"How to find the limitlimn→∞∏j=1n(1+1j(cos⁡(tjn)−1))" was answered by our experts

Travis Caldwell

Answered question

2022-04-23

How to find the limit

lim_{n \to \infty} \prod_{j = 1}^{n} (1 + \frac{1}{j} (\cos (\frac{t j}{n}) - 1))

Tatairfzk

Beginner2022-04-24Added 12 answers

Put

a_{n} = \prod_{j = 1}^{n} (1 + \frac{1}{j} (\cos (\frac{t j}{n}) - 1))

. Then

b_{n} = {\ln a}_{n} = \sum_{j = 1}^{n} \ln (1 + \frac{1}{j} (\cos (\frac{t j}{n}) - 1))

Using the inequality

x - \frac{x^{2}}{2} \leq \ln (1 + x) \leq x

for

x ≻ 1

, we get

\sum_{j = 1}^{n} \frac{1}{j} (\cos (\frac{t j}{n}) - 1) - {\frac{12}{\sum}}_{j = 1}^{n} \frac{1}{j^{2}} {(\cos (\frac{t j}{n}) - 1)}^{2} \leq b_{n} \leq \sum_{j = 1}^{n} \frac{1}{j} (\cos (\frac{t j}{n}) - 1)

Put

c_{n} = \sum_{j = 1}^{n} \frac{1}{j} (\cos (\frac{t j}{n}) - 1)

c_{n} = \frac{1}{n} \sum_{j = 1}^{n} \frac{n}{j} (\cos (\frac{t j}{n}) - 1) = \frac{1}{n} \sum_{k = 1}^{n} f (\frac{k}{n})

with

f (x) = \frac{\cos (t x) - 1}{x}

, which can be extended by continuity to

[0, 1]

. Hence we get

lim_{n \to \infty} c_{n} = \int_{0}^{1} f (x) d x = \int_{0}^{1} \frac{\cos (t x) - 1}{x} d x

. Now, since

\sum_{j = 1}^{n} \frac{1}{j^{2}} {(\cos (\frac{t j}{n}) - 1)}^{2} = \frac{1}{n^{2}} \sum_{j = 1}^{n} \frac{n^{2}}{j^{2}} {(\cos (\frac{t j}{n}) - 1)}^{2}

\frac{1}{n}

times a Riemann sum of a continuous function (after extension), it's limit

n \to \infty

is 0, and we conclude by the squeeze theorem.

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