The condition of convergence for series $\sum_{n=0}^{\infty} x^2_n$

Question

I am reading some paper and the author stated as the following.

Let $\lambda$ be a positive number and choose a sequence $\left\{x_n\right\}_{n=0}^{\infty}$ defined by

$$x_0=1 \quad \text{and}\quad x_n = \frac{1}{\lambda^n} \prod_{j=0}^{n-1} \alpha_j \quad \left(n = 1, 2, \dots\right).$$

Consider $\sum c_n z^n$, where $c :=1$ and $c_n = \prod_{j=0}^{n-1}\alpha_j $ $\left( n= 0, 1, 2, \dots \right)$. If $\rho$ is the radius of convergence of this power series, then $\sum_{n=0}^{\infty}x_n^2$ will converge whenever $\frac{1}{\lambda} < \rho$, or $\lambda > \frac{1}{\rho}=:R$.

I do not understand why the can deduce the last condition

• then $\sum_{n=0}^{\infty}x_n^2$ will converge whenever $\frac{1}{\lambda} < \rho$, or $\lambda > \frac{1}{\rho}=:R$*

Could you please show me the reason? Thank you for your time.

Answer 1

Consider any (real or complex) power series $\Sigma c_nz^n$. In your question you have $c_n = \Pi_{j=0}^{n-1} \alpha_j$, but this is completely irrelevant (it yields $c_0 = 1$ which may be considered as the definition of a product with $0$ factors, but the following arguments work for any $c_0$).

You define $x_n = c_n/\lambda^n$ (with $c_0= 0$ this yields $x_0 = 1$, but again this is irrelevant). Now consider the power series

$$\Sigma c_n^2 u^n .$$

Its radius of convergence is $\rho' = 1/\limsup \sqrt[n]{\lvert c_n^2 \rvert} = 1/\limsup \sqrt[n]{\lvert c_n \rvert}^2 =1/(\limsup \sqrt[n]{\lvert c_n \rvert})^2 = \rho^2$.

Now let $u_\lambda = 1/\lambda^2$. Then $\Sigma c_n^2u_\lambda^n$ converges if $1/\lambda^2 < \rho^2$, i.e. if $1/\lambda < \rho$. But $\Sigma c_n^2u_\lambda^n = \Sigma x_n^2$ which yields the desired result.

Answer 2

$$\infty>\sum c_n \left(\frac{1}{\lambda}\right)^n =\sum \frac{1}{\lambda^n}\prod_{j=0}^{n-1} \alpha_j =\sum x_n $$ Thus the series $\sum x_n $ is absolutely convergent. But $x_n^2 \leq |x_n |$ for sufficiently large $n's$ thus $\sum x_n^2 $ is convergent