5

Prove that the series $$\sum_{n=2}^{\infty} \frac{\cos(n)}{n}$$ is conditionally convergent?

I tried to prove that it is not absolutely convergent series by trying to prove that $\sum_{n=2}^{\infty} \frac{\vert\cos(n)\vert}{n}$ is divergent, but I did not succeed. Is there another way?

Martin Sleziak
  • 53,687
Jasmine
  • 319
  • 2
    Hint: Dirichlet's test. – achille hui Jun 17 '16 at 02:30
  • I used Dirichlet's test to prove the convegence of $\sum_{n=2}^{\infty} \frac{\cos(n)}{n}$ now I need to prove divergence of $\sum_{n=2}^{\infty} \frac{\vert\cos(n)\vert}{n}$. – Jasmine Jun 17 '16 at 02:35
  • if you make a plot of $\max(|\cos(x)|,|\cos(x+1)|)$, you will notice it is always $\ge 0.4$. This means for any $m$, at least one of $|\cos(2m)|$ and $|\cos(2m+1)|$ is $\ge 0.4$. So the sums $\sum_{n=2}^\infty \frac{|\cos n|}{n}$ diverges as least logarithmically. – achille hui Jun 17 '16 at 02:44

3 Answers3

4

Dirichlet's test does the job pretty nicely. Let we prove that the partial sums of $\cos n$ are bounded: $$ \left|\sum_{n=1}^{N}\cos(n)\right| = \left|\frac{\sin\left(N+\frac{1}{2}\right)-\sin\left(\frac{1}{2}\right)}{2\sin\left(\frac{1}{2}\right)}\right|\leq\frac{1+\sin\frac{1}{2}}{2\sin\frac{1}{2}}<\frac{14}{9} $$ and we are done, since $\left\{\frac{1}{n}\right\}_{n\geq 1}$ is a sequence monotonically decreasing to zero.

We may even compute the value of the series: by summation by parts,

$$\sum_{n\geq 1}\frac{\cos n}{n}=\text{Re}\left(-\log(1-e^i)\right)=\color{red}{-\log\left(2\sin\frac{1}{2}\right)}=0.0420195\ldots $$

On the other hand, the series $\sum_{n\geq 1}\frac{\left|\cos n\right|}{n}$ is clearly divergent by a simple combinatorial lemma: given three consecutive integers $n,n+1,n+2$, for at least one of them the value of $\left|\cos(\cdot)\right|$ is greater than $\sin(1)$, hence:

$$ \sum_{n=1}^{3N}\frac{\left|\cos n\right|}{n}\geq \sum_{k=1}^{N}\frac{\sin(1)}{3k} $$ and the RHS is divergent as $N\to +\infty$.

Jack D'Aurizio
  • 353,855
  • Did you see my comment. I really need your help. If you don't feel like checking it out I totally understand. – Ahmed S. Attaalla Jun 17 '16 at 02:38
  • @AhmedS.Attaalla: your question actually asks for proving the conditional convergence, and my answer does. If you want a proof that the series is absolutely divergent, please suitably modify your question. – Jack D'Aurizio Jun 17 '16 at 02:41
  • Huh it's not this specific question. I'm asking if you may please check my answer to another question here: http://math.stackexchange.com/questions/1828884/prove-lim-a-to-infty-left-ln-a-sum-n-geq-1-fracannn-ri. Thanks. If you scroll down you will see it. Again, you don't have to but I feel strongly if you say it's correct then it's correct. – Ahmed S. Attaalla Jun 17 '16 at 02:46
  • 1
    @Jasmine: same old story: http://math.stackexchange.com/questions/297452/sines-and-cosines-of-angles-in-arithmetic-progression – Jack D'Aurizio Jun 17 '16 at 02:57
  • 1
    Jack, please, could you write in a detailed way how it is possible to get the sum of the series $\sum\limits_{n\geqslant1}\dfrac{\cos n}n;?;$ I am also interested in the proof of combinatorial lemma. – Angelo Dec 26 '23 at 10:15
  • Jack, please read my answer and tell me if my proof of combinatorial lemma is ok or if it can be improved in some way. How would you prove it? – Angelo Dec 26 '23 at 13:30
1

We can use Dirichlet's test to prove that $\sum_{n=1}^\infty \frac{\cos n}n$ is convergent. Note that $$ \sum_{n=1}^\infty \frac{|\cos n|}n > \sum_{n=1}^\infty \frac{\cos ^2n}n = \sum_{n=1}^\infty \frac 1{2n} + \sum_{n=1}^\infty \frac{\cos 2n}{2n}, $$ where $\sum_{n=1}^\infty \frac 1{2n}$ is obviously divergent and $\sum_{n=1}^\infty \frac{\cos 2n}{2n}$ is convergent by Dirichlet's test. Hence $\sum_{n=1}^\infty \frac{|\cos n|}n$ is divergent, implying $\sum_{n=1}^\infty \frac{\cos n}n$ is conditionally convergent.

John Smith
  • 39
-1

Too long for a comment.

Jack in his answer used the following combinatorial lemma, so I asked him to write the proof. Since he has not written it yet, I have tried to write one by myself.
If someone knows how to get a better and shorter proof, please give me a hint about it.

Combinatorial lemma: Given three consecutive integers $\,n\,,\,$ $\,n+1\,,\,$ $\,n+2\,,\,$ for at least one of them the value of $\,\left|\cos(\cdot)\right|\,$ is greater than $\,\sin1\,.$

Proof: Since $\,n\,$ is an integer which is a rational number, there does not exist any $\,k\in\Bbb Z\,$ such that $\,n\in\left\{1-\frac\pi2+k\pi,\,\frac\pi2-1+k\pi,\,\frac\pi2+k\pi\right\}.$
Moreover, for any $\,n\in\Bbb Z\,$ there exists $\,k\in\Bbb Z\,$ such that
$1-\frac\pi2+k\pi<n<\frac\pi2-1+k\pi\;$ or $\;\frac\pi2-1+k\pi<n<\frac\pi2+k\pi\;$ or $\;\frac\pi2+k\pi<n<1-\frac\pi2+(k+1)\pi\,.$

If $\;1-\frac\pi2+k\pi<n<\frac\pi2-1+k\pi\,,\;$ then $\,-\left(\frac\pi2-1\right)<n-k\pi<\left(\frac\pi2-1\right)\,,\;$ so $\,\cos(n-k\pi)>\cos\left(\frac\pi2-1\right)=\sin1\,,\;$ hence $\,|\cos n|=|\cos(n-k\pi)|>\sin1\,.$

If $\;\frac\pi2-1+k\pi<n<\frac\pi2+k\pi\,,\;$ then $\,(k+1)\pi-\left(\frac\pi2-1\right)<n+2<(k+1)\pi+\left(\frac\pi2-1\right)\,,\;$ so $\,-\left(\frac\pi2-1\right)<(k+1)\pi-(n+2)<\left(\frac\pi2-1\right)\,,\;$ therefore $\,\cos[(k+1)\pi-(n+2)]>\cos\left(\frac\pi2-1\right)=\sin1\,,\;$ hence $\,|\cos(n+2)|=|\cos[(k+1)\pi-(n+2)]|>\sin1\,.$

If $\;\frac\pi2+k\pi<n<1-\frac\pi2+(k+1)\pi\,,\;$ then $\,(k+1)\pi-\left(\frac\pi2-1\right)<n+1<(k+1)\pi+\left(\frac\pi2-1\right)\,,\;$ so $\,-\left(\frac\pi2-1\right)<(k+1)\pi-(n+1)<\left(\frac\pi2-1\right)\,,\;$ therefore $\,\cos[(k+1)\pi-(n+1)]>\cos\left(\frac\pi2-1\right)=\sin1\,,\;$ hence $\,|\cos(n+1)|=|\cos[(k+1)\pi-(n+1)]|>\sin1\,.$

Consequently, in any case, for any $\,n\in\Bbb Z\,,\,$ it results that $\,|\cos n|>\sin1\,$ or $\,|\cos(n+1)|>\sin1\,$ or $\,|\cos(n+2)|>\sin1\,.$

Angelo
  • 12,328