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