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

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?

• "If $|x|<1$ then $\frac{(2x)^n}{n}>\frac{(2x)^{n+1}}{n+1}$ " Are you sure? Mar 2, 2018 at 18:13
• @LordSharktheUnknown $|x|<1$, when the power increases the numerator decreases. Regarding the denominator it is sure that it is greater then 1 once $n\in\mathbb{N}$. Mar 2, 2018 at 18:19
• Try $x=0.9$ and $n=2$. Mar 2, 2018 at 18:21

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.