Determine whether the sequence converges or diverges. If it converges, find the limit. If it diverges, enter NONE
A(n)= arctan(2n)
2007-09-28 10:37:22 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
It really depends if range of arctangent is limited or not. If the range is limited to -pi/2 to pi/2, then yes, the sequence does exist and converges to pi/2.
But generally, if it's limited, it's written as Arctan with a capital A. If it's not capitalized, it sometimes is restricted and sometimes isn't, based on the context.
2007-09-28 10:47:23 · answer #1 · answered by np_rt 4 · 0⤊ 0⤋
Who the F are you to ask people to do your homework and then DEMAND to show work? Here's your work you demanded *sling middle finger* go study, freak!!
2016-11-09 22:55:45 · answer #2 · answered by Junk 1 · 0⤊ 0⤋
As x --> + infinity A(n) --> pi/2
As x --> - infinity A(n) --> - pi/2
y = +/- pi/2 are horizontal asymptotes
2007-09-28 10:53:26 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋