Question

Convergence and divergence.... (see photo)

Answer 1

diverges because there is going to be a point where that denominator is going to 0

Answer 2

somewhere from
1 to infinity
there is a number that is a multiple of pi
and cos(pi)=-1

Answer 3

furthermore, u know it is never below -1 so there is no change of a negative infinity popping up anywhere, and there is only 1 real number from 1 to infinity that is a perfect multiple of pi... (really never happens, or happens once overinfinity) so cosn>=-1 ,
however there are going many multiples that are multiples of pi, to the 1st decimal.. 2nd decimal.. 3rd decimal to infinity decimal, so you will infact me adding many large numbers

Answer 4

That is similar as 1/n right? which is harmonic test (approaching 0 o.o?)so it's always divergent :o

Answer 5

it would help if the terms went to zero...

Answer 6

\[0\le1+\cos(n)\le2\\\frac{1}{2}\le\frac{1}{1+\cos(n)}\]
the series \(\sum_{n=1}^{\infty}\frac{1}{2}\) is divergent.
by comparison test, the given series is divergent

Answer 7

@dan815, \(\cos n+1\neq0\) if \(n\in\mathbb{N}\).