Infinite sequences of a(little n)?

Question

an = integral from 0 to 1 of xe^(-nx)dx, n = 1,2...

a) use integration by parts to evaluate this integral explicitly

b) use the formula an obtained in part a to show that the limit as n approaches infinity of an = 0

Answer 1

By parts: Let u = x, then du = dx

Let dv = e^(-nx) dx, then v = (-1/n) * e^(-nx)

∫ xe^(-nx) dx

= (-x/n)*e^(-nx) + [ (1/n) * ∫ e^(-nx) dx ]

= (-x/n)*e^(-nx) - (1/n^2)*e^(-nx)

= [ (-x/n) - (1/n^2) ] * e^(-nx)

= (1/n) * (-x - (1/n)) * e^(-nx).

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For the definite integral, evaluate the above on [0,1] to get

a(sub n) = (1/n) * { [ (-1 - (1/n)) * e^(-n) ] + (1/n) }

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limit as n -> infinity (1/n) = 0.

limit as n -> infinity e^(-n) = 0, so

lim as n -> infinity [ (-1 - (1/n)) * e^(-n) ] = 0

Taken together, this implies that

lim as n -> infinity a(sub n) = 0.