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Comparison Test or Limit Comparison Test
In the preceding section we saw how to relate a series to an improper integral to find out the convergence of a series. When the integral test is a good test, it does force us to do improper integrals that aren't all time easy and in some cases may be not possible to find out the convergence of.
For example consider the subsequent series.
To make use of the Integral Test we would have to integrate
∫∞0 1/ (3x + x) dx
and I'm not even make sure if it's possible to do this integral. Nicely sufficient for us there is other test that we can use on this series that will be very much easier to use.
1st, let's note that the series terms are positive. Like, with the Integral Test that will be significant in this section. Next let us note that we should have x > 0 as we are integrating on the interval 0 ≤ x < ∞. Similarly, regardless of the value of x we will all time have 3x > 0. Thus, if we drop the x from the denominator the denominator will get smaller and therefore the whole fraction will obtain larger.
Prove that a m + n + a m - n =2a m Ans: a m + n = a 1 + (m + n - 1) d a m-n = a 1 + (m - n -1) d a m = a 1 + (m-1) d Add 1 & 2 a m+n + a m-n =
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The sum of the series 1+1/2+1/4,..is
n(aubuc)..
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