Determine if this series : $\sum_{n=1}^\infty \frac{2n-1}{(\sqrt2)^n}$ converges or diverges.

Question

I need to determine if this series converge or diverge using the d'Alembert's ratio test

Using the ratio test I have that:

$\lim\limits_{n \to \infty} \frac{2(n+1)-1}{(\sqrt2)^{n+1}}\frac{(\sqrt2)^n}{2n-1}=\lim\limits_{n \to \infty} \frac{2(n+1)-1}{(\sqrt2)^{n}2n-1}$

How can I proceed from there? or if I was wrong tell me the mistake please.

Answer 1

\begin{align*} \dfrac{1}{\sqrt{2}}\cdot\dfrac{2n+1}{2n-1}\rightarrow\dfrac{1}{\sqrt{2}}\cdot 1=\dfrac{1}{\sqrt{2}}<1, \end{align*} so it is convergent by the ratio test.

Answer 2

$$\lim\limits_{n \to \infty} \frac{2(n+1)-1}{(\sqrt2)^{n+1}}\frac{(\sqrt2)^n}{2n-1}=\lim\limits_{n \to \infty} \frac{(2n+1)}{\sqrt2 (2n-1)} = \frac{1}{\sqrt 2}<1$$

Thus the series converges.