How do you use the Integral test on the infinite series #sum_(n=1)^oo1/sqrt(n+4)# ?

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Answer 1 · 2014-09-20T19:55:03

Since the integral

#int_1^infty 1/sqrt{x+4} dx#

diverges, the series

#sum_{n=1}^infty1/sqrt{n+4}#

also diverges by Integral Test.

Let us evaluate the integral.

#int_1^infty 1/sqrt{x+4} dx#

by the definition of improper integral,

#=lim_{t to infty}int_1^t 1/sqrt{x+4} dx#

by taking the antiderivative,

#=2lim_{t to infty}[sqrt{x+4}]_1^t#

#=2lim_{t to infty}(sqrt{t+4}-sqrt{5})=infty#