0
$\begingroup$

My series is $$ \frac{x}{1+x}-\frac{x^2}{1+x^2}+ \frac{x^3}{1+x^3}-..... $$

Given: $0<x<1$

I see that my nth term is $(-1)^{n+1} (\frac{x^n}{1+x^n})$

My approach was to use Dirichlet's test. I see that $\frac1{1+x^n}$ is a monotone decreasing sequence converging to $0$.

So, I need to show that the partial sum sequences of the series $(-1)^{n+1} (x^n)$ is bounded. Is it true?

If not, how do I approach this problem?

$\endgroup$
  • $\begingroup$ I'm sorry, is the general term $\frac{x^k}{1+x^k}$? $\endgroup$ – Alex Aug 26 '14 at 15:29
  • $\begingroup$ Aren't you working with another alternating series again? $\endgroup$ – Quang Hoang Aug 26 '14 at 15:31
  • $\begingroup$ Yes, I've been working with a lot of infinite series, lol. I got the last one, I'm stuck with this one. $\endgroup$ – Diya Aug 26 '14 at 15:32
  • $\begingroup$ Learn to use LaTeX $\endgroup$ – Shahar Aug 26 '14 at 15:33
  • $\begingroup$ Just show the terms are going to zero and apply the alternating series test. $\endgroup$ – Ian Mateus Aug 26 '14 at 15:33
6
$\begingroup$

Your infinite series is absolutely convergent: $$ \left| \frac{x}{1+x}-\frac{x^2}{1+x^2}+ \frac{x^3}{1+x^3}-....\right|\leq\sum_{k=1}^{\infty}\left|\frac{x^k}{1+x^k}\right|\leq\sum_{k=1}^{\infty}x^k=\frac{x}{1-x}, \quad 0<x<1. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.