Determine whether the series is absolutely convergent, conditionally convergent, or divergent.
Sigma{n=1 to infinity} of cos((n*pi)/7)/(tan^-1 n)^n
Sorry for posting this, but I have been stuck on this for an hour.
2007-04-12 16:52:18 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
I don't have a calculator here, but I can make a suggestion.
The top of the fraction will always be somewhere in the range of -1 to +1 no matter what n is
The term tan^-1 (n) will always be between pi/4 and pi/2
So if you were to show that cos(npi/7) / (tan-1(n))^n
is less than or more than some simpler series, then you can use the comparison test to show that the series must converge or diverge.
2007-04-12 17:30:32 · answer #1 · answered by z_o_r_r_o 6 · 0⤊ 0⤋
I a_nI < =1/((tan^-1n)^n
tan^-1n has limit pi/2 and is increasing as the function
tan^-1 x has a derivative 1/(1+x^2) always positive
so from some n on
tan^-1n >pi/3 and 1/ (tan^-1n)^n <(3/pi)^n
so Ia_n<(3/pi)^n and (3/pi)^n is a geometric series with
r= 3/pi< 1 convergent
so the given series is absolutely convergent
2007-04-13 01:13:26 · answer #2 · answered by santmann2002 7 · 0⤊ 0⤋