Show $\sum_{k=1}^{\infty}\left(\frac{1+\sin(k)}{2}\right)^k$ diverges Show$$\sum_{k=1}^{\infty}\left(\frac{1+\sin(k)}{2}\right)^k$$diverges.

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.
$\lim \limits_{k\to\infty}\left|\dfrac{a_{k+1}}{a_k}\right|&gt;1$—Limit does not converge; inconclusive.
$\lim \limits_{k\to\infty}\sqrt[k]{|a_k|}&gt;1$—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?

EDIT

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

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.

EDIT AGAIN

The proof above is no good. More help pls. I can't figure this one out.
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.
It's clearly convergent because of

$$ 0 &lt; \frac{1+sin(n)}{2} &lt; 1 $$

$$ \sum_{n=1}^{\infty} a^n \ , \ 0&lt;a&lt;1 $$