Convergence of the series $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{x^n}$.

Question

Show that $$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{x^n}$$ converges for every $x>1$.

let $a(x)$ be the sum of the series. does $a$ continious at $x=2$? differentiable?

I guess the first part is with leibniz but I am not sure about it.

Answer 1

Hint:

Use the geometric series:

$$a(x) = \frac1{x}\sum_{n=0}^\infty \frac{(-1)^n}{x^n} = \frac{1}{x\left(1+\frac1x\right)} = \frac{1}{x+1}$$

Answer 2

Using root test $$\lim_{n\to\infty}\sqrt[n]{\left|\dfrac{(-1)^n}{x^n}\right|}=\dfrac{1}{|x|}<1$$ then the series is converge for $|x|>1$.

Answer 3

Hint:

What if you use the ratio test?$$\lim_{n\to\infty}|\dfrac{a_{n+1}}{a_n}|$$

Answer 4

Let's look at the partial sums, and let $y = -1/x$ so $-1 < y < 0$..

$\begin{array}\\ s_m(y) &=\sum_{n=1}^{m} (-1)^{n-1}(-y)^n\\ &=\sum_{n=1}^{m} (-1)^{n-1}(-1)^ny^n\\ &=-\sum_{n=1}^{m} y^n\\ &=-y\sum_{n=0}^{m-1} y^n\\ &=-y\dfrac{1-y^m}{1-y}\\ &=\dfrac{-y}{1-y}-\dfrac{-y^{m+1}}{1-y}\\ \end{array} $

so

$\begin{array}\\ s_m(y)+\dfrac{y}{1-y} &=\dfrac{y^{m+1}}{1-y}\\ \end{array} $

Therefore $\sum_{n=1}^{m} \frac{(-1)^{n-1}}{x^n}+\dfrac{-1/x}{1+1/x} =\dfrac{1}{(-x)^{m+1}(1+1/x)} $ or $\sum_{n=1}^{m} \frac{(-1)^{n-1}}{x^n}-\dfrac{1}{x+1} =\dfrac{(-1)^{m+1}}{x^{m}(x+1)} $.

What is needed now is to show that $\lim_{m \to \infty} \dfrac{1}{x^{m}(x+1)} =0$.

(This is from "What is Mathematics")

Since $x > 1$, $x = 1+z$ where $z > 0$.

By Bernoulli's inequality, $x^m =(1+z)^m \ge 1+mz \gt mz =m(x-1)$, so $ \dfrac{1}{x^{m}(x+1)} \lt \dfrac{1}{m(x-1)(x+1)} =\dfrac{1}{m(x^2-1)} $, so to make $\dfrac{1}{x^{m}(x+1)} \lt \epsilon $ it is enough to take $m \gt \dfrac1{\epsilon(x^2-1)} $.

This is certainly not the best $m$, but it is completely elementary.