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?

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

1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$ 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 $x$. $\endgroup$ Aug 18, 2019 at 12:09
  • $\begingroup$ @BaroqueFreak See added details... $\endgroup$ Aug 18, 2019 at 12:20

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.