Find all values of α such that series ∑ n = 1 ∞ ( 1...

Frederick Kramer

Frederick Kramer

Answered

2022-07-07

Find all values of α such that series
n = 1 ( 1 n sin ( 1 / n ) cos ( 1 n ) ) α
converges.
I used Maclaurin for sin and cos and got:
a n = ( 1 1 1 3 ! n 2 + 1 + 1 2 ! n 2 1 4 ! n 4 + ) α
Put it together in one fraction seems to be a hard thing to do.

Answer & Explanation

Franco Cohen

Franco Cohen

Expert

2022-07-08Added 8 answers

HINT:
You are on the right track. Note that
1 1 1 6 n 2 + O ( 1 n 4 ) = 1 + 1 6 n 2 + O ( 1 n 4 )
Hence, we see that
1 n sin ( 1 / n ) cos ( 1 / n ) = 2 3 n 2 + O ( 1 n 4 )
skynugurq7

skynugurq7

Expert

2022-07-09Added 3 answers

Hint
n sin ( 1 n ) = 1 1 6 n 2 ( 1 + ϵ 1 ( n ) )
1 n sin ( 1 n ) = 1 + 1 6 n 2 ( 1 + ϵ 2 ( n ) )
cos ( 1 n ) = 1 1 2 n 2 ( 1 + ϵ 3 ( n ) )
thus, when n + , the general term of your series u n , satisfies
u n ( 2 3 n 2 ) α
and by the limit comparison test,
u n converges α > 1 2

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