Determine whether the following series is convergent or divergent.

**TWiX** · January 8th 2010, 11:10 AM

Hi !
I have this series
Sigma from 1 to infinity ( 1 - n*sin(1/n) )
nth term test fails
and rewrtting it as
Sigma(1) - Sigma( n*sin(1/n) )
will not solve it.
since both series diverges .. and the convergence of the difference of two divergent series is unknown .. it maybe converges or diverges
i want only a hint
I dont want full solution .. because its my favorite chapter =D

**TWiX** · January 8th 2010, 11:44 AM

OK Let me thinking more
N.T.T Failed
I cant apply A.S.T here since there is no (-1)^n or (-1)^(n-1) .. etc
Root and Ratio Failed since its algebric function
its not telescoping << am not sure about this .. maybe it needs some algebric operations to make it telescoping.
i have the comparison tests
but the problem i must prove that ( 1 - n*sin(1/n) ) is positive for all positive integers n > 1
Clearly n*sin(1/n) is positive
since n is positive and (1/n) are angles in the first quadrant for all n > 1
and sine is positive in the first quadrant
but the problem here i have ( 1 - n*sin(1/n) ) !!
if i proved that n*sin(1/n) < 1 for all n then i can use the comparison tests ..
but i cant prove it !
n*sin(1/n) ---> 1 n-->infinity
n*sin(1/n) = sin(1) as n=1
ohhh .. did i prove n*sin(1/n) belongs to [sin(1) , 1) for all n > 1
or this is wrong ?

Any mistakes here?

**Unbeatable0** · January 8th 2010, 10:23 PM

Originally Posted by TWiX

n*sin(1/n) ---> 1 n-->infinity
n*sin(1/n) = sin(1) as n=1
ohhh .. did i prove n*sin(1/n) belongs to [sin(1) , 1) for all n > 1
or this is wrong ?
Any mistakes here?

Although your assertion that $n \sin{\frac{1}{n}} \in [\sin(1), 1)$ for $n\ge 1$ is correct, your proof is not, as you did not prove it for $n \in (1,\infty)$ . To show that from $1 \sin{\frac{1}{1}} = \sin(1)$ and $\lim_{n\rightarrow \infty} n \sin{\frac{1}{n}} = 1$ , $n \sin{\frac{1}{n}} \in [\sin(1), 1)$ follows, you may try to show that the function $f(x) = x \sin{\frac{1}{x}}$ is increasing in the interval $(1,\infty)$ . Another way is to use the inequality that holds for all $x \in \mathbb{R}\setminus\{0\}$ : $|\sin{x}|<|x|$

**Drexel28** · January 8th 2010, 10:30 PM

Originally Posted by TWiX

Hi !
I have this series
Sigma from 1 to infinity ( 1 - n*sin(1/n) )
nth term test fails
and rewrtting it as
Sigma(1) - Sigma( n*sin(1/n) )
will not solve it.
since both series diverges .. and the convergence of the difference of two divergent series is unknown .. it maybe converges or diverges
i want only a hint
I dont want full solution .. because its my favorite chapter =D

I just did this one for someone recently.
Hint!:

Spoiler:

P.S. As anyone here that has posted for a while can attest, I too love infinite series! PM me if you have any questions or just want some suggestions of areas of study!

**Drexel28** · January 8th 2010, 10:54 PM

It just occurred to me that you may not understand the hint I gave you (if you are currently studying the convergence of infinite series). So, here is another hint:

Spoiler:

**Drexel28** · January 8th 2010, 11:23 PM

Alternatively, alternatively.

Spoiler:

**CaptainBlack** · January 8th 2010, 11:30 PM

Originally Posted by TWiX

Hi !
I have this series
Sigma from 1 to infinity ( 1 - n*sin(1/n) )
nth term test fails
and rewrtting it as
Sigma(1) - Sigma( n*sin(1/n) )
will not solve it.
since both series diverges .. and the convergence of the difference of two divergent series is unknown .. it maybe converges or diverges
i want only a hint
I dont want full solution .. because its my favorite chapter =D

Big-O or Landau notation and operations always makes these things seem simple:

$a_n=1-n\sin(1/n) =$ $1-n\left(\frac{1}{n} + \frac{1}{6n^3}+O(n^{-5}) \right)=\frac{1}{6n^2}+O(n^{-4})=O(n^{-2})$

Which means that there exists a constants $C>0$ and $n_0\in \mathbb{N}$ such that:

$|a_n|<C n^{-2}$

for all $n>n_0$ .

(you can put in explicit remainder terms if you wish and get the same result)

CB

**TWiX** · January 9th 2010, 07:20 AM

Originally Posted by Drexel28

It just occurred to me that you may not understand the hint I gave you (if you are currently studying the convergence of infinite series). So, here is another hint:

Spoiler:

Thanks all but all soultions are too advanced to me.

this one is good
but my experience did not help me

how did you make it $\lim_{x\to0}\frac{1-\cos(x)}{x^2}$ ??

**TWiX** · January 9th 2010, 09:29 AM

Thanks.
I got it

**Drexel28** · January 9th 2010, 10:36 AM

Originally Posted by TWiX

Thanks.
I got it

Are you good then with this problem?

**TWiX** · January 9th 2010, 10:48 AM

Originally Posted by Drexel28

Are you good then with this problem?

Yeah, but using the known tests.
not O notation and complicated solutions

I love testing series for convergence.
In meduim level.
not the series like 1/(ln n)^9 from 2 to infinity
I hate it

Can you test it?

Sorry I should open new thread, but anyway I can ask 2 question in 1 thread .
Limit comparison test with $\frac{1}{ n^{ \frac{9}{11} } }$ will show it diverges.
But I want another solution.

**Drexel28** · January 9th 2010, 11:00 AM

Originally Posted by TWiX

Yeah, but using the known tests.
not O notation and complicated solutions

I love testing series for convergence.
In meduim level.
not the series like 1/(ln n)^9 from 2 to infinity
I hate it

Can you test it?

Sorry I should open new thread, but anyway I can ask 2 question in 1 thread .
Limit comparison test with $\frac{1}{ n^{ \frac{9}{11} } }$ will show it diverges.
But I want another solution.

Cauchy's condensation test. Relatively easy to prove. It says that under certain conditions (that this series satisfies) we have that $\sum_{n\in\mathbb{N}}a_n\text{ converges }\Longleftrightarrow \sum_{n\in\mathbb{N}}2^n a_{2^n}\text{ converges}$

P.S. Yout might be interested in knowing that the integral brought up in the integral test post actually has a nice solution. $\int_0^{\infty}\left\{1-n\sin\left(\tfrac{1}{n}\right)\right\}\text{ }dn=\frac{\pi}{4}$ . Not sure if you cared...you can prove it a couple of ways...if you're interested.

**Drexel28** · January 9th 2010, 11:05 AM

Also, notice alternatively that $\lim_{n\to\infty}\frac{\ln(n)}{n^{\frac{1}{9}}}=0$ so it follows that eventually $\ln(n)\leqslant n^{\frac{1}{9}}\implies \frac{1}{n^{\frac{1}{9}}}\leqslant\frac{1}{\ln(n)} \implies \frac{1}{n}\leqslant\frac{1}{\ln^9(n)}$

**TWiX** · January 9th 2010, 11:09 AM

Originally Posted by Drexel28

Cauchy's condensation test. Relatively easy to prove. It says that under certain conditions (that this series satisfies) we have that $\sum_{n\in\mathbb{N}}a_n\text{ converges }\Longleftrightarrow \sum_{n\in\mathbb{N}}2^n a_{2^n}\text{ converges}$

P.S. Yout might be interested in knowing that the integral brought up in the integral test post actually has a nice solution. $\int_0^{\infty}\left\{1-n\sin\left(\tfrac{1}{n}\right)\right\}\text{ }dn=\frac{\pi}{4}$ . Not sure if you cared...you can prove it a couple of ways...if you're interested.

Although I dont know this test, but it leads to (1/ln2)Sigma (2^n/n)
which diverges by Ratio,B.C.T and maybe L.C.T
Sorry but I dont care about the first one.
Although this integral seems not easy to solve.
maybe x=1/n can solve it but am not sure.
Forget it!
Do you have another solution for my second series?

**Drexel28** · January 9th 2010, 11:10 AM

Originally Posted by TWiX

Although I dont know this test, but it leads to (1/ln2)Sigma (2^n/n)
which diverges by Ratio,B.C.T and maybe L.C.T
Sorry but I dont care about the first one.
Although this integral seems not easy to solve.
maybe x=1/n can solve it but am not sure.
Forget it!
Do you have another solution for my second series?

I already posted a second solution, son.

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