I was wondering whether a sequence like[tex]x_n=n\sin

[dalcde]

I was wondering whether a sequence like
x_n=n\sin n
converges* to infinity or diverges.

I'm pretty sure it goes to infinity but it still oscillates.

*Let's say we are in the extended real number system where we can converge to infinity

EDIT: I mean
x_n=n+\sin n

 

[NeroKid]

first of all there is no such thing as convergence to infinity
and infinity is not really a number

 

[Neesan]

converges is it goes towards a certain value like a limit.

Yea i agree with daclde.

The sin n part will just oscillate from -1 to 0 to 1

So yea it diverges . No clue about extended real system but

 

[Yuqing]

Divergence doesn't really have points where it must diverge towards. Rather anything that doesn't converge to a specific real point is defined as divergent. Divergence towards infinity is just a popular saying for when something grows without bound. Just like oscillations, divergence towards infinity is not heading towards any definite point.

In the extended real system, this sequence is still divergent as it oscillates between positive and negative.

 

[dalcde]

Sorry, I wanted to say
x_n=n+\sin n.
Does it converge or diverge in this case?

 

[micromass]

Sorry, I wanted to say
x_n=n+\sin n.
Does it converge or diverge in this case?

This will diverge (or converge to infinity in the extended reals). the reason is that n+sin(n) becomes arbitrarily large. That is, we have

n-1\leq n+sin(n)

and the sequence n-1 goes to infinity.

 

[Yuqing]

It converges to infinity in this case since n-1 is a lower bound.

edit: micromass explained it better

 

[dalcde]

Sorry, what about
0.5n + \sin n
since the previous sequence is actually always increasing?

 

[NeroKid]

it still diverges

 

[HallsofIvy]

You can put any (positive) number you like in front of the "n". Eventually, n will be so large that even the product is far larger than sin(n).