Can you help me with finding x values where series is convergent and not convergent

Question

Given series:$$\sum_{n=1}^\infty (-1)^n\frac{x^n}{x^n+1}$$ for $x>0$

So what I want is the interval of $x$ where given series is convergent and the interval of $x$ where series given is not convergent.

I tried ratio test and the root test. But I could not manage to get through it.

Any explanation is highly appreciated. Thank you very much.

Answer 1

For $x \ge 1$ we have: $|(-1)^n\frac{x^n}{x^n+1}| \ge 1/2$, hence $((-1)^n\frac{x^n}{x^n+1})$ does not converge to $0$.

Conclusion ?

For $0<x<1$ we have $|(-1)^n\frac{x^n}{x^n+1}| \le x^n$.

Conclusion ?

Answer 2

Fix $x > 0$, and let

$$a_n = (-1)^n\frac{x^n}{x^n+1}$$

A necessary condition for convergence of the given series is that $$\lim_{n \to \infty} a_n = 0$$ If $x > 1$, $$\lim_{n \to \infty} |a_n| = \lim_{n \to \infty} \frac{x^n}{x^n+1} = \lim_{n \to \infty} 1-\frac{1}{x^n+1} = 1 - 0 = 1$$ so the series diverges.

If $x = 1$, then $\displaystyle{|a_n|= \frac{1^n}{1^n + 1} =\frac{1}{2}}$ (for all $n$), so the series diverges.

For $0 < x < 1$,

• The terms alternate in sign.
• $\displaystyle{\lim_{n \to \infty} |a_n|= \lim_{n \to \infty}\frac{x^n}{x^n+1} = \frac{0}{0+1} = 0}$, so $\displaystyle{\lim_{n \to\infty}a_n =0}$.
• $\displaystyle{|a_{n}| - |a_{n+1}| = \frac{x^n}{x^n+1} - \frac{x^{n+1}}{x^{n+1}+1} = \frac{x^n - x^{n+1}}{(x^n+1)(x^{n+1}+1)}} > 0$

hence, by the alternating series test, the series converges.

Answer 3

Let $a_n = \displaystyle \frac{x^n}{x^n+1} = \displaystyle \frac{1}{1+\frac{1}{x^n}}$, then it is clear that $a_n$ is clearly monotonically decreasing.

• If $|x| < 1$, then $\frac{1}{x^n} \rightarrow \infty$, leaving $a_n \rightarrow 0$, as required and so we satisfy the alternating series test. And have convergence there.
• It is clear that if $|x|> 1$, then that definitely isn't the case as $\frac{1}{x^n} \rightarrow 0$, leaving $a_n \rightarrow 1$ as $n$ gets large. The alternating series test doesn't show divergence, but what does show divergence is that $(-1)^n a_n$ does not tend to 0! So we cannot have convergence here.

So we have convergence $|x| < 1$ and divergence $|x|>1$. What about at $x=1$? Well you have to test that separately. I'll leave you to deal with that case.