# Thread: Determine whether the following series is convergent or divergent.

1. ## Determine whether the following series is convergent or divergent.

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

2. 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?

3. 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 $x \in \mathbb{R}\setminus\{0\}$: $|\sin{x}|<|x|$

4. 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:

$\sin\left(\frac{1}{n}\right)=\sum_{\jmath=0}^{\inf ty}\frac{(-1)^{\jmath}}{n^{2\jmath+1}\left(2\jmath+1\right)!}$

Bigger Hint!

Spoiler:

Write our just the first two terms of this series, multiply by $n$ and subtract one. It should be obvious from there.

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!

5. 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:
Conjecture that your series shares convergence/divergence with some other series $\sum_{n\in\mathbb{N}}\frac{1}{n^\lambda}$. Find this lambda.

Here is how:

Spoiler:

We need $\lim_{n\to\infty}\frac{1-n\sin\left(\tfrac{1}{n}\right)}{\frac{1}{n^{\lambd a}}}=C$. Make the substitution $\frac{1}{n}=z$ to get the limit $\lim_{z\to0}\frac{1-\frac{\sin(z)}{z}}{z^{\lambda}}$. Or, equivalently $\lim_{z\to0}\frac{z-\sin(z)}{z^{\lambda+1}}$. Applying L'hopital's (since I assume that is your most familiar means of computing limits) we see that $\frac{1-\cos(z)}{\left(\lambda+1\right)z^{\lambda}}=C$. Now, your experience with limits should bring the limit $\lim_{x\to0}\frac{1-\cos(x)}{x^2}=\frac{1}{2}$ to let you conclude that $\lambda=2$. Lo and behold $\lim_{n\to\infty}\frac{1-n\sin\left(\tfrac{1}{n}\right)}{\frac{1}{n^2}}=\fr ac{1}{6}$. Now, just apply the limit comparison test

6. Alternatively, alternatively.

Spoiler:

Show that you may apply the integral test. And then,

$\int_1^{\infty}\left\{1-n\sin\left(\tfrac{1}{n}\right)\right\}$, from where the substitution $\frac{1}{n}=z$ leads to $\int_0^1\left\{\left(1-\frac{\sin(z)}{z}\right)\cdot\frac{1}{z^2}\right\} =\int_0^1\sum_{\jmath=1}^{\infty}\frac{(-1)^{\jmath}z^{2\jmath-2}}{(2\jmath+1)!}=$ $\sum_{\jmath=1}^{\infty}\int_0^{1}\frac{(-1)^{\jmath+1}z^{2\jmath-2}}{(2\jmath+1)!}=\sum_{\jmath=1}^{\infty}\frac{(-1)^{\jmath+1}z^{2\jmath-1}}{(2\jmath-1)(2\jmath+1)!}$ the last of which I know you know converges. Thus, we may draw our conclusion.

7. 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|

for all $n>n_0$.

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

CB

8. 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:
Conjecture that your series shares convergence/divergence with some other series $\sum_{n\in\mathbb{N}}\frac{1}{n^\lambda}$. Find this lambda.

Here is how:

Spoiler:

We need $\lim_{n\to\infty}\frac{1-n\sin\left(\tfrac{1}{n}\right)}{\frac{1}{n^{\lambd a}}}=C$. Make the substitution $\frac{1}{n}=z$ to get the limit $\lim_{z\to0}\frac{1-\frac{\sin(z)}{z}}{z^{\lambda}}$. Or, equivalently $\lim_{z\to0}\frac{z-\sin(z)}{z^{\lambda+1}}$. Applying L'hopital's (since I assume that is your most familiar means of computing limits) we see that $\frac{1-\cos(z)}{\left(\lambda+1\right)z^{\lambda}}=C$. Now, your experience with limits should bring the limit $\lim_{x\to0}\frac{1-\cos(x)}{x^2}=\frac{1}{2}$ to let you conclude that $\lambda=2$. Lo and behold $\lim_{n\to\infty}\frac{1-n\sin\left(\tfrac{1}{n}\right)}{\frac{1}{n^2}}=\fr ac{1}{6}$. Now, just apply the limit comparison test

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 $\lim_{x\to0}\frac{1-\cos(x)}{x^2}$ ??

9. Thanks.
I got it

10. Originally Posted by TWiX
Thanks.
I got it
Are you good then with this problem?

11. 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.

12. 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.

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

14. 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?

15. 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.