Just going down the list, the following tests don't work (or I failed at using them correctly) because:

  • $\lim \limits_{k\to\infty}a_k\neq0$ — The limit is hard to evaluate.
  • $\lim \limits_{k\to\infty}\left|\dfrac{a_{k+1}}{a_k}\right|>1$—Limit does not converge; inconclusive.
  • $\lim \limits_{k\to\infty}\sqrt[k]{|a_k|}>1$—Limit does not converge; inconclusive.
  • $\int \limits_{1}^{\infty}a_k\,dk$—How do I even do this.
  • There exists a $|b_k|\geq|a_k|$ and $\sum b_k$ converges—Cannot think of any such a $b_k$.
  • $\lim \limits_{k\to\infty}\cfrac{a_k}{b_k}$ exists and $\sum b_k$ converges — Cannot think of any such $b_k$.

Can I have some help?


We will show that there is no $\delta$ or corresponding $\epsilon$ so that $k>\delta$ implies$$\left|\left(\frac{1+\sin(k)}{2}\right)^k-\lim_{n\to\infty}\left(\frac{1+\sin(n)}{2}\right)^n\right|<\epsilon$$

Let the first term be denoted $a_k$ and the second $L$ (for limit). Pick the the smaller number of $|a_{\frac{3\pi}2}-L|$ or $|a_{\frac\pi2}-L|$. Whichever was chosen (denote choice as $\epsilon$), it is clear that setting $k:=k+\pi$, which is greater than $\delta$, will produce a result such that $|a_k-L|\geq\epsilon$.

Therefore, the limit does not exist, and the sum diverges.


The proof above is no good. More help pls. I can't figure this one out.

  • $\begingroup$ Your edit doesn't make sense - $0\lt\frac{1}{(k+2)^2}\lt 1$ for all $k$, but that series converges. $\endgroup$ – Steven Stadnicki Jun 10 '14 at 0:05
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    $\begingroup$ @user156190 Before you can say that it must diverge, you have to show the statement in your edit (that the limit doesn't exist, and more specifically that it's bounded away from zero infinitely often); that proof is (as my answer shows) pretty non-trivial in its own right. $\endgroup$ – Steven Stadnicki Jun 10 '14 at 0:35
  • $\begingroup$ Is this a better solution? I don't know what to do about the situation where $L=\frac12$ though, since no matter how large I increase $k>\delta$, I'll always wind up with $|a_k-L|<\epsilon$. $\endgroup$ – user156190 Jun 10 '14 at 0:50
  • $\begingroup$ You can't set $k=k+\pi$ - the sum (and thus the limit) is only over integer $k$, so that operation is invalid. $\endgroup$ – Steven Stadnicki Jun 10 '14 at 0:56
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    $\begingroup$ Word of mouth. My roommate tutors math, and got this question from some kid. I got it from my roommate. You may also be amused to learn there is a part 2 to this question. (Don't have it on me at the moment, maybe will edit it in later when I get home.) $\endgroup$ – user156190 Jun 10 '14 at 1:41

The trick is to use the classic theorem of Hurwitz on rational approximation of irrationals, that there are infinitely many $m,n$ with $|\frac\pi2-\frac{m}{n}|\lt\frac1{n^2}$ (more specifically, you need infinitely many such pairs with $n\equiv 1\bmod 4$, but this result is known; see below). Any such pair will have $|m-\frac{n\pi}{2}|\lt\frac1n\lt\frac2m$; but then $\sin(\frac{n\pi}{2}) = 1$ (this is where the restriction on $n$ comes from), and by quick application of either the addition identity for sin or the mean value theorem, we get $\sin(m)\gt1-\frac2m$ for any such pair. This means that infinitely often we have $\frac{1+\sin(m)}{2}\gt1-\frac1m$, and $\left(\frac{1+\sin(m)}{2}\right)^m\gt\left(1-\frac1m\right)^m$, and you should be able to bound the RHS of this away from zero. This means that you've found infinitely many $a_m$ bounded away from zero, and that's enough to show divergence.

IMPORTANT CAVEAT: as @robjohn points out in the comments, it's not as easy as I thought at first to get the Hurwitz result in an arbitrary residue class! The paper "On The Approximation Of Irrational Numbers With Rationals Restricted By Congruence Relations" gives the following result:

For any irrational number $\xi$, any $s\geq 1$, and any integers $a$ and $b$, there are infinitely many integers $u,v$ satisfying $\left|\xi-\frac uv\right|\lt \frac{2s^2}{v^2}$ with $u\equiv a\pmod s$ and $v\equiv b\pmod s$.

We can take $\xi=\frac\pi2$, $s=4$, $b=1$ here and get infinitely many $u,v$ with $\left|\frac\pi2-\frac uv\right|\lt\frac{8}{v^2}$ and $v\equiv 1\pmod 4$; this is a factor of 8 away from the result above, but it's still good enough to give infinitely many $m$ with $\left(\frac{1+\sin(m)}{2}\right)^m\gt\left(1-\frac8m\right)^m$ and that's good enough to give a bound away from zero and prove divergence.

  • $\begingroup$ What is the easiest way to modify the result so that the denominator is $1$ mod $4$? $\endgroup$ – robjohn Jun 10 '14 at 0:40
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    $\begingroup$ @robjohn That's a very good question! My (very sloppy) initial thinking was that you could just take $n=n+1,2,3$ as needed, since $\frac1n-\frac1{n+1}\in O(n^{-2})$, but of course because of the multiplier of $m$ the actual difference there is in $\Theta(n^{-1})$. I'm pretty sure I've seen a modification of the Hurwitz theorem before that gives the odd/even case, but I'd have to walk through the 'bucket' style proof to see whether it can be extended to arbitrary residue classes. $\endgroup$ – Steven Stadnicki Jun 10 '14 at 1:00
  • $\begingroup$ Can't say I'm happy about a solution using a paper's result, but thanks for the writeup. $\endgroup$ – user156190 Jun 10 '14 at 17:29

There is a "power saving" of $\sqrt{n}$, in our favor, compared to the usual Diophantine approximation problems, because of the quadratic flatness of $\sin(x)$ near its maxima. [Updated: but this might not be a large enough saving to get around the use of sophisticated arguments. Analysis is still in progress.]

If $k \mod 2\pi = \frac{\pi}{2} + u$ for small $u$, then $1 - \sin(k)$ is of order $u^2$. In fact $\frac{1+\sin k}{2} \sim (1 - \frac{u^2}{4})$, but the constant $1/4$ is irrelevant. For a divergence proof what we would like to know is if $u = u_k$ is infinitely often small enough so that $(1 - u_k^2)^k$ is bounded below, i.e., is $ku_k^2$ bounded. The equivalent Diophantine question is

does $\frac{\pi}{2}$ have infinitely many rational approximations $\frac{k}{4n+1}$ accurate (up to a constant factor) to within $\frac{1}{n\sqrt{n}}$?

This is weaker than the $O(n^{-2})$ approximation from continued fractions and pigeonhole arguments and suggests the possibility of a more direct construction.

Originally I wrote $1/\sqrt{n}$ as the accuracy requirement, which is much too easy. Anything $1/n$ or larger has a simple construction. Whether $n^{-u}$ with $u \in (1,2)$ can be achieved without exponentially thin subsequences (such as continued fractions) seems like a very interesting question.


You can show that it diverges using the first condition, the limit of $a_k$ as $k\to\infty$. You don't need to compute the limit. You need to show that either it does not exist or if exists it is not 0. In this case the limit does not exist(? see the comments below)

  • $\begingroup$ I can't seem to prove that the limit doesn't exist for $k\in\mathbb Z$ using $\delta,\epsilon$. Could you help me a bit further? $\endgroup$ – user156190 Jun 10 '14 at 1:20
  • $\begingroup$ At first glance, that seemed like something provable in a straightforward manner. But I realized that it is not, which is incidentally what makes your question so interesting. $\endgroup$ – DirkGently Jun 10 '14 at 1:41

It's clearly convergent because of

$$ 0 < \frac{1+sin(n)}{2} < 1 $$

$$ \sum_{n=1}^{\infty} a^n \ , \ 0<a<1 $$

  • $\begingroup$ Yes, each $\frac{1+\sin n}2$ is smaller than $1$, but you cannot deduce from that that there is a number $a<1$ such that $(\forall n\in\mathbb N):\frac{1+\sin n}2<a^n$. $\endgroup$ – José Carlos Santos Mar 2 at 10:13

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