1
$\begingroup$

I tried to use Weierstrass M-test for checking if $$\sum_{k=2}^{\infty}\left(\cos\frac{x}{k}-\cos\frac{x}{k-1}\right)$$ converges uniformly on $(-\infty, \infty)$ and I got $\left|\cos\frac{x}{k}-\cos\frac{x}{k-1}\right|\leq\left|\cos\frac{x}{k}\right|+\left|\cos\frac{x}{k-1}\right|\leq1+1=2$,

$\sum_{k=2}^{\infty}2$ diverges $\Rightarrow \sum_{k=2}^{\infty}\left(\cos\frac{x}{k}-\cos\frac{x}{k-1}\right)$ doesn't converge uniformly on $(-\infty, \infty)$.

Is this correct?

$\endgroup$
2
  • $\begingroup$ You have shown that your series is less than a divergent series. That is not enough to show convergence $\endgroup$ – whpowell96 May 24 '20 at 21:03
  • $\begingroup$ This is a telescoping series. Use that to obtain the partial sums in closed form. This will tell you whether the series converges and whether the convergence is uniform for all $x$. $\endgroup$ – Hans Engler May 24 '20 at 21:05
1
$\begingroup$

Hint: since the series is telescoping, we can find an expression for the partial sums relatively easily

$$\sum_{k=2}^n \cos\left(\frac{x}{k}\right) - cos\left(\frac{x}{k-1}\right) = cos\left(\frac{x}{n}\right) - \cos x \to 1 - \cos x$$

With the formula and limit in hand, can you take it from here?

$\endgroup$
0

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.