A calculus problem by Rindell Mabunga

Calculus Level 4

Find the value of

lim z [ x = 0 y = 1 z y x + 1 x ! ( z x + 2 ) ] \lim_{z \to \infty} \left[\sum_{x = 0}^{\infty} \sum_{y = 1}^{z} \frac{y^{x + 1}}{x!(z^{x + 2})}\right]


The answer is 1.000.

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1 solution

Rindell Mabunga
Oct 8, 2015

lim z x = 0 [ y = 1 z y x + 1 x ! ( z x + 2 ) ] \huge \lim _{ z\rightarrow \infty }\sum _{ x=0 }^{ \infty }[{ \sum _{ y=1 }^{ z }\frac { y^{x + 1} }{ x!(z^{x+2}) } }]

= lim z y = 1 z [ x = 0 y x + 1 x ! ( z x + 2 ) ] \huge =\lim _{ z\rightarrow \infty } \sum _{ y=1 }^{z}[\sum _{ x=0 }^{ \infty }{\frac { y^{x + 1} }{ x!(z^{x+2}) } }]

= lim z y = 1 z [ x = 0 ( y z 2 × ( y z ) x x ! ) ] \huge =\lim _{ z\rightarrow \infty } \sum _{ y=1 }^{ z }[\sum _{ x=0 }^{ \infty }{(\frac{y}{z^2} \times \frac{(\frac{y}{z})^x}{x!}) }]

= lim z y = 1 z [ ( y z 2 ) x = 0 ( ( y z ) x x ! ) ] \huge =\lim _{ z\rightarrow \infty } \sum _{ y=1 }^{ z }[(\frac{y}{z^2})\sum _{ x=0 }^{ \infty }{(\frac{(\frac{y}{z})^x}{x!}) }]

= lim z y = 1 z [ ( y z 2 ) ( e y z ) ] \huge =\lim _{ z\rightarrow \infty } \sum _{ y=1 }^{ z}[(\frac{y}{z^2})(e^{\frac{y}{z}})]

= lim z y = 1 z [ ( 1 z ) ( y z ) ( e y z ) ] \huge =\lim _{ z\rightarrow \infty } \sum _{ y=1 }^{ z }[(\frac{1}{z})(\frac{y}{z})(e^{\frac{y}{z}})]

= lim z ( 1 z ) y = 1 z [ ( y z ) ( e y z ) ] \huge =\lim _{ z\rightarrow \infty }(\frac{1}{z}) \sum _{ y=1 }^{ z }[(\frac{y}{z})(e^{\frac{y}{z}})]

= 0 1 y e y d y \huge = \int _{ 0 }^{ 1 }{ y{ e}^{ y }dy }

= y e y e y ] 0 1 \huge = ye^y - e^y]_{0}^1

= 1 \huge = \boxed{1}

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