The unusual infinity

Calculus Level 5

lim ⁡ x → ∞ x − x 2 [ ( x + 1 ) ( x + 1 2 ) ( x + 1 2 2 ) … ( x + 1 2 x − 1 ) ] x = ? \lim_{x\to\infty} x^{-x^2} \bigg[ (x+1)\left(x+\frac12\right)\left(x+\frac1{2^2} \right)\ldots \left(x + \frac1{2^{x-1}} \right) \bigg]^x = \ ?

Give your answer to 3 decimal places.


The answer is 7.389.

Pawan Dogra by pawan dogra , 21, India

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2 solutions

Ayush Verma
Jul 23, 2015

L = l i m x → ∞ x − x 2 [ ( x + 1 ) ( x + 1 2 ) ( x + 1 2 2 ) . . . ( x + 1 2 x − 1 ) ] x = l i m x → ∞ [ ( 1 + 1 x ) ( 1 + 1 2 x ) ( 1 + 1 2 2 x ) . . . ( 1 + 1 2 x − 1 x ) ] x = l i m x → ∞ [ ∏ r = 1 r = x ( 1 + 1 2 r − 1 x ) ] x = l i m x → ∞ [ ∏ r = 1 r = x ( 1 + 1 2 r − 1 x ) x ] ∵ l i m x → ∞ ( 1 + 1 2 r − 1 x ) x = e 1 2 r − 1 ∴ L = l i m x → ∞ [ ∏ r = 1 r = x e 1 2 r − 1 ] = l i m x → ∞ e 1 + 1 2 + 1 2 2 + . . . + 1 2 x − 1 = e 1 1 − 1 2 = e 2 = 7.389 L=\underset { x\rightarrow \infty }{ lim } { x }^{ { -x }^{ 2 } }{ \left[ \left( x+1 \right) \left( x+\cfrac { 1 }{ 2 } \right) \left( x+\cfrac { 1 }{ { 2 }^{ 2 } } \right) ...\left( x+\cfrac { 1 }{ { 2 }^{ x-1 } } \right) \right] }^{ x }\\ \\ =\underset { x\rightarrow \infty }{ lim } { { \left[ \left( 1+\cfrac { 1 }{ x } \right) \left( 1+\cfrac { 1 }{ 2x } \right) \left( 1+\cfrac { 1 }{ { 2 }^{ 2 }x } \right) ...\left( 1+\cfrac { 1 }{ { 2 }^{ x-1 }x } \right) \right] } }^{ x }\\ \\ =\underset { x\rightarrow \infty }{ lim } { \left[ \prod _{ r=1 }^{ r=x }{ \left( 1+\cfrac { 1 }{ { 2 }^{ r-1 }x } \right) } \right] }^{ x }=\underset { x\rightarrow \infty }{ lim } { \left[ \prod _{ r=1 }^{ r=x }{ { \left( 1+\cfrac { 1 }{ { 2 }^{ r-1 }x } \right) }^{ x } } \right] }\\ \\ \because \underset { x\rightarrow \infty }{ lim } { \left( 1+\cfrac { 1 }{ { 2 }^{ r-1 }x } \right) }^{ x }={ e }^{ \cfrac { 1 }{ { 2 }^{ r-1 } } }\\ \\ \therefore L=\underset { x\rightarrow \infty }{ lim } \left[ \prod _{ r=1 }^{ r=x }{ { e }^{ \cfrac { 1 }{ { 2 }^{ r-1 } } } } \right] =\underset { x\rightarrow \infty }{ lim } { e }^{ 1+\cfrac { 1 }{ 2 } +\cfrac { 1 }{ { 2 }^{ 2 } } +...+\cfrac { 1 }{ { 2 }^{ x-1 } } }\\ \\ ={ e }^{ \cfrac { 1 }{ 1-\cfrac { 1 }{ 2 } } }={ e }^{ 2 }=7.389

8 Helpful 0 Interesting 0 Brilliant 0 Confused

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Unfortunately, this is not true. When x x approaches infinity, x + 1 , x + 1 2 , x + 1 4 , … x + 1, x + \frac12, x + \frac14, \ldots are STILL all distinct values. So you have shown that the upper bound is e^2, but you need to prove that the lower bound is e^2 as well. I think Power Mean (QAGH) should help. Then by Squeeze Theorem, we can only conclude that the limit is e^2.

Pi Han Goh - 5 years, 6 months ago

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