What is the limit of the following series and is it conditionally or absolutely convergent?

Question

∞

∑(-2/arctan(n))^n

n=1

Answer 1

It diverges. You can do the root test:

r = |a_n|^1/n with r <1 converges

r is 2/atan(n) which is approaching 2/(pi/2) = 4/pi as n goes to infinity (>1)

For conditional convergence, you need to show that ( -2/arctan(n))^n is decreasing as it alternates sign, but the -2/arctan term has an absolute value >1, so this is not so.

Please consider a tutor. Take care.

Answer 2

The conditional sum is an alternating series of positive and negative terms, Make (-2)ⁿ into (-1)ⁿ2ⁿ.

These series to infinity only converge if terms tend to zero as n increases.

Since arctan(n) is going to valued approx. 1.57 for large n and numerator is

growing, there is no limit. The sum swings back and forth positive to negative.

Using absolute value positive terms has sum diverging to infinity.