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Tuesday, 26 March 2013

LOCAL BEHAVIOR of FUNCTIONS PART 5


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 sequencedisplaymath131
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 integralsdisplaymath175
are all divergent while the improper integrals
displaymath177
are all convergent.
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