Using Dirichlet series test I proved that the series <mstyle displaystyle="true" scriptlevel="0">

Logan Wyatt 2022-07-04 Answered
Using Dirichlet series test I proved that the series n = 2 sin n x n log n converges for all x R
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engaliar0l
Answered 2022-07-05 Author has 13 answers
Let { a n } be a decreasing sequence of real numbers such that n a n 0. Then the series n 2 a n sin ( n x ) is uniformly convergent on R
Thanks to Abel transform, we can show that the convergence is uniform on [ δ , 2 π δ ] for all δ > 0. Since the functions are odd, we only have to prove the uniform convergence on [ 0 , δ ]. Put M n := sup k n k a k , and R n ( x ) = k = n a k sin ( k x ). Fix x 0 and N such that 1 N x < 1 N 1 . Put for n < N
A n ( x ) = k = n N 1 a k sin k x  and  B n ( x ) := k = n + a k sin ( k x ) ,
and for A n ( x ) := 0
Since | sin t | t for t 0 we have
| A n ( x ) | k = n N 1 a k k x M n x ( N n ) N n N 1 M n ,
so | A n ( x ) | M n
If N > n, we have after writing D k = j = 0 k sin j x on ( 0 , δ ] for some constant c. Indeed, we have | D k ( x ) | 1 2 ( 1 cos x ) and cos x = 1 x 2 2 ( 1 + ξ ) where | ξ | 1 2 so 2 ( 1 cos x ) x 2 2 and | D k ( x ) 2 x . Therefore
| B n ( x ) | 2 x k = N + ( a k a k + 1 ) + a N 2 x = 2 2 x a N 2 2 N a N 2 2 M n .

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