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Question Number 44473 by abdo.msup.com last updated on 29/Sep/18

l e t A_{n} = \int_{0}^{\infty} s i n (n [t]) e^{- t} d t

2) c a l c u l a t e A_{n} a n d {l i m}_{n \to + \infty} n A_{n}

3) s t u d y t h e c o n v e r g e n c e o f \sum_{n} A_{n}

Commented by maxmathsup by imad last updated on 30/Sep/18

1) w e h a v e A_{n} = \sum_{k = 0}^{\infty} \int_{k}^{k + 1} s i n (n k) e^{- t} d t

= \sum_{k = 0}^{\infty} s i n (k n) \int_{k}^{k + 1} e^{- t} d t = \sum_{k = 0}^{\infty} s i n (k n) {[- e^{- t}]}_{k}^{k + 1}

= \sum_{k = 0}^{\infty} s i n (k n) {e^{- k} - e^{- (k + 1)}} = \sum_{k = 0}^{\infty} e^{- k} s i n (k n) - \sum_{k = 0}^{\infty} e^{- (k + 1)} s i n (k n)

b u t \sum_{k = 0}^{\infty} e^{- k} s i n (k n) = I m (\sum_{k = 0}^{\infty} e^{- k + i k n}) a n d

\sum_{k = 0}^{\infty} e^{- k + i k n} = \sum_{k = 0}^{\infty} e^{(- 1 + i n) k} = \frac{1}{1 - e^{- 1 + i n}} = \frac{1}{1 - e^{- 1} (c o s (n) + i s i n (n))}

= \frac{1}{1 - e^{- 1} c o s n - {i e}^{- 1} s i n (n)} = \frac{1 - e^{- 1} c o s (n) + {i e}^{- 1} s i n (n)}{{(1 - e^{- 1} c o s (n))}^{2} + e^{- 2} {s i n}^{2} n} \Rightarrow

\sum_{k = 0}^{\infty} e^{- k} s i n (k n) = \frac{e^{- 1} s i n (n)}{{(1 - e^{- 1} c o s n)}^{2} + e^{- 2} {s i n}^{2} n} a l s o

\sum_{k = 0}^{\infty} e^{- (k + 1)} s i n (k n) = e^{- 1} \sum_{k = 0}^{\infty} e^{- k} s i n (k n) = \frac{e^{- 2} s i n (n)}{{(1 - e^{- 1} c o s (n))}^{2} + e^{- 2} {s i n}^{2} n}

A_{n} = (1 - e^{- 1}) \frac{e^{- 1} s i n (n)}{1 - 2 e^{- 1} c o s n + e^{- 2}} = \frac{(e^{- 1} - d^{- 2}) s i n (n)}{1 - 2 e^{- 1} c o s (n) + e^{- 2}}

A_{n} = \frac{(e - 1) s i n (n)}{e^{2} - 2 e c o s (n) + 1} .