4
$\begingroup$

Does the following series $$\sum_{n=1}^{\infty}\frac{\cos(n+x)}{n}$$ converge uniformly?

I know the series converges pointwise since $\sum_{n}\frac{\cos n}{n}$ and $\sum_{n}\frac{\sin n}{n}$ converge. From desmos, it seems the series converges to some sort of sine wave and is infinitely differentiable.

I have tried rewriting the series into $$\sum_{n=1}^{\infty}\frac{\cos n\cos x - \sin n\sin x}{n}$$ in order to use the Weierstrass M-Test. However, I'm not sure how to get a sequence of constants $C_{n}$ such that $$\sup_{x\in\mathbb{R}}\left|\frac{\cos n\cos x - \sin n\sin x}{n}\right|\leq C_{n}$$ and where $\sum_{n=1}^{\infty}C_{n}$ converges. I tried using the triangle inequality but this gives me something like $$\frac{|\cos n| + |\sin n|}{n}$$ This doesn't appear to help because it negates cancellation of positive and negative terms so my intuition tells me $\sum_{n=1}^{\infty}\frac{|\cos n| + |\sin n|}{n}$ would diverge as the harmonic series diverges. Is it possible to use the Weierstrass M-Test here to prove the series $\sum_{n=1}^{\infty}\frac{\cos(n+x)}{n}$ converges uniformly?

$\endgroup$
1
  • 6
    $\begingroup$ $cos(n+x)=cosn cosx-sinn sinx$ $\endgroup$
    – Korra
    Commented Aug 18, 2019 at 11:11

1 Answer 1

4
$\begingroup$

Yes, it converges uniformly. No, the M-test will not work - if you could find those $C_n$ that would show the series converges absolutely, which is not so. But:

Define $$s_n(x)=\sum_{j=1}^n\frac{\cos(j+x)}{j},$$ $$t_n=\sum_{j=1}^n\frac{\cos(j)}{j},$$ $$r_n=\sum_{j=1}^n\frac{\sin(j)}{j}.$$Then $$s_n(x)-s_m(x)=\cos(x)(t_n-t_m)-\sin(x)(r_n-r_m),$$so $$|s_n(x)-s_m(x)|\le|t_n-t_m|+|r_n-r_m|.$$Since $t_n-t_m\to0$ and $r_n-r_m\to0$ as $n,m\to\infty$ this shows that $s_n(x)-s_m(x)\to0$ uniformly; hence your series converges uniformly.

Details added, in reply to a comment: Let $\epsilon>0$. There exists $N$ so that $$|t_n-t_m|+|r_n-r_m|<\frac\epsilon2+\frac\epsilon2=\epsilon\quad(n,m>N).$$So the inequality above shows that $$|s_n(x)-s_m(x)|<\epsilon\quad(n,m>N),$$hence $$|s_n(x)-s(x)|=\lim_{m\to\infty}|s_n(x)-s_m(x)|\le\epsilon\quad(n>N),$$which says precisely that $s_n(x)\to s(x)$ uniformly.

$\endgroup$
3
  • $\begingroup$ I have seen this on more than one answer but do not understand - why does the $|s_{n}(x)-s_{m}(x)|\to0$ prove uniform convergence? How does this relate to the $\lVert s_{n}-s\rVert_{\infty}$, where $s$ is the infinite sum? Is this using the fact that any Cauchy sequence in $\mathbb{R}$ is convergent? $\endgroup$ Commented Aug 18, 2019 at 11:56
  • 1
    $\begingroup$ @BaroqueFreak Yes, if a sequence of functions $(f_n)$ is uniformly Cauchy then it is uniformly convergent. The fact that any Cauchy sequence converges shows there exists $f(x)$ such that $f_n(x)\to f(x)$. This means that for every $\epsilon>0$ there exists $N(x)$ such that $|f_n(x)-f(x)|<\epsilon$ for all $n>N(x)$. And now the proof that every Cauchy sequence converges shows that you can choose $N(x)=N$, independent of $x$. $\endgroup$ Commented Aug 18, 2019 at 12:09
  • $\begingroup$ @BaroqueFreak See added details... $\endgroup$ Commented Aug 18, 2019 at 12:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .