# $\sum_{n=1}^\infty (1 - \cos \frac1n)^\alpha\log n$

How do I find for which values of $\alpha \in \mathbb{R}$ the sum converges?

$$\sum_{n=1}^\infty (1 - \cos \frac1n)^\alpha\log n$$

I have tried using the following techniques:

• Comparison test. Hard to compare with $\sum_{n=1}^\infty\frac1{n^\alpha}$ as there is no similarity.
• Limit comparison test. Comparing with $\sum_{n=1}^\infty \frac1{n^\alpha}$, I get no information, as the limit is zero and $\sum_{n=1}^\infty \frac1n$ diverges. (Also tried for small cases of $n$.)

The other convergence criteria do not seem to be handy in this particular exercise at all.

This should be very easy, what am I doing wrong?

Thanks!

• Do you see how to bound $\cos(1/n)$ in terms of $n^{-2}$? – Gerry Myerson Aug 9 '13 at 11:46
• @GerryMyerson, I do not. Maybe the same way Harald suggested in his post? – David Aug 9 '13 at 11:59

Hint: $$1-\cos\frac1n \sim \frac1{2n^2}$$ as $n\to\infty$. So you can compare your series with $$\sum_{n=1}^\infty\frac{\log n}{n^{2\alpha}}.$$ And you can compare that from one side with $$\sum_{n=1}^\infty\frac{1}{n^{2\alpha}},$$ and from the other with $$\sum_{n=1}^\infty\frac{1}{n^{2\alpha-\varepsilon}}$$ for any $\varepsilon>0$.

(Edited to include more detail after the discussion in the comments.)

Yet another edit, adding more detail: For the last bit, you need to know that $\log n<n^\varepsilon$ for large $n$, if $\varepsilon>0$. If you put $x=n^\varepsilon$, this can be rewritten as $\log x^{1/\varepsilon}<x$ for large $x$, which in turn simplifies to $\log x<\varepsilon x$ when $x$ is large enough. But that follows from $$\lim_{x\to\infty}\frac{\log x}{x}=0.$$ This very fundamental equality can easily be shown using l'Hôpital's rule, but it is so important you should know it by heart.

• Missing a $1/2$? – Gerry Myerson Aug 9 '13 at 11:52
• I got this; this comes from the fact that $\lim_{n \to\infty} \frac{1 - \cos\frac1n}{\frac1{n^2}} = \frac12$. I have noticed this, but comparing the sum with both $\sum_{n=1}^\infty \frac1{n^\alpha}$ and $\sum_{n=1}^\infty \frac1{n^{2\alpha}}$, I got nowhere. – David Aug 9 '13 at 11:58
• The point is, you can use this to compare your sum to $\sum n^{-2a}\log n$. Then the question is, can you tell which values of $a$ make $\sum n^{-2a}\log n$ converge? – Gerry Myerson Aug 9 '13 at 12:03
• That seems to be the problem. When I reduced the problem to the convergence of $\sum n^{-2\alpha}\log n$, it seemed to be improper form for appropriate discussion. I know that $\sum \frac1{n^\alpha}$ will converge $\forall \alpha > 1$. But I have to compensate for $\log n$ in the numerator. I could divide both the numerator and the denominator by $n$ getting $\sum \frac{\frac{\log n}n}{n^{2\alpha - 1}}$, but I could also divide by $\sqrt{n}$ getting different constraints for the values of $\alpha$ for which the series converges. – David Aug 9 '13 at 12:12
• Try thinking about it this way: $\log(n)$ is eventually smaller than any positive power of $n$. So, if $2\alpha>1$, write $2\alpha=1+\epsilon$, $\epsilon>0$, and note that $\log(n)$ is much smaller than $n^{\epsilon/2}$. Then, you have to consider the case when $2\alpha\leq 1$; what can you do comparison-wise with this? – Nick Peterson Aug 9 '13 at 12:42

In this answer, all the bounds will be explicit. There are no "for large enough $x$" type statements.

Since $\cos(2x) = \cos^2(x)-\sin^2(x) =1-2\sin^2(x)$, $1-\cos(x) =2\sin^2(x/2)$.

So $S(\alpha) =\sum_{n=1}^\infty (1 - \cos \frac1n)^\alpha\log n =\sum_{n=1}^\infty (2\sin^2\frac1{2n})^\alpha\log n =2^\alpha\sum_{n=1}^\infty (\sin\frac1{2n})^{2\alpha}\log n$.

There are two simple ways to bound $\sin(\frac1{2n})$.

First: For $0 \le x \le \pi/2$, $2x/pi \le \sin(x) \le x$, so $1/(\pi n) \le\sin(1/2n) \le 1/(2n)$.

Second: Since $x \ge \sin(x) \ge x-x^3/6$, for $n \ge 1$, $\sin(\frac1{2n}) \ge \frac1{2n}-(\frac1{2n})^3/6 = \frac1{2n}-\frac1{48n^3} \ge \frac1{n}(\frac1{2}-\frac1{48}) = \frac{23}{48n}$.

The first bound shows that $S(\alpha)$ is between $2^\alpha\sum_{n=1}^\infty (\frac1{\pi n})^{2\alpha}\log n =\left(\frac{2}{\pi^2}\right)^\alpha\sum_{n=1}^\infty \frac1{ n^{2\alpha}}\log n$ and $2^\alpha\sum_{n=1}^\infty (\frac1{2 n})^{2\alpha}\log n =\left(\frac{1}{2}\right)^\alpha\sum_{n=1}^\infty \frac1{ n^{2\alpha}}\log n$, so that $S(\alpha)$ converges exactly when $\sum_{n=1}^\infty \frac1{ n^{2\alpha}}\log n$ converges.

I will now show that, if $x > e^{2/\epsilon}(1/(\epsilon\ d))^{2/(\epsilon\ln 2)}$ then $\ln x/x^\epsilon < d$.

This is, of course, far from the best (or even a good) bound, but it will be good enough to show what we want and derive the convergence result for $S(\alpha)$.

Since $e^x > x^m/m!$ for any positive integer $m$, $x/e^x < m!/x^{m-1} \le m^{m-1}/x^{m-1} = (m/x)^{m-1}$.

Therefore, if $x > 2m$, $x/e^x < (1/2)^{m-1}$. Note that the "$2$" here is moderately arbitrary: it can be replaced by any value $> 1$. This is made explicit later on, but I will use $2$ for now.

Replacing $x$ by $\ln x$, if $x > e^{2m}$, $\ln x/x < (1/2)^{m-1}$.

Putting $x^c$ for $x$ (I write $c$ for $\epsilon$ since I'm lazy), for $c > 0$, if $x^c > e^{2m}$, $\ln x^c/x^c = c\ln x/x^c < (1/2)^{m-1}$.

Therefore, for any $c > 0$, if $x > e^{2m/c}$, $\ln x/x^c < (1/2)^{m-1}/c$.

To make $(1/2)^{m-1}/c < d$, for any $d > 0$, we want $2^{m-1} > \frac1{c\ d}$ or $m>1+\ln(\frac1{c\ d})/\ln 2$.

This gives $e^m > e^{1+\ln(1/(c\ d))/\ln 2} = e (1/(c\ d))^{1/\ln 2}$ so $e^{2m/c} > e^{2/c}(1/(c\ d))^{2/(c\ln 2)}$.

Replacing $c$ by $\epsilon$, we get this:

If $x > e^{2/\epsilon}(1/(\epsilon\ d))^{2/(\epsilon\ln 2)}$, then $\ln x/x^\epsilon < d$.

The following makes explicit what I said above about the "$2$" being arbitrary. It also simplifies things.

Choose any $r > 1$.

If $x > rm$, replacing $x$ by $\ln x$, if $x > e^{rm}$, $\ln x/x < (1/r)^{m-1}$.

To make $(1/r)^{m-1} < d$, where $1 > d > 0$, we want $r^{m-1} > 1/d$ or $m > 1+\ln(1/d)/\ln r$. Putting this in the bound for $x$, above, if $x > e^{r(1+\ln(1/d)/\ln r)}$, $\ln x/x < d$. Note that $e^{r(1+\ln(1/d)/\ln r)} =e^r (1/d)^{r/\ln r}$, so the bound for $x$ becomes $x > e^r (1/d)^{r/\ln r}$.

Letting $r = e$, this becomes if $x > (e/d)^e$, $\ln x/x < d$.

To get a bound on $x$ when $\ln x/x^c <d$, where $c > 0$, this becomes $\ln(x^c)/x^c =c\ln x/x^c < cd$. Putting $x^c$ for $x$, the bound is $x^c > e^r (1/d)^{r/\ln r}$ or $x > e^{r/c} (1/d)^{r/(c\ln r)}$. Letting $r = e$, the bound is $x > e^{e/c} (1/d)^{e/c} = (e/d)^{e/c}$.

• Good. Maybe you'll edit out the formatting typos --- I'd do it myself, but I'm worried I'd get the math wrong. – Gerry Myerson Aug 10 '13 at 0:49
• I'll do it. Thanks for the approval. OK, I fixed a missing $\$\$. I didn't see any others. Feel free to edit it. – marty cohen Aug 10 '13 at 1:37