arctan(1-3n): Does it converge or diverge? Why?

Question

I have no idea where to begin to figure out whether any arctan function diverges or converges -- how would I even take its limit? Can I use l'hospital's rule somehow, or is there a better way?

Answer 1

As n -> ∞, 1-3n -> -∞ and arctan (1-3n) -> -π/2.

Arctan is rather nice in that it takes the whole real line and compresses it into the open interval (-π/2, π/2) while still keeping it 1-1. So if f(x) has a limit which is a real number, +∞ or -∞, arctan f(x) will have a limit in [-π/2, π/2].

That doesn't mean that arctan f(x) will always converge. If f(x) doesn't converge to a (finite or infinite) limit, arctan f(x) won't either. So, for instance, lim(x->0) arctan (1/x) does not exist because it goes to -π/2 on one side and +π/2 on the other. Also, lim(x->∞) arctan (sin x) does not exist because sin x oscillates between -1 and 1, so arctan (sin x) oscillates between -π/4 and +π/4.

Answer 2

you have a ) lacking, yet besides . . . replace x for n! Then (n+a million)! = x(n+a million) and (n+2)! = x(n+a million)(n+2) So the sequence turns into the sum of all . . . 3x/x*x(n+a million)*x(n+a million)(n+2) or 3/(x(n+a million)*x(n+a million)(n+2)) merely on the face of it, you are able to discover the ingredient converges immediately on account that x and n the two develop at recent.

Answer 3

Try graphing it.

arctan( ) goes to -pi /2, as goes to +infinity.

Answer 4

lim{x-->+inf} arctan(1-3x) = arctan(-inf) = -pi/2
lim{x-->-inf} arctan(1-3x) = arctan(inf)=pi/2

Converge

saludos.