alternating series

[renegade05]

\(\displaystyle \sum_{n=1}^{\infty}(-1)^nnsin(1/n)\)

That is the series in question.

I am thinking AST and than L'Hôpital to evaluate the limit as n goes to infinity. After this I get \(\displaystyle \lim_{n \to \infty}\cos{\frac{1}{n}}\) which equals 1.

So this series diverges?

Looking for a confirmation.

Thanks!

 

[Deleted member 4993]

\(\displaystyle \sum_{n=1}^{\infty}(-1)^nnsin(1/n)\)

That is the series in question.

I am thinking AST and than L'Hôpital to evaluate the limit as n goes to infinity. After this I get \(\displaystyle \lim_{n \to \infty}\cos{\frac{1}{n}}\) .....How? which equals 1.

So this series diverges?

Looking for a confirmation.

Thanks!

\(\displaystyle \frac{d}{dx}[x * sin(\frac{1}{x})] \ = \ sin(\frac{1}{x}) + x*[cos(\frac{1}{x})]*[-\frac{1}{x^2}] \)

you could write:

\(\displaystyle \lim_{x\to \infty}[x * sin(\frac{1}{x})] = \lim_{x\to \infty}[\frac{sin(\frac{1}{x})}{\frac{1}{x}}] \ = \ \lim_{z\to 0}[\frac{sin(z)}{z}] \ = \ 1 \)

That does not mean that the alternating series diverges - it only means that the alternating series is not absolutely convergent.

You have not told us what you were supposed to do - or what question you were supposed to answer......

 

[renegade05]

The question asks if the series converges or diverges.

I got cos(1/n) by using l'hopitals rule.

\(\displaystyle \lim_{x\to \infty}[\frac{sin(\frac{1}{x})}{\frac{1}{x}}] \) =L'H \(\displaystyle \lim_{x\to \infty} \cos(1/x) = 1\)

So does this series converge, not absolutely?

 

[Deleted member 4993]

Does series pass or fail AST?

 

[renegade05]

Does series pass or fail AST?

It fails since it does not equal zero, so it diverges ?

 

[galactus]

On an interesting note, for fun I ran this through my calculator, Maple, Mathematica.

The calculator and Maple shows it converging, with increasing values, to around .1293....

That is, I used increasing values up to 100,000 and it levels off around .1293787629

Yet, when I plug the sum into Maple with the upper index of infinity, it says it converges to -.3706

I ran it through WolframAlpha and it says it diverges by the limit test.

I have never seen such diverse answers from different software. Strange.

 

[Deleted member 4993]

It fails since it does not equal zero, so it diverges ?

Failing AST - only means that we cannot ascertain that it converges.

 

[renegade05]

And it fails because the following is not true: 0<a_n+1<a_n ?

a_n\(\displaystyle =nsin(1/n)\)

??

And Subhotosh Khan: give me an answer ! haha. Does this converge or diverge?????

 

[galactus]

It would appear this is "conditionally convergent". Hence the conflicting answers gotten from the respective tech sources.

Check that \(\displaystyle \sum|a_{n}|\) diverges.

Yet, \(\displaystyle \sum a_{n}\) converges.

\(\displaystyle \displaystyle\sum_{k=1}^{\infty}(-1)^{n}sin(1/n)\) is conditionally convergent as well.

 

[Deleted member 4993]

And it fails because the following is not true: 0<a_n+1<a_n ?

a_n\(\displaystyle =nsin(1/n)\)

??

And Subhotosh Khan: give me an answer ! haha. Does this converge or diverge?????

It fails AST because:

\(\displaystyle \displaystyle\lim_{n\to \infty}[n*sin(\frac{1}{n})] \ \ne 0 \ [ = \ 1]\)