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.
- $\lim \limits_{k\to\infty}\left|\dfrac{a_{k+1}}{a_k}\right|>1$—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?
EDIT
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.
EDIT AGAIN
The proof above is no good. More help pls. I can't figure this one out.