Lim Log cosx/2 .cosx/4…cosx/2^n x…>+∞?

The (n)th term of this sequence is

A(n) = log ( cos θ/2 )( cos θ/2² )( cos θ/2³ ) ... ( cos θ/2ⁿֿ¹ )( cos θ/2ⁿ )

. . . .= log [ 1/ ( 2 sin θ/2ⁿ ) ] (...)(...)(...) ... ( cos θ/2ⁿֿ¹ ) • [ 2. sin θ/2ⁿ. cos θ/2ⁿ ]

. . . .= log [ 1/ ( 2 sin θ/2ⁿ ) ] (...)(...)(...) ... ( cos θ/2ⁿֿ¹ ) • [ sin ( 2 · θ/2ⁿ ) ]

. . . .= log [ 1/( 2 sin θ/2ⁿ ) ] (...)(...)(...) ... ( cos θ/2ⁿֿ¹ ) • ( sin θ/2ⁿֿ¹ )

. . . .= ... continue in this manner from Right to Left ...

. . . .= log { ( sin θ ) / [ 2ⁿ · sin (θ/2ⁿ) ] } ........................................… (1)

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Note that, as n → ∞, θ is to be treated as a constant.

Also, ∵ | (1/2) | = (1/2) < 1 ∴ (1/2)ⁿ = 1/2ⁿ → 0 as n → ∞

∴ (θ/2ⁿ) → 0 as n → ∞ ........................................… (2)

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From (1) and (2), then,

... lim ( n → ∞ ) A(n)

= lim (n→∞) log { ( sin θ ) / [ 2ⁿ · sin (θ/2ⁿ) ] }

= log (( sin θ ) • 1/ { lim (n→∞) [ sin (θ/2ⁿ) ] / (1/2ⁿ) })

= log (( sin θ ) • 1/ { lim (n→∞) [ sin (θ/2ⁿ) / (θ/2ⁿ) ] } • 1/ θ )

= log ([ ( sin θ ) / (θ) ] • 1/ [ lim ( h → 0 ) ( sin h ) / h ]), ............. where h = (θ/2ⁿ)

= log ([ ( sin θ ) / (θ) ] • 1/ [ 1 ])

= log (( sin θ ) / (θ)) ........................ Ans.

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Happy To Help !

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not sure if you mean the argument is divided by 2^n or the expression cosx..or do you mean rpoduct or do u mean lim of cosx/2^n as n > infinity...anyway, suggest you look at eulers identity or if not , use series expansion of sinx but i think e^inx = cosnx + isinnx is the key here if you are allowed to use complex numbers....sorry cant help you more .....also there is the fact that lim x>infinity of sinx/x is 1