Can you do it ?
i have some opinions about the second one
sum from k=2 to infinity a^(logb k)
It is equal to sum from k=2 to infinity k^(logb a)
= sum from k=2 to infinity k^-(logb 1/a)
= sum from k=1 to infinity k^-(logb 1/a) - 1
= Z[ logb(1/a) ] - 1 where Z is Riemann Zeta Function
The answer to your question is "no, I can't".
$\displaystyle \sum_{k=2}^{\infty}\left(\frac{\sin k +2}{3}\right)^k$
All I can say is that the root test shows that it converges, because $\displaystyle \frac{1}{3}<\lim_{k\to\infty}\frac{\sin k +2}{3}<1$
Beyond that, I can't say much.
It's not an easy question elementary test learned in ordinary textbooks fail to break this question.
The as k ---> infinity sink --> we don't know .
You arrived at
$\displaystyle \frac{1}{3}<\lim_{k\to\infty}\frac{\sin k +2}{3}<1$
But what if sin K = 1
Then the root test fails. Because the limit =1
I think this question reduces to finding out if there is a finite number of times such that sink = 1 for every k E I
We can't conclude that, i thought the same thing but you have to prove it.
It could come really close to 1. For example sin 55 = -0.999755
As you go higher it gets closer to 1.
But i would like to see some sound mathematical prove that sink =/= 1
Ofcourse this is probably true but how close can it go.
We would probably agree thaty 0.999999 = 1
Assume $\displaystyle \exists~k$ such that $\displaystyle \frac{k}{\pi}\in\mathbb{Q}$ (i.e. $\displaystyle \frac{k}{\pi}=\frac{a}{b})$. Thus, by cross-multiplying and such, we have that $\displaystyle \pi = \frac{bk}{a}$, implying that $\displaystyle \pi\in\mathbb{Q}$, which is clearly false, creating a contradiction. Thus, $\displaystyle \nexists~k$ such that $\displaystyle \frac{k}{\pi}\in\mathbb{Q}$ and therefore $\displaystyle \sin k\neq 1~\forall k$.
The root test is inconclusive here. Although it is true that $\displaystyle \frac{\sin k+2}3<1$ for all k, this fraction can get arbitrarily close to 1. Therefore $\displaystyle \limsup_{k\to\infty}\frac{\sin k+2}3=1$, and the root test does not cover this case.
It certainly looks as though the series $\displaystyle \sum\Bigl(\frac{\sin k+2}3\Bigr)^k$ ought to converge, but I don't know of a test that would prove that. It's not even clear that the terms of this series converge to 0. We know that $\displaystyle \frac{\sin k+2}3$ can be very close to 1. But can it be so close to 1 that its k'th power is bounded away from 0, for some arbitrarily large k?
I forgot that the root test was $\displaystyle \lim\sup a_n$; I had thought it was just $\displaystyle \lim a_n$. Graphing this seems to indicate that k gets large enough to counteract $\displaystyle \sin(k)$ approaching 1, but obviously that doesn't constitute a proof.
That what i was getting at that before.
But then someguy confused me saying that
If $\displaystyle k = |\frac{n+1}{\frac{pi}{2}} | = 1 $
But this is a contradition since K is an integer.
But Opalg is right you can get really really close to 1, but how close?
I think this question boils down to :
Is there a finite number of time such that sink =~ 1 ?
sink oscillates so regardless of how small sink gets such that $\displaystyle sin^{k}k\to 0$. It eventually become 1 again.
That why i say that is there must be a finite number of times that sink =1.
If it happens infinitely many times then you are also saying that we have infinitely many +1's then it would follow that our series diverges.