$\frac{2x}{1}-\frac{(2x)^2}{2}+\frac{(2x)^3}{3}-...$ Alembert (Ratio test) or Leibniz?

Question

Theorem (Leibniz): If an alternating series:

$u_1-u_2+u_3-u_4+...\:\:\:(u_n>0)$

if the terms are decreasing

$u_1>u_2>u_3>...>u_n...$ and if $\lim_{n\to\infty}u_n=0$, the series converge and its sum is positive and not superior to the first term.

Problem: Show that $\frac{2x}{1}-\frac{(2x)^2}{2}+\frac{(2x)^3}{3}-...$ converges.

I tried to solve the question using both the Leibniz Theorem and Alembert's(ratio criterion).

If $|x|<1$ then $\frac{(2x)^n}{n}>\frac{(2x)^{n+1}}{n+1}$ and $\lim_{n\to\infty}\frac{(2x)^n}{n}=\lim_{n\to\infty}\frac{(2x)^n\ln(2)}{1}=0$.

So the series converge in $|x|<1$

However using Alembert theorem (ratio test):

$\lim_{n\to\infty}|\frac{\frac{(2x)^{n+1}}{n+1}}{\frac{(2x)^{n}}{n}}|=\lim_{n\to\infty}|\frac{{n}}{n+1}||2x|=|2x|$

The series converge when $|2x|<1\iff|x|<\frac{1}{2}$.

Question:

Why are these two methods delivering different results regarding the convergence dominion? Which one is wrong? Why is that one wrong?

Thanks in advance!

Answer 1

The problem with your application of the test is that the sequence $(2x)^n / n$ is not a decreasing sequence in all cases. In particular, the ratio of terms is

$$\left|\frac{(2x)^{n + 1} / (n + 1)}{(2x)^n / n}\right| = \frac{n}{n + 1} \cdot 2|x|.$$

If $|x| > \frac 1 2$, then once $n$ is large enough (recalling that $\lim_{n \to \infty} \frac n {n + 1} = 1$), this ratio is strictly greater than $1$. As such, the series does not converge.

In fact, this suggests a simpler proof of divergence: If $|x| > \frac 1 2$, then the terms don't even tend to zero.