# Does the series $\sum_{n=1}^{\infty}\left( 1-\cos\big(\frac{1}{n} \big) \right)$ converge?

I'm having trouble determining whether the series:

$$\sum_{n=1}^{\infty}\left[1-\cos\left(1 \over n\right)\right]$$ converges.

I have tried the root test:

$$\lim_{n\rightarrow\infty}\sqrt[n]{1-\cos\frac{1}{n}}=\lim_{n\rightarrow\infty}\left(1-\cos\frac{1}{n}\right)^{1/n}=\lim_{n\rightarrow\infty}\mathrm{e}^{\frac{\log(1-\cos\frac{1}{n})}{n}}=\mathrm{e}^{\lim_{n\rightarrow\infty}\frac{\log(1-\cos\frac{1}{n})}{n}}$$

Now by applying the Stolz–Cesàro theorem, that upper limit is equal to:

\begin{align} \lim_{n\rightarrow\infty}\frac{\log(1-\cos\frac{1}{n+1})-\log(1-\cos\frac{1}{n})}{(n+1)-n}&=\lim_{n\rightarrow\infty}\left(\log(1-\cos\frac{1}{n+1})-\log(1-\cos\frac{1}{n})\right) \\&=\lim_{n\rightarrow\infty}\log{\frac{1-\cos{\frac{1}{n+1}}}{1-\cos{\frac{1}{n}}}} \end{align}

Now I'm totally stuck, unless that quotient is actually 1, in which case the limit would be 0, the Root test result would be $$\mathrm{e}^0=1$$ and all this would have been to no avail.

I'm not sure this method was the best idea, the series sure seems way simpler than that, so probably another method is more appropriate?

• The quotient is almost $1$, the root test is inconclusive. The best method is always to see whether you can get a grip on the size of the terms first. Here a Taylor expansion of $\cos$ immediately yields $$1 - \cos \frac1n = 1 - \left(1 - \frac{1}{2n^2} + \frac{1}{4!n^4} +\dotsc\right) = \frac{1}{2n^2} + O\left(\frac{1}{n^4}\right).$$ Jan 2, 2014 at 19:08
• Oh true that's pretty intuitive, wouldn't it be a substraction though? (I haven't worked with Landau notation yet so I'm not sure about it). Jan 2, 2014 at 19:25
• Sign doesn't matter in Landau notation. That's all absorbed in the big unknown constant. Note that also the $4!$ vanished in the $O(\cdot)$. Jan 2, 2014 at 19:33
• Ah ok thanks for the info Daniel ;) Jan 2, 2014 at 19:36
• The general term $\large\sim{1 \over 2n^{2}}$ when $\large n \gg 1$ like the Basel Problem. Jan 28, 2014 at 10:27

Note that $$0\le 1-\cos\frac{1}{n}=2\sin^2\frac{1}{2n}\le 2\cdot\left(\frac{1}{2n}\right)^2=\frac{1}{2n^2}.$$ We have used above that $$1-\cos (2x)=2\sin^2 x,$$ and also that $0 \le \sin x\le x$, whenever $x\in [0,\pi/2]$.

• Oh that yields the convergence pretty straightforwardly, but I'm unsure on how you found that expression for $1-\cos{\frac{1}{n}}$. Jan 2, 2014 at 19:02
• $1-\cos(2x)=2\sin^2 x$. Jan 2, 2014 at 19:04
• Oh I see now, that was pretty elegant. Thanks Yiorgos. Jan 2, 2014 at 19:13
• (+1) Your approach is the one I would have used. Jun 24, 2018 at 17:49

We can apply the Limit Test with $\rho =2$, $$\lim_{k \to \infty} k^2\left( 1 - \cos{\frac{1}{k}}\right) = \lim_{u \to 0}\frac{1 - \cos{u}}{u^2} = \frac{1}{2}$$ and thus the series converges absolutely.

• Wow that seems really simple, I hadn't heard about that test before. Searching for limit test doesn't yield anything interesting on Google, could you give me some info about it? Or maybe some other name that test is also known as? Jan 2, 2014 at 19:35
• Limit test for positive series: If the limit $\;\lim\frac{a_n}{b_n}\;$ exists and it is finite and non-zero, then the series $\;\sum a_n\;$ converges iff $\;\sum b_n\;$ converges. In this case, Ayesha is using $\;b_n=\frac1{n^2}\;$ Jan 2, 2014 at 19:38
• Oh it's that test, true I hadn't realized, thanks for your help DonAntonio. Also thank you Ayesha of course, the solution is pretty elegant as well, I love those. Jan 2, 2014 at 19:44
• The one I had in mind says that for $\sum u_k$, if $\lim_{k \to \infty}\,\,k^\rho u_k < \infty$ for any $\rho > 1$, then $\sum u_k$ converges absolutely. I read about in in a text from 1947, and it was called the Limit Test, so I apologize if that causes confusion - it follows easily from what DonAntonio mentioned. Jan 2, 2014 at 20:51
• What Ayesha is calling the "limit test" is a consequence of what is often called the "limit comparison test" in modern-day US calculus classrooms. Nov 17, 2022 at 19:33

A small addendum. It has already been pointed out that the series is clearly convergent, since $1-\cos\frac{1}{n}$ behaves like $\frac{C}{n^2}$ for large values of $n$, hence asymptotic comparison and the p-test are conclusive. In explicit terms, $$\sum_{n\geq 1}\left(1-\cos\tfrac{1}{n}\right)=2\sum_{n\geq 1}\sin^2\tfrac{1}{2n}$$ can be efficiently approximated through Bhaskara's $\cos y\approx \frac{\pi^2-4y^2}{\pi^2+y^2}$ and Poisson's summation formula, leading to $$S=\sum_{n\geq 1}\left(1-\cos\tfrac{1}{n}\right)\approx \frac{5}{e^2-1}$$ whose relative approximation error is already less than one part in two hundreds.
An exact representation is provided by $$S = \sum_{m\geq 1}\frac{(-1)^{m+1}}{(2m)!}\zeta(2m)=\sum_{m\geq 1}\frac{(-1)^{m+1}}{(2m)!(2m-1)!}\int_{0}^{+\infty}\frac{z^{2m-1}}{e^z-1}\,dz\\= -\int_{0}^{+\infty}\frac{\text{ber}_1(2\sqrt{z})+\text{bei}_1(2\sqrt{z})}{(e^z-1)\sqrt{2z}}\,dz$$ where $\text{ber}$ and $\text{bei}$ are Kelvin functions. The last integrand function is concentrated in a right neighbourhood of the origin, where it behaves like $\frac{1}{2}\exp\left(-\frac{x}{2}-\frac{x^2}{18}\right)$. The error of the resulting approximation $$S\approx \frac{3}{2} e^{9/8} \sqrt{\frac{\pi }{2}} \text{Erfc}\left[\frac{3}{2 \sqrt{2}}\right]$$ essentially has the same magnitude of the previous one.

I am a student too so I hope my answer is correct.

$$1-\cos(1/n)=\cos(0)-\cos(1/n)=-2\sin(1/2n)\sin(-1/2n)=2\sin^2(1/2n)$$
Now as $$\sin(1/2n)<1/2n \; \forall n \in \mathbb{N} \Rightarrow \sin^2(1/2n)<1/4n^2$$

Hence $$|\sum_{n=1}^{\infty }(1-\cos(1/n))| \leq \sum_{n=1}^{\infty }|(1-\cos(1/n))|=\sum_{n=1}^{\infty }|2\sin^2(1/2n)| \leq \sum_{n=1}^{\infty }|(2/4n^2)|< \infty$$. Because we all ready know that $$\sum_{n=1}^{\infty }1/n^2$$ converge

We know: $$\cos t \sim 1-\dfrac{t^2}{2}$$ Hence, for $$n \to +\infty$$ $$\sum_{n=1}^{+\infty}\left(1-\cos\left(\dfrac{1}{n}\right)\right)\sim \sum_{n=1}^{+\infty}\dfrac{1}{2n^2}=\frac{\pi^2}{12}$$ Thus it converges