# Is there such a theorem about uniform convergence?

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

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

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.