Does <munderover> <mo movablelimits="false">&#x2211;<!-- ∑ --> <mrow class="MJX-TeXAto

arridsd9

arridsd9

Answered question

2022-06-22

Does n = 1 cos 2 ( n + 1 ) n converge?

Answer & Explanation

humusen6p

humusen6p

Beginner2022-06-23Added 22 answers

n = 1 cos 2 ( n + 1 ) n clearly diverges. For any two consecutive integers, at least one of them is distant from a (non-integer) half-integer multiple of π by at least 1 2 , which is more than π 8 . That number of the pair will have a cosine greater than cos ( 3 π 8 ) in absolute value, and cos ( 3 π 8 ) = sin ( π 8 ) > 1 2 sin ( π 4 ) = 2 4 > 1 3 . Thus, any two consecutive terms cos 2 ( j + 1 ) j + cos 2 ( ( j + 1 ) + 1 ) j + 1 will contribute at least 1 9 ( j + 1 ) to the sum. Since j 1, we have 1 j + 1 1 4 j + 1 4 ( j + 1 ) , so 1 9 ( j + 1 ) 1 36 j + 1 36 ( j + 1 ) , i.e.
cos 2 ( j + 1 ) j + cos 2 ( ( j + 1 ) + 1 ) j + 1 1 36 j + 1 36 ( j + 1 )
But this means
n = 1 cos 2 ( n + 1 ) n 1 36 n = 1 1 n
which diverges.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?