1
$\begingroup$

If we have $\sum (-1)^nx_n$ and if $x_n>0$, and $\lim_{n\rightarrow +\infty }x_n=0$ and $x_n$ is a decreasing sequence then

$\left | \sum_{n=q+1}^{p} (-1)^nx_n\right |\leq x_{q+1}$

And if $\underset{x}{sup}(x_{q+1})\rightarrow0 $ then the series converges uniformly

$\endgroup$
  • 1
    $\begingroup$ the first statement is true. The second makes no sense, what do you mean by supremum over all $x$'s? $\endgroup$ – Yanko Mar 30 at 17:21
  • $\begingroup$ To prove uniform convergence, I took supremum of both sides, if the R.H.S tends to zero ,then we're finshed. $\endgroup$ – Pedro Alvarès Mar 30 at 17:52
  • $\begingroup$ Maybe I'm wrong, it should be $x_{q+1}$ that tends to zero ? $\endgroup$ – Pedro Alvarès Mar 30 at 17:53
  • $\begingroup$ It is given that $x_n$ goes to zero. $\endgroup$ – Yanko Mar 30 at 18:01
  • $\begingroup$ Oh okay, but the thing I didn't understad is why $\left | \sum_{n=q+1}^{p} (-1)^nx_n\right |\leq x_{q+1}$ $\endgroup$ – Pedro Alvarès Mar 30 at 18:33
2
$\begingroup$

Presumably $x_n$ is a function of $x$ in some domain $D$ as you are concerned with uniform convergence.

Since $x_n$ is a decreasing sequence, $x_{n} - x_{n+1} > 0$ and

$$\left | \sum_{n=q+1}^{p} (-1)^nx_n\right| = \left |(-1)^{q+1}(x_{q+1} - x_{q+2} + x_{q+3} - x_{q+4} + \ldots) \right| \\= x_{q+1} - (\, x_{q+2} - x_{q+3}\, ) - (\, x_{q+3} - x_{q+4}\, ) - \ldots \leqslant x_{q+1} \leqslant \sup_{x \in D} x_{q+1} $$

Thus, if the RHS tends to $0$ as $q \to \infty$ then uniform convergence follows as the uniform Cauchy criterion is satisfied.

$\endgroup$

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.