Question

Show that \[a_n=\sin(n)n^{-1/2}\] and \[b_n=\sin(\frac{ n \pi }{ 2 }+\frac{ \pi }{ 3 })n^{-1/2}\]
converges or diverges

Answer 1

I assume that it is convergent. Now I have:\[\sin(n)n^{-1/2}=\frac{ \sin(n) }{ \sqrt(n) }\]
And then I want to use the definition for convergence, so I let \[\epsilon >0\] and now I want to find a \[N \in \mathbb{N}:|a_n-a|\le \epsilon \]. From here I am having rouble proving or dis-proving this.

Answer 2

Does it make sens to split \[a_n\] into 2:
\[sin(n), n^{-1/2}\]
And then I could do:
\[n^{-1/2}=1/sqrt(n)<1/n<epsilon\] Now here we know 1/n converges towards 0, hereby we see that \[n^{-1/2}\] also converges towards 0. And since \[sin(n)\] is limited:
\[-1<=sin(n)<=1\] The whole sequence is convergent towards 0? - using the multiplication rule for sequences.

Answer 3

\(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ n^{-1/2}\sin(n) }\)


\(\large\color{black}{ \displaystyle \sin(n) =\displaystyle \sum_{ i=1 }^{ \infty }\frac{(-1)^kn^{1+2k}}{(1+2k)!} }\)

\(\large\color{black}{ \displaystyle n^{-1/2}\sin(n) =\displaystyle \sum_{ i=1 }^{ \infty }\frac{(-1)^kn^{1+2k~~-1/2 }}{(1+2k)!} }\)

\(\large\color{black}{ \displaystyle n^{-1/2}\sin(n) =\displaystyle \sum_{ i=1 }^{ \infty }\frac{(-1)^kn^{2k+1/2 }}{(1+2k)!} }\)

\(\large\color{black}{ \displaystyle \int n^{-1/2}\sin(n) {\rm ~d}n=\displaystyle \sum_{ i=1 }^{ \infty }\frac{(-1)^kn^{2k+3/2 }}{(2k+3/2)(1+2k)!} }\)

Answer 4

So the first series converges byu the integra test

Answer 5

Correction: (grammar)

integral test**

Answer 6

I haven't learned about the integral test yet. I know the definition for convergence, and cauchy sequence.

Answer 7

Use the Riemann Series:
\[a _{n}=\frac{ 1 }{ n ^{q} }\]
The series is:
Divergent when q<=1
Convergent when q>1

Answer 8

Wel, the integral test is that you integrate A_n.

If the integral converges, then series also converges.
If integral diverges, then series also diverges.

(You can take integral with limits ∞ and 1, or ∞ and 100, doesn't matter)

Answer 9

Loser66, I checked the sries A_n does converge.
http://www.wolframalpha.com/input/?i=summation+from+n%3D1+to+n%3D%E2%88%9E+of+sin%28n%29%2F%E2%88%9An

Answer 10

Ignore that, not gonna work here. Your series has to be inferior to a converging series, but here 1/sqrt(n) is divergent.

Answer 11

Doesnt in converge towards 0? Since \[1/sqrt(n)\] converges towards 0, and then if you multiply it with a limited sequence: \[-1<=sin(n)<=1\] You would get 0

Answer 12

the series 1/sqrt(n) is not convergent

Answer 13

it meets alternating series test conceptualy....

Answer 14

Terms decrease, there is a non-integer alternation, if you want....
Try partial sums to confirm

Answer 15

But without the sin(n), it would indeed diverge, yes.

Answer 16

because sin(n)/n is not 1 as n->∞

Answer 17

it is 0

Answer 18

yes:)

Answer 19

\(\large\color{black}{ \displaystyle \sum_{n=1}^{\infty}\sin\left(\frac{ n \pi }{ 2 }+\frac{ \pi }{ 3 }\right) n^{-1/2} }\)


\(\large\color{black}{ \displaystyle \int_{n=1}^{n=\infty} n^{-1/2}\sin\left(\frac{ n \pi }{ 2 }+\frac{ \pi }{ 3 }\right) {\rm ~d}n=}\)

\(\large\color{blue}{ \displaystyle u=\left(\frac{ n \pi }{ 2 }+\frac{ \pi }{ 3 }\right) }\)

\(\large\color{blue}{ \displaystyle du=\frac{ \pi }{ 2 } dn\quad \Longrightarrow\frac{ 2 } { \pi } du=dn}\)

\(\large\color{blue}{ \displaystyle n=1\quad \Longrightarrow u=\left(\frac{ \color{red}{1\cdot } \pi }{ 2 }+\frac{ \pi }{ 3 }\right)=\frac{5\pi}{6}}\)

\(\large\color{blue}{ \displaystyle n=\infty \quad \Longrightarrow u=\left(\frac{ \color{red}{\infty\cdot } \pi }{ 2 }+\frac{ \pi }{ 3 }\right)=\infty}\)

\(\large\color{black}{ \displaystyle \frac{ 2 } { \pi } \int_{u=\frac{5\pi}{6}}^{u=\infty} n^{-1/2}\sin\left(u\right) {\rm ~d}u=}\)

Answer 20

By the same tocken, the second series goes by the integral test.
Converges :)

Answer 21

(because the integral would converge 9since it is even smaller than the integral by the A_n)

Answer 22

Just in case, here is a link to integral test.
One possible resource for it.

http://tutorial.math.lamar.edu/Classes/CalcII/IntegralTest.aspx

Answer 23

Thanks, ill get a look at it

Answer 24

Sure enjoy ...

Answer 25

One more correction, in my last line of the first reply,
the "i=1" index under the sigma is supposed to be "k=1".

Answer 26

Great job guys :D Thanks for helping