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

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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... –  Elias Jan 19 '13 at 13:56
    
@Elihu: thanks for clarifying. –  Ron Gordon Jan 19 '13 at 14:03
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4 Answers

up vote 2 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 $m\rightarrow \infty$.

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  • 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.
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Davide Giraudo, $\sum_{n\in\mathbb{Z}}\exp{[1+2\cdot M]}=\infty$ this no help me. –  Elias Jan 19 '13 at 14:05
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@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
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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.

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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? –  Elias 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
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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 $n$.

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