Improper Integrals and Series: The Integral Test

Improper integrals and series have a lot in common. The integral test bridges the two notions. Notice that series do possess tools which are not available for improper integrals (such as the ratio and root tests) and the improper integrals possess other tools not available for series (such as the techniques of integration). So depending on the nature of the problem, you may switch from one to the other one via the integral test.

The Integral Test. Consider a decreasing function tex2html_wrap_inline89 . Hence for any tex2html_wrap_inline91 , we have

displaymath93

which implies

displaymath95

Set tex2html_wrap_inline97 , then we have

displaymath99

for tex2html_wrap_inline91 . If we add these inequalities from n=1 to n=N, we get

displaymath107

If tex2html_wrap_inline109 is the sequence of partial sums associated to the series tex2html_wrap_inline111 , then we have

displaymath113

or equivalently

displaymath115

Since tex2html_wrap_inline117 for any tex2html_wrap_inline119 , then we know that

tex2html_wrap_inline121
the series tex2html_wrap_inline111 is convergent if and only if the sequence tex2html_wrap_inline109 is bounded;
tex2html_wrap_inline121
the improper integral tex2html_wrap_inline129 is convergent if and only if the sequence

displaymath131

is bounded.

Using the above inequalities, we conclude:

The Integral test

The improper integral tex2html_wrap_inline133 is convergent if and only if series tex2html_wrap_inline135 is convergent

Remark. Note that it may happen that f(x) is not decreasing on the entire interval tex2html_wrap_inline139 but only on some subinterval tex2html_wrap_inline141 (where A > 1). The above conclusion is still valid.

Example. Establish convergence or divergence of

displaymath145

Answer. Set

displaymath147

Then we have

Clearly the function tex2html_wrap_inline151 for tex2html_wrap_inline153 . Hence f(x) is decreasing on tex2html_wrap_inline141 . So the integral test implies that the improper integral

displaymath159

is convergent if and only if the series

displaymath161

is convergent. We recognize Bertrand's series. So we conclude that the improper integral

displaymath159

is convergent if and only if

tex2html_wrap_inline121
tex2html_wrap_inline167 or
tex2html_wrap_inline121
tex2html_wrap_inline171 and tex2html_wrap_inline173 .

So for example, the improper integrals

displaymath175

are all divergent while the improper integrals

displaymath177

are all convergent.

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Mohamed A. Khamsi
Tue Dec 3 17:39:00 MST 1996

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