ for $n = 1$,

$S_2 = \displaystyle\Bigg(\frac{y + zx^2}{y + zx}\Bigg)^2 + \displaystyle\Bigg(\frac{y + zx^3}{y + zx^2}\Bigg)^2$ for $n = 2$,

$S_3 = \displaystyle\Bigg(\frac{y + zx^2}{y + zx}\Bigg)^2 + \displaystyle\Bigg(\frac{y + zx^3}{y + zx^2}\Bigg)^2 + \displaystyle\Bigg(\frac{y + zx^4}{y + zx^3}\Bigg)^2$ for $n = 3$,

so the $n^{\text{th}}$ partial sum is can be defined as $S_n = \displaystyle \sum_{k=1}^{n} \displaystyle\Bigg(\frac{y + zx^{k+1}}{y + zx^k}\Bigg)^2 $

How can I prove the convergence (or not) of the defined series? If it is convergent, what is $\lim_{n \rightarrow \infty} S_n$ ?

Answer 1

If $x=y=0$ or $y=z=0$, none of the summands is even defined. So assume $y\ne 0$, or that $y=0$ and neither $z$ nor $x$ is $=0$.

• $y\ne 0$: Then as $|x|<1$, the $n$th summand tends to $\left(\frac{y+0}{y+0}\right)^2=1$ and thus the series diverges
• $y=0$, $x\ne 0$, $z\ne 0$: Then the $n$th summand is $x^2$ and again the series diverges.