As 𝜶 varies in ℝ, study the behaviour of this series

[DottZakapa]

Homework Statement

As 𝜶 varies in ℝ study the behaviour of the series
$\sum_{n=1}^\infty \frac {(sin 𝜶)^n}{2n}$

Relevant Equations

series tests

$\sum_{n=1}^\infty \frac {(sin 𝜶)^n}{2n}$

I apply the root test and i get

$\lim_{n \rightarrow +\infty} \frac {sin 𝜶}{2n^\frac 1 n}$

at this point i don't know how to treat the denominator.

 

[Mark44]

Homework Statement:: As 𝜶 varies in ℝ study the behaviour of the series
$\sum_{n=1}^\infty \frac {(sin 𝜶)^n}{2n}$
Homework Equations:: series tests
$\sum_{n=1}^\infty \frac {(sin \alpha)^n}{2n}$

I apply the root test and i get

$\lim_{n \rightarrow +\infty} \frac {sin 𝜶}{2n^\frac 1 n}$

The denominator would be $\sqrt[n]{2n}$, or $(2n)^{1/n}$, which is different from what you have.

at this point i don't know how to treat the denominator.

One thing to keep in mind is that for any $\alpha, -1 \le \sin(\alpha) \le 1$. Also, can you evaluate this limit? $\lim_{n \to \infty} n^{1/n}$?

 

[DottZakapa]

The denominator would be $\sqrt[n]{2n}$, which is different from what you have.

sorry just a typing mistake i meant $\left (2n \right)^\left (\frac 1 n \right )$ guess is the same as $\sqrt[n]{2n}$

I did

$\left|\sin \alpha \right |\leq 1$

then i guess i should divide both sides by
$\sqrt[n]{2n}$

$\frac {\left|\sin \alpha \right |} {\sqrt[n]{2n}} \leq \frac 1 {\sqrt[n]{2n}}$

then i am stuck

 

[Mark44]

Since $\lim_{n \to \infty} \sqrt[n]{2n} = 1$, the root test is no help, so I would try another test, such as the following.

• Ratio test
• Comparison test
• Limit comparison test
• n-th term test for divergence
• Alternating series test

Also, keep in mind that $\alpha$ is fixed, so for some values of it, the series is alternating.

 

[DottZakapa]

Since $\lim_{n \to \infty} \sqrt[n]{2n} = 1$, the root test is no help, so I would try another test, such as the following.

• Ratio test
• Comparison test
• Limit comparison test
• n-th term test for divergence
• Alternating series test

Also, keep in mind that $\alpha$ is fixed, so for some values of it, the series is alternating.

in the solution of the exercise uses the root test

with this conclusions

- if $\left|sin \alpha \right |\lt 1$ the series converges
-if $\sin\alpha$=1 the series is the divergent series $\sum_{n=1}^\infty \frac {1} {2n}$
- if $\sin\alpha$=-1 we get the convergent ,not absolutely convergent series$\sum_{n=1}^\infty (-1)^n \frac {1} {2n}$

but I'm not understanding it

 

[Mark44]

in the solution of the exercise uses the root test

with this conclusions

- if $\left|sin \alpha \right |\lt 1$ the series converges

For $|\sin(\alpha)| < 1$, what is the value of the limit $\lim_{n \to \infty}\frac {\sin(\alpha)}{(2n)^{1/n}}$? Under what conditions does the root test say a series converges?

-if $\sin\alpha$=1 the series is the divergent series $\sum_{n=1}^\infty \frac {1} {2n}$

This should be pretty clear.

- if $\sin\alpha$=-1 we get the convergent ,not absolutely convergent series$\sum_{n=1}^\infty (-1)^n \frac {1} {2n}$

This should be almost as clear. If $\sin(\alpha) = -1$, the series is alternating. Can you determine that the alternating series converges?

but I'm not understanding it

 

[DottZakapa]

For $|\sin(\alpha)| < 1$, what is the value of the limit $\lim_{n \to \infty}\frac {\sin(\alpha)}{(2n)^{1/n}}$? Under what conditions does the root test say a series converges?
This should be pretty clear.
This should be almost as clear. If $\sin(\alpha) = -1$, the series is alternating. Can you determine that the alternating series converges?

where is the 1/n gone?
$\lim_{n \to \infty}\frac {\sin(\alpha)}{(2n)^{1/n}}$ I am not understanding the behaviour of the limit at the denominator.

 

[Mark44]

where is the 1/n gone?

Look at the original series -- $\sum_{n=1}^\infty \frac {(sin 𝜶)^n}{2n}$.
What do you have if $\sin(\alpha) = 1$? What is $\sin^2(\alpha)$? $\sin^3(\alpha)$? Etc.

 

[DottZakapa]

where is the 1/n gone?
$\lim_{n \to \infty}\frac {\sin(\alpha)}{(2n)^{1/n}}$ I am not understanding the behaviour of the limit at the denominator.

I might understood
so

for $|\sin\alpha|$<1 $\Rightarrow$ $\sum_{n=1}^\infty \frac {|\sin\alpha|^n}{2n}$, by the root test

$\sum_{n=1}^\infty \frac {|\sin\alpha|}{\sqrt[n]2n} ~\Rightarrow~\lim_{n \rightarrow +\infty} \frac {|\sin\alpha|} {e^0}~=\lim_{n \rightarrow +\infty} \frac {|\sin\alpha|} {1}~=|\sin\alpha|<1~$hence convergent.

then

supposing $\sin \alpha=1$ the series becomes$\sum_{n=1}^\infty \frac {1}{2n}~\Rightarrow$ convergent as $\frac 1 {n}$

and

if $\sin \alpha=-1$ the series becomes$\sum_{n=1}^\infty (-1)^n \frac {1}{2n}~\Rightarrow \sum_{n=1}^\infty \left| \frac {1}{2n}\right|~\Rightarrow$ divergent, hence not absolutely convergent, but, by leibniz test it converges

Did I get the reasoning correctly?

 

[Mark44]

I might understood
so

for $|\sin\alpha|$<1 $\Rightarrow$ $\sum_{n=1}^\infty \frac {|\sin\alpha|^n}{2n}$, by the root test

$\sum_{n=1}^\infty \frac {|\sin\alpha|}{\sqrt[n]2n} ~\Rightarrow~\lim_{n \rightarrow +\infty} \frac {|\sin\alpha|} {e^0}~=\lim_{n \rightarrow +\infty} \frac {|\sin\alpha|} {1}~=|\sin\alpha|<1~$hence convergent.

Your reasoning is correct, but you are misusing the implication symbol. This symbol ($\rightarrow$k) should be used between statements such as equations or inequalities.
For example, $x = 2 \Rightarrow x^2 = 4$. It should not be used between expressions such as what you wrote above:
$\sum_{n=1}^\infty \frac {|\sin\alpha|}{\sqrt[n]2n} ~\Rightarrow~\lim_{n \rightarrow +\infty} \frac {|\sin\alpha|} {e^0}$
In fact, these two expressions shouldn't be connected at all. You're investigating the series by looking at the limit.

then

supposing $\sin \alpha=1$ the series becomes$\sum_{n=1}^\infty \frac {1}{2n}~\Rightarrow$ convergent as $\frac 1 {n}$

No. The harmonic series, $\sum_{n=1}^\infty \frac 1 n$ is divergent. The same is true for $\sum \frac 1 {2n}$, which can be shown by using the Limit Comparison Test or maybe the Ratio Test.

and

if $\sin \alpha=-1$ the series becomes$\sum_{n=1}^\infty (-1)^n \frac {1}{2n}~\Rightarrow \sum_{n=1}^\infty \left| \frac {1}{2n}\right|~\Rightarrow$ divergent, hence not absolutely convergent, but, by leibniz test it converges

Your conclusion here is correct, but your work is a bit confusing, as it implies that the alternating series is divergent. $\sum (-1)^n \frac 1 {2n}$ is convergent but not absolutely convergent.

Did I get the reasoning correctly?

 

[DottZakapa]

Now is all clear, thank you for being so patient.