1. ## another series...is my answer right?

$\displaystyle \sum(-1)^{n}[1-nsin(1/n)]$ determine if A.convergent, C.convergent or divergent

I chose C.Convergent

I saw that nsin(1/n) goes to 1 because of some other question I solved earlier and that would make everything 0

which means that $\displaystyle a_k \rightarrow 0$ making the series conditionally convergent?

if this is wrong, could you please point out what my error of thinking about this is? from what you see how I analyzed this atleast...

thanks

2. Originally Posted by akhayoon
$\displaystyle \sum(-1)^{n}[1-nsin(1/n)]$ determine if A.convergent, C.convergent or divergent

I chose C.Convergent

I saw that nsin(1/n) goes to 1 because of some other question I solved earlier and that would make everything 0

which means that $\displaystyle a_k \rightarrow 0$ making the series conditionally convergent?

if this is wrong, could you please point out what my error of thinking about this is? from what you see how I analyzed this atleast...

thanks
this is wrong...is this statement true here $\displaystyle a_{n+1}<a_n,\forall{n}\in\mathbb{Z^{+}}$?

3. I don't even know what that means.....is this relevant to solving the question?

4. Originally Posted by akhayoon
I don't even know what that means.....is this relevant to solving the question?
Yes...it means for every positive integer is the next term in the sequence less than the one before it...that is necesarry in proving an alternating series convergent

5. how would i go about doing that with something that has something like nsin(1/n) in it?

6. Originally Posted by akhayoon
how would i go about doing that with something that has something like nsin(1/n) in it?
you evaluate it...check it...does any value of sin(1/x)<0?

7. I guess not, but I still can't make the connection, and I'm getting I'm getting kind of confused actually...

8. Originally Posted by akhayoon
I guess not, but I still can't make the connection, and I'm getting I'm getting kind of confused actually...
Ok before we continue...do you actually know how to apply the alternating series test?

9. all I know is that the series must be positive and that it must be decreasing, my main problem is I don't know how to actually apply this to a series with sin, ln etc...

10. Originally Posted by akhayoon
all I know is that the series must be positive and that it must be decreasing, my main problem is I don't know how to actually apply this to a series with sin, ln etc...
Ok...the theorem states that if $\displaystyle a_{n+1}<a_n$ and $\displaystyle \lim_{n \to {\infty}}a_n=0$...now do both those things apply to your series? If so it is conditonally convergent

11. well I found out before that $\displaystyle a_k goes to 0$

however, how do I show that

$\displaystyle (-1)^{n+1}[1-(n+1)sin(1/n+1)]< (-1)^{n}[1-nsin(1/n)]$

do I just plug in 1 for the value of n or something...in this course I can't use a calculator...

12. Originally Posted by akhayoon
well I found out before that $\displaystyle a_k goes to 0$

however, how do I show that

$\displaystyle (-1)^{n+1}[1-(n+1)sin(1/n+1)]< (-1)^{n}[1-nsin(1/n)]$

do I just plug in 1 for the value of n or something...in this course I can't use a calculator...
First of all you dont include the $\displaystyle (-1)^{n}$ in the test for the decreasing terms...and I will tell you that it does decrease...so now that we know both those things what can we conclude?

13. it must mean that it is conditionally convergent

14. P.S if this shows up on the exams I will say that MathStud28 told me that it was decreasing and I will refer them to the link of this post lol

15. Originally Posted by akhayoon
P.S if this shows up on the exams I will say that MathStud28 told me that it was decreasing and I will refer them to the link of this post lol