Limit with a summation and sine: how to calculate lim n → ∞ n 2...

opepayflarpws

opepayflarpws

Answered

2022-06-26

Limit with a summation and sine: how to calculate lim n n 2 k = 0 n 1 sin ( 2 π k n ) ?

Answer & Explanation

laure6237ma

laure6237ma

Expert

2022-06-27Added 27 answers

k = 0 n 1 sin ( 2 π k / n ) is the imaginary part of k = 0 n 1 e i 2 π k / n . We can evaluate the latter sum directly since it's a truncated power series:
k = 0 n 1 z k = z n 1 z 1
provided that z 1, so
k = 0 n 1 e i 2 π k / n = e i 2 π 1 e i 2 π / n 1 = 1 1 e i 2 π / n 1 = 0
Hence also
k = 0 n 1 sin ( 2 π k / n ) = 0
for every positive integer n. So your limit is
lim n n 2 k = 0 n 1 sin ( 2 π k / n ) = lim n ( n 2 0 ) = lim n 0 = 0

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