Theorem (Leibniz): If an alternating series:


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):


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


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

Thanks in advance!

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

1 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.


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