# Math Help - Challenge problem ( Infinite series)

1. ## Challenge problem ( Infinite series)

Can you do it ?

2. 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

3. Originally Posted by simplependulum
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

That's a really good step.
from here
= sum from k=2 to infinity k^-(logb 1/a)

A even better step is to recoqnize this as a P-series which converges for

logb a <-1
lnb < -lna
b< 1/a
ab<1
.

Good job though now someone should try the second one.

It's a tough one.

4. Originally Posted by ╔(σ_σ)╝

Can you do it ?
The answer to your question is "no, I can't".

$\sum_{k=2}^{\infty}\left(\frac{\sin k +2}{3}\right)^k$

All I can say is that the root test shows that it converges, because $\frac{1}{3}<\lim_{k\to\infty}\frac{\sin k +2}{3}<1$

Beyond that, I can't say much.

5. Originally Posted by redsoxfan325
The answer to your question is "no, I can't".

$\sum_{k=2}^{\infty}\left(\frac{\sin k +2}{3}\right)^k$

All I can say is that the root test shows that it converges, because $\frac{1}{3}<\lim_{k\to\infty}\frac{\sin k +2}{3}<1$

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
$\frac{1}{3}<\lim_{k\to\infty}\frac{\sin k +2}{3}<1$

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

6. $\sin k\neq 1~\forall k$ because $k\in\mathbb{N}$ and thus is never an odd multiple of $\frac{\pi}{2}$. So the root test does work.

7. Isn't it safe to assume $sin(k) \ne 1$ because $k=1,2,3,...$, or in other words, $k$ is only ever an integer?

Edit: redsoxfan beat me to it.

8. Originally Posted by redsoxfan325
$\sin k\neq 1~\forall k$ because $k\in\mathbb{N}$ and thus is never an odd multiple of $\frac{\pi}{2}$. So the root test does work.

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

9. Originally Posted by ╔(σ_σ)╝
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 $\exists~k$ such that $\frac{k}{\pi}\in\mathbb{Q}$ (i.e. $\frac{k}{\pi}=\frac{a}{b})$. Thus, by cross-multiplying and such, we have that $\pi = \frac{bk}{a}$, implying that $\pi\in\mathbb{Q}$, which is clearly false, creating a contradiction. Thus, $\nexists~k$ such that $\frac{k}{\pi}\in\mathbb{Q}$ and therefore $\sin k\neq 1~\forall k$.

10. Originally Posted by redsoxfan325
Assume $\exists~k$ such that $\frac{k}{\pi}\in\mathbb{Q}$ (i.e. $\frac{k}{\pi}=\frac{a}{b})$. Thus, by cross-multiplying and such, we have that $\pi = \frac{bk}{a}$, implying that $\pi\in\mathbb{Q}$, which is clearly false, creating a contradiction. Thus, $\nexists~k$ such that $\frac{k}{\pi}\in\mathbb{Q}$ and therefore $\sin k\neq 1~\forall k$.

That does it.

11. The root test is inconclusive here. Although it is true that $\frac{\sin k+2}3<1$ for all k, this fraction can get arbitrarily close to 1. Therefore $\limsup_{k\to\infty}\frac{\sin k+2}3=1$, and the root test does not cover this case.

It certainly looks as though the series $\sum\Bigl(\frac{\sin k+2}3\Bigr)^k$ 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 $\frac{\sin k+2}3$ 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?

12. I forgot that the root test was $\lim\sup a_n$; I had thought it was just $\lim a_n$. Graphing this seems to indicate that k gets large enough to counteract $\sin(k)$ approaching 1, but obviously that doesn't constitute a proof.

13. That what i was getting at that before.

But then someguy confused me saying that

If $k = |\frac{n+1}{\frac{pi}{2}} | = 1$

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 ?

14. Well, I'm pretty sure that for any $\epsilon>0$, there will be infinitely many k such that $|1-\sin(k)|<\epsilon$, so I think the question really whether eventually k gets so large that $\sin^k(k)\to 0$.

15. Originally Posted by redsoxfan325
Well, I'm pretty sure that for any $\epsilon>0$, there will be infinitely many k such that $|1-\sin(k)|<\epsilon$, so I think the question really whether eventually k gets so large that $\sin^k(k)\to 0$.
sink oscillates so regardless of how small sink gets such that $sin^{k}k\to 0$. 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.

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