Convergence and limit of <munderover> &#x2211;<!-- ∑ --> <mrow class="MJX-TeXAtom-ORD"

veirarer

veirarer

Answered question

2022-06-10

Convergence and limit of j = 1 n sin ( j 1 n ) { cos ( j n ) cos ( j 1 n ) }

Answer & Explanation

mar1nerne

mar1nerne

Beginner2022-06-11Added 20 answers

We have:
lim n + j = 1 n sin ( j 1 n ) [ cos ( j n ) cos ( j 1 n ) ] = sin ( 2 ) 2 4 .
Sketch of proof: we have:
cos ( j n ) cos ( j 1 n ) = j 1 n j n ( sin x ) d x = 1 n sin ( j 1 n ) + O ( 1 n 2 ) ,
hence our limit is the same as:
lim n + j = 1 n 1 n sin 2 ( j 1 n ) = 0 1 sin 2 ( x ) d x .

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