Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Does the following series converge or diverge? $$ \sum_{n\in\mathbb{Z}} \exp\left(-\sum_{k=-n}^{+n}\cos(k)\right) $$

share|cite|improve this question
What is $\mathbb{Z^2}$? – Ron Gordon Jan 19 '13 at 13:54
Use \exp instead of e^ when the exponent is big. – Ayman Hourieh Jan 19 '13 at 13:56
@rlgordonma I corrected may mistak. I play $\mathbb{Z}^2$ to $\mathbb{Z}$. It's the Ising model... – MathOverview Jan 19 '13 at 13:56
@Elihu: thanks for clarifying. – Ron Gordon Jan 19 '13 at 14:03
up vote 3 down vote accepted

Sure don't look convergent.

Don't look like it one bit.

To prove it concretely, notice that $-3$ is a lower bound for $-\cos{n} - \cot{\frac{1}{2}}\sin{n}$, so $$\sum_{n=-m}^me^{-3}\leq\sum_{n=-m}^me^{-\cos{n} - \cot{\frac{1}{2}}\sin{n}}=\sum_{n=-m}^me^{-\sum_{k=-n}^n\cos{k}}$$ which obviously diverges as $m\rightarrow \infty$.

share|cite|improve this answer
  • The term in the exponential is $1+2\sum_{j=1}^n\cos j$ as the cosine is even.
  • We can find a constant $M$ such that for all $n$, $\left|\sum_{j=1}^n\cos j\right|\leqslant M$, which can be seen computing the sum.
share|cite|improve this answer
Davide Giraudo, $\sum_{n\in\mathbb{Z}}\exp{[1+2\cdot M]}=\infty$ this no help me. – MathOverview Jan 19 '13 at 14:05
@אליהוצלע Your sum is estimated from below by the divergent $\sum_{n \in \mathbb{Z}} \exp[-(1+2M)] = \infty$. – Martin Jan 19 '13 at 14:10

Note that

$$ \sum_{k=-n}^{n} \cos{k} = \cos{n} + \cot{\left (\frac{1}{2} \right )} \sin{n} $$

Because of the oscillatory nature of this sum, i.e. each term is bounded, the sum in question does not converge.

share|cite|improve this answer
But the serie $\sum_{n=1}^\infty \exp{[-\sum_{k=1}^{\;+n}\left|\frac{\sin(k)}{k}\right|]}$ converge and $\left|\frac{\sin(k)}{k}\right|$ is oscillatory. This does not contradict their arguments? – MathOverview Jan 19 '13 at 14:13
@Elihu: that is different. I do not know what this sum evaluates to. Further, it may have an oscillatory term, but it is a uniformly increasing function of $n$. So the terms in the exponential are increasing rather than oscillating within a bound, as is the case for the stated problem. – Ron Gordon Jan 19 '13 at 14:18

Clearly your sum is equal to $e^{-1}+2\sum_{n=1}^\infty\exp\bigl(-1-2\sum_{k=1}^n\cos(k)\bigr)=e^{-1}+2\sum_{n=1}^\infty\exp\bigl(1-2\sum_{k=0}^n\cos(k)\bigr)$. On the other hand, $$\sum_{k=0}^n\cos(k)=\Re\biggl(\sum_{k=0}^ne^{ik}\biggr)=\Re\biggl(\frac{e^{(n+1)i}-1}{e^i-1}\,\frac{e^{-i}-1}{\overline{e^i-1}}\biggr)=\frac{\Re(e^{ni}-e^{(n+1)i}-e^{-i}+1)}{|e^i-1|^2}=\frac{\cos(n)-\cos(n+1)+1-\cos(1)}{2\bigr(1-\cos(1)\bigl)}$$ and so $$1-2\sum_{k=0}^n\cos(k)=\frac{\cos(n+1)-\cos(n)}{1-\cos(1)}\,.$$ Now consider $$b_n=\sqrt[n]{\exp\bigl(-1-2\sum_{k=1}^n\cos(k)\bigr)}=\exp\biggl(\frac1{1-\cos(1)}\,\frac{\cos(n+1)-\cos(n)}{n}\biggr)\,.$$ In order to show divergence of the original series it is suffices to show that $b_n\geq1$ for infinitely many $n$ (see here, page 370, Proposition 9.15), or, equivalently, $\cos(n+1)\geq\cos(n)$ for infinitely many $n$. But this is true: in fact, it is well-known that the set $\{n+2\pi m: n, m\in\mathbb Z\}$ is dense in $\mathbb R$, so the image of this set under cosine function is dense in $[-1,1]$. But $\cos(n+2\pi m)=\cos(|n|)$, which shows that the set $\bigl(\cos(n)\bigr)_{n\geq1}$ is dense in $[-1,1]$.

Since the function $\arccos:[-1,1]\to[0,\pi]$ is continuous, then there exists $\epsilon\in(0,1/2)$ such that $\pi-\arccos(x)<1/2$ for all $x$ with $-1\leq x<-1+\epsilon$. Moreover, for infinitely many $n$ we have $\cos(n)<-1+\epsilon$. If $n=2k\pi+\theta_n$, with $k\in\mathbb Z$ and $0\leq\theta_n<2\pi$, then $\cos(\theta_n)<-1+\epsilon$, so necessarily $\theta_n\in(\pi/2,3\pi/2)$. If $\theta_n\leq\pi$ then $\arccos\bigl(\cos(\theta_n)\bigr)=\theta_n>\pi-1/2$; otherwise we have $\arccos\bigl(\cos(\theta_n)\bigr)=2\pi-\theta_n>\pi-1/2$, that is $\theta_n<\pi+1/2$. Consequently $\pi-1/2<\theta_n<\pi+1/2$, which implies $\cos(\theta_n)=\cos(n)<\cos(\pi+1/2)$, and similarly $\pi+1/2<\theta_n+1<2\pi$, and since $\cos$ is increasing on the interval $[\pi+1/2,2\pi]$, it follows that $\cos(n+1)=\cos(\theta_n+1)>\cos(\pi+1/2)$. This proves that $\cos(n+1)>\cos(n)$ for infinitely many $n$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.