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 as k ---> infinity sink --> we don't know .
You arrived at
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
The root test is inconclusive here. Although it is true that for all k, this fraction can get arbitrarily close to 1. Therefore , and the root test does not cover this case.
It certainly looks as though the series 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 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?
That what i was getting at that before.
But then someguy confused me saying that
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 ?
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.