Determine whether the series $\sum_{n=1}^\infty \ln\left(\cos \frac{1}{n}\right)$ is convergent.

Question

Attempt: Notice that the $p-$ series $\sum_{n=1}^\infty \frac{1}{n^2}$ with $p=2$ is convergent. Then, we have \begin{align*} \lim\limits_{n \to \infty} \frac{\ln\left(\cos \frac{1}{n} \right)}{\frac{1}{n^2}} &= -\frac{1}{2} \cdot \lim\limits_{n \to \infty} \frac{\sin \frac{1}{n}}{\frac{1}{n}} \cdot \frac{1}{\cos \frac{1}{n}} \\ &= -\frac{1}{2}. \end{align*} Hence, the limit is equal to $-\frac{1}{2} \ne 0$. Since the $p-$ series is convergent with $p=2$, by the limit comparison test, the series is convergent.

Am I true?

How do you justify $\lim\limits_{n \to \infty} \frac{\ln\left(\cos \frac{1}{n} \right)}{\frac{1}{n^2}} = -\frac{1}{2} \cdot \lim\limits_{n \to \infty} \frac{\sin \frac{1}{n}}{\frac{1}{n}} \cdot \frac{1}{\cos \frac{1}{n}} $? — mathcounterexamples.net, Mar 22 '21 at 13:28
Looks good. Maybe note that $\log\cos\frac 1n < 0$, so the test applies. — Milten, Mar 22 '21 at 13:28
What you have is good. I think Milten is saying that it’s worth pointing out that you don’t have any kind of oscillation so it’s valid to use the limit comparison test. — Clayton, Mar 22 '21 at 13:34
@user795084 The test applies by definition to series with positive terms only, but it could be extended to series with same sign terms. So what you've done is correct. — CHAMSI, Mar 22 '21 at 13:34
I do not think that if the limit appeared to be $0$ it would change the conclusion. — user, Mar 22 '21 at 13:44
@user: Yes, it is valid to use the comparison test. There is no reason to use the limit comparison test. — robjohn, Mar 22 '21 at 17:59

VIVID · Answer 1 · 2021-03-22T13:36:38.973

2

Alternative way using Taylor series: $$\ln\left(\cos \frac{1}{n}\right) = \ln\left(1-O\left(\frac{1}{n^2}\right) \right) = O\left(\frac{1}{n^2}\right)$$

edited Mar 22 '21 at 13:36

answered Mar 22 '21 at 13:33

VIVID

11,604

If by the comparison test, how can it approach with? – lap lapan Mar 22 '21 at 13:34
1

It should be $n^2$ and not $n^3$. – Gary Mar 22 '21 at 13:36
@Gary Thanks, fixed the typo. – VIVID Mar 22 '21 at 13:36

score 2 · Answer 2 · answered Mar 22 '21 at 13:54

For every $n\geqslant 1$, $\cos \dfrac{1}{n}$ gets closer and closer to $1$. That is,

$$\cos \dfrac{1}{n}\approx 1-\dfrac{1}{2n^2}+\dfrac{1}{24n^4}-\dfrac{1}{720n^6}$$ and for large enough $n$, the larger terms go to zero, making the approximation more accurate; as $n$ approaches $\infty$, our polynomial, and the cosine, go to $1$. Thus in the sum

$$\sum_{n\geqslant 1}\ln\left(\cos\dfrac{1}{n}\right)$$ the terms

$$\ln\left(\cos\dfrac{1}{n}\right)\longrightarrow 0,$$ since $\ln 1 = 0$. The approximate value of $$\ln\cos \dfrac{1}{n}\approx -\dfrac{1}{2n^2} - \dfrac{1}{12n^4} - \dfrac{1}{45n^6} - \dfrac{17 }{2520n^8}$$ and the terms go to zero for sufficiently large $n$. Therefore your series converges and is (for fun) approximately $$-\left[\frac{1}{2}\zeta(2)+\dfrac{1}{12}\zeta (4)+\dfrac{1}{45}\zeta(6)+\dfrac{17}{2520}\zeta(8)\right]$$ though this agrees only to two decimal places with the correct figure $-0.945369$.

score 1 · Answer 3 · answered Mar 22 '21 at 15:51

Just for the fun.

In the same spirit as @Alexander Conrad, we have the infinite series representation $$\log \left(\cos \left(\frac{1}{n}\right)\right)=\sum_{k=1}^\infty (-1)^k\frac{ 2^{2 k-3} (E_{2 k-1}(1)-E_{2 k-1}(0)) }{k (2 k-1)!}\,n^{-2 k}$$ where appear Euler polynomials.

Truncated to any order, we can transform the Taylor series into a $[2k,2]$ Padé approximant $P_k$.

For example $$P_2=-\frac {3(10n^2-1) } {4n^2(15n^2-4) }=-\frac{3}{16 n^2}-\frac{75}{16 \left(15 n^2-4\right)}$$

$$\sum_{n=1}^\infty P_2=-\frac{1}{128} \left(75+4 \pi ^2-10 \sqrt{15} \pi \cot \left(\frac{2 \pi}{\sqrt{15}}\right)\right)\sim -0.943375$$ Repeating for various values of $k$, we have more and more accurate values as shown below $$\left( \begin{array}{cc} 1 & -0.9295780173 \\ 2 & -0.9433749128 \\ 3 & -0.9449845778 \\ 4 & -0.9452801699 \\ 5 & -0.9453463832 \\ 6 & -0.9453628819 \\ 7 & -0.9453672911 \\ 8 & -0.9453685316 \\ 9 & -0.9453688948 \\ 10 & -0.9453690046 \end{array} \right)$$

robjohn · Answer 4 · 2021-03-24T19:27:52.720

Yes, your approach is valid.

More Fun with Numeric Approximations

Integrating the negative of the series for $\tan(x)$, derived in $(5)$ of this answer $$ \tan(x)=\sum_{k=1}^\infty\frac{(-1)^{k-1}B_{2k}\left(4^k-1\right)4^kx^{2k-1}}{(2k)!}\tag1 $$ where $B_{2k}$ are the Bernoulli Numbers, we get the Taylor series $$ \begin{align} \log(\cos(x)) &=\sum_{k=1}^\infty\frac{(-1)^kB_{2k}\left(4^k-1\right)4^kx^{2k}}{2k(2k)!}\tag{2a}\\ &=-\frac{x^2}2-\frac{x^4}{12}-\frac{x^6}{45}-\frac{17x^8}{2520}-\frac{31x^{10}}{14175}-\frac{691x^{12}}{935550}+O\!\left(x^{14}\right)\tag{2b} \end{align} $$ Applying the Euler-Maclaurin Sum Formula, we get that $$ \begin{align} \sum_{k=1}^n\log\left(\cos\left(\frac1k\right)\right) &=C+\frac1{2n}-\frac1{4n^2}+\frac1{9n^3}-\frac1{24n^4}+\frac7{450n^5}-\frac1{90n^6}\\ &+\frac{89}{8820n^7}-\frac{17}{5040n^8}-\frac{482}{127575n^9}-\frac{31}{28350 n^{10}}\\[6pt] &+\frac{46577}{3430350 n^{11}}-\frac{691}{1871100 n^{12}}+O\left(\frac1{n^{13}}\right)\tag3 \end{align} $$ The error term is bounded by $\frac1{20n^{13}}$, so for a given $n$, we should get about $p=13\log_{10}(n)+1$ places of precision: $$ \begin{array}{r|c|l} n&p&C\\\hline 10&14&-0.94536905472633\color{#AAA}{705334}\\ 100&27&-0.945369054726332935266095215\color{#AAA}{8385884}\\ 1000&40&-0.9453690547263329352660952154082701981170\color{#AAA}{39142468} \end{array} $$

Determine whether the series $\sum_{n=1}^\infty \ln\left(\cos \frac{1}{n}\right)$ is convergent.

4 Answers4