L19seriesTest.mws
Calculus II
Lesson 19: Convergence Tests for Infinite Series
We have seen what is meant by saying that an infnite series
converges, with sum s. (You should review that definition now if you do not remember it.) Unfortunately, there are very few series to which the definition can be applied directly; the most important is certainly the
Geometric Series
. In the same section, it was shown that if
converges, then
.
The converse of this statement is false:
if
, it does
not
follow that
converges! (What example have you seen that shows this?) The idea to bear in mind is that
will converge if
goes to 0 "quickly enough". It is not possible to give an exact meaning to "quickly enough" in the last sentence, but you have been or will be shown in class that
diverges, but
converges for any positive
(the

series
), so
just fails to be fast enough. (You may recall a similar situation with improper integrals.)
We now study some basic tests we can use to determine whether a series converges or diverges.
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A. The Integral Test & P Series
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There are many occasions where a series is closely related to an integral. In these cases,the integral will converge or diverge if and only if the corresponding sum converges or diverges. This result is calle the Integral Test. Remember a sequence is function defined on the domain of natural numbers. If it makes sense to define the same function for all positive real numbers, and that functioin is integrable ( able to be intergrated) , then the integral test will apply. Lets look at an example.
Here is a function, the sequence it defines, and a plot of the sequence and function.
>
f := x > 1/x^(3/2);
>
seq( evalf( f(k)), k = 1..12);
>
plot( [f(x), f( floor(x)), f(floor(x+1))], x = 1..12, thickness = [4,2,2], color = [blue, coral, red]);
Note that the blue curve is between the two step functions colored yellow and red. If curve encloses infinite area then so does the larger step function because at all times it is greater than or equal to the curve. On the other hand, if the curve encloses a finite area, then so does the smaller step function. Simple enough. However, the upper step function is the lower one shifted one unit to the right. In other words, the upper one is the same as the lower one plus one additional term of a1 = 1. Consequently, the series converge or diverge as the integral does.
>
Sum( f(k), k = 1..infinity): % = evalf(value(%));
>
Int( f(x), x = 1..infinity): % = value(%);
>
Sum( f(k+1), k = 1..infinity): % = evalf(value(%));
In this case, we can actually compute the integral and the sums. Note that all three converge, however they are not equal. The value of the integral is between the value of the sums from k = 1.. infinity and k = 2..infinity  just as we saw on the graph above. Thus the integral tells us about convergence & divergence but not exact values.
One important result that follows directly from the integral test, is that series of the form 1/k^p converge only for
p > 1, and diverges otherwise.
>
assume( q > 0); p := q + 1; Int( 1/x^p, x = 1..infinity ): % = value(%);
>
assume( q <= 0); p := q + 1; Int( 1/x^p, x = 1..infinity ): % = value(%);
Note that integral is finite for p > 1, and finite for p =< 1.
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B. The Limit Comparison Test
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Another important test is the Limit Comparison Test. A simplified version of this test asserts that if the ratio of the terms of two different sequences have a finite, nonzero limit, then both converge or both diverge. Lets examine why test works from a geometric point of view. Consider these three sequences.
bk is known to converge while ck is known to diverge. For now we are not sure if ak converges or diverges.
>
a := n > (3 + sqrt(n))/(n^2  sqrt(n) + 5);
>
b := n > n^(3/2);
>
c := n > 1/n;
Lets look at what is happening graphically. The red graph is subject ak, while bk is blue and ck is green.
>
plot( [a(floor(x)), b(floor(x)), c(floor(x)) ], x = 2..5, color = [red, blue,green]);
From this graph it appears that ak is larger than both bk and ck. And since ck diverges, ak might diverge also!
>
plot( [a(floor(x)), b(floor(x)), c(floor(x)) ], x = 5..20, color = [red, blue,green]);
That theory was premature. We now see that things are changing.
Ultimately, ak and bk are more or less in sync. We can see this we apply the Limit Comparison test and take the limits of the terms. This does prove the limit comparison test, but makes it plausible.
>
limit( a(n)/b(n), n = infinity);
>
limit( b(n)/c(n), n = infinity);
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C. The Ratio Test
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More generally, the ratio of consecutive terms is an expression. The ratio test requires us to take the limit of the absolute value of this ratio. When this limit is strictly less than 1, the series converges absolutely.
Another important test is the Ratio test. In this test, we take the limit of the absolute value of consecutive terms. If
>
restart;
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Sum( a*r^n, n = 0..infinity): % = value(%);
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a := n > (6/7)^n;
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a(n+1)/a(n);
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simplify(%);
>
Sum( a(n), n = 0..infinity): % = value(%);
Here we create a generic geometric series, then look at an example for an = (6/7)^n. We take the ratio of consecutive terms, which is a constant of course. We then verify that the series converges.
>
a := n > (2^n) / n!;
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a(n+1)/a(n);
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abs( simplify(%));
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R := limit( %, n = infinity);
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Sum( a(n), n = 0..infinity): % = value(%);
If the limit of the ratio is 1, the test is inconclusive.
The ratio limit is 1 in this case, and the series diverges. However this is not always the case.
>
a := n > 1/sqrt(n);
>
abs( a(n+1)/a(n) );
>
simplify(%);
>
limit( %, n = infinity);
>
Sum ( a(n), n=1..infinity): % = evalf( value(%));