The question is to compute: ( 1 + cos &#x2061;<!-- ⁡ --> A ) + 2 (

Leah Pope

Leah Pope

Answered question

2022-06-09

The question is to compute:
( 1 + cos A ) + 2 ( 1 + cos A ) 2 + 3 ( 1 + cos A ) 3 + = k = 1 k ( 1 + cos A ) k .
I tried by setting 1 + cos A = y, then the serie becomes
y + 2 y 2 + 3 y 3 + = k = 1 k y k
It's not a geometric progression as the coefficients are not in the series.
How can I go further?

Answer & Explanation

Judovh0

Judovh0

Beginner2022-06-10Added 16 answers

Using Ratio test,
lim n ( n + 1 ) ( 1 + cos A ) n + 1 n ( 1 + cos A ) n = ( 1 + cos A ) lim n ( 1 + 1 n ) = 1 + cos A
So, the series can only converge if | 1 + cos A | < 1 1 + cos A < 1 cos A < 0
For | y | < 1 ,, let
S n = r = 1 n r y r
y S n = r = 1 n r y r + 1
( 1 y ) S = y + y 2 + y 3 + = y 1 y

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