TESTING INFINTE SERIES FOR CONVERGENCE 
TABLE OF CONTENTS
C/ The Test for Divergence: (n th term test)
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Our Integral Calculus course began with summations: more precisely, Riemann sums. Remember breaking up an area into a bunch of rectangles with as the base and f (x) for the height and then summing these areas?
We ended up with a sum limited to infinity that looked like this:
Then we learned the Fundamental Theorem of Calculus which turned our summations into integrals. The Sigma turned into an integral sign but we were still finding sums. They looked like this:
Many of them were over a closed interval for finding area between 2 curves or the volume of a solid of revolution. And then, just before we started infinite series, we covered l'Hopital's Rule (dealing with indeterminate forms) and Improper Integrals ( infinite limits of integration or infinite discontinuities on the interval of integration). This was the first time we tested for Convergence.
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An improper integral or an infinite series is CONVERGENT if we can assign a number value to the integral or to the sum of an infinite number of terms in the series.
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This lesson covers categories of infinite series and the tests we use to determine if they converge or diverge. Think of these tests as different colored litmus papers  like in chemistry when we want to find out if a liquid is an acid. We dip blue litmus paper in it  if it turns pink, we know we have an acid. We'll do the same with infinite series when we test them for convergence. Problem is, with litmus paper, there's only 2 colors  with tests for infinite series, there's a bunch. Learn their names, as well as how and when to apply them.
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The basic form for this convergent series is:
We will decompose the fraction and then express a number of terms to see what happens when we add them all up. Using partial fraction decomposition we get:
Now we write some of the terms setting n = 1, 2 , 3 ...
Notice how the second term of each bracket cancels the first term in the next bracket.
When we cancel all these terms, we're left with the first and last terms. This is why it's called the telescoping series. It's like one of those foldup telescopes the pirates had. When folded, it shows only the first and last segment of the tube.
Now we take so the series converges; the sum of infinite terms = 1.
Say we have to test the series .
When we write out some of the terms, it's obvious that this is the same series as above but it's missing the first 2 terms. This series is therefore convergent.
The sum of this series is 1 – (the sum of the first 2 terms):
The same thinking can be applied to other known series. For instance, we know that the harmonic series, is divergent, (we'll prove it later), so anything of that form such as will also be divergent.
C/ The Test for Divergence: (n th term test)
This is one of the first tests we should apply since it's simple and direct. We take the limit at infinity of the nth term. If that limit is not equal to zero, the series diverges.
This doesn't say that a limit of zero makes the series convergent.
It says if the limit is NOT ZERO the series is divergent and that's all it says!!!
If the limit at infinity = 0 we need to apply another test.
Examples: Apply the test for divergence and state your conclusion:
a)  b)  c)  d) 
lim at = 2 series diverges 
lim at = 0 we know nothing 
lim at = 0 we know nothing 
lim at series diverges 
For parts a) b) and c) we divide through by the highest power of n then set it = .
For part d), we just evaluate. Remember that e^{ – n} = 1 / e^{ n }.
Memory Tweak: In a fraction where the degree of top = degree of bottom, the limit at is the ratio of the coefficients of the highest power of the variable.
So the limit at infinity for
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This family of series is of the form
a + ar + ar² + ar³ + ...... + ar^{ n  1}
with a as first term and r the ratio of two consecutive terms.
Notice that, if we divide any term by the term that comes before, we get r.
Also notice that since we have r^{ 0} in the first term, we have r^{ n} – 1 in the last term.
A recent discovery revealed that Archimedes was doing infinite sums 300 before J.C.
Here is how he proved that the infinite sum of (1/2)^{n} = 1.
Say I have a pie which I cut in half. I give you one half, then cut the remaining half in half and give you half of that. You would then have ½ + ¼ . If I continue the process, you will end up with the whole pie and I'll be left with nothing. So the sum of the infinite series is 1.
Let's approach in the same way. Only now, I'm dishing out powers of 1/3 to two other people. This time, once there's no pie left to cut in thirds, 2 people will each have a half of the whole pie and I'll be left with nothing. So the sum of an infinite number of terms from this series must be ½.
it is a Geometric Series. If  r  < 1, the series converges to . 
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Examples: Write the 1st 4 terms of these Geometric series.
State the value of a and r and find the sum if it converges.
a)  b)  c) 
= 1 + 2/5 + 2²/5² + 2³/5³  
a = 1, r = 2/5 < 1 series converges

a = 3/e, r = 3/e, > 1 diverges.  a = 7/25, r = 7/5 > 1 series diverges 
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A pseries looks like this:
When p = 2 the terms are :
When p = ½ the terms are :
A pseries Converges if p > 1 ; Diverges if p œ 1. They are useful for comparison tests. 
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Since an infinite series is a sum of an infinite number of terms, and the area under a continuous function found by integration, is the limit at infinity of a sum of areas, the Integral test says this:
If function f is positive, continuous and decreasing for x 1, then the infinite series f (1) + f (2) + f (3) + ..... f (n) + converges if converges. 
If the series converges, the sum of infinite terms equals the integral's value.
Example
Use the Integral test to show that the harmonic series 1/n is divergent.
(We could use the pseries theorem, since p = 1).
We'll set which is positive for x 1, continuous (discontinuity is at x = 0),
and it is decreasing.
So we evaluate which of course is ln x which goes to infinity so diverges.
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Note: When using this test, we are required to state the 3 conditions: that the function is positive, continuous and decreasing. There are times that the limit at infinity will be an indeterminate form and we will have to apply l'Hopital's rule to evaluate it. For a memory tweak on improper integrals, see the lesson on that topic.
Example
Use the Integral test to determine the convergence or divergence of .
Since which is positive, continuous and decreasing for x 1.
Now, if we make a substitution of u = x², then du = 2x dx.
So the Integral becomes:
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The integral converges therefore so does the series.
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Test these infinite series for convergence.
State the test you use and if the series converges or diverges.
If it converges, find the sum.
1)  2) 
3)  4) 
5)  6) 
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1) nth term test, lim at inf. = ½, diverges.  2) Geometric series: a = 3, r = 1/5, converges to 15/4 = 3.75 
3) pseries: p = 5 > 1, converges. sum = integral value = 1/12 
4)use integral test on substitute u = x² + 1, so du = 2x dx Integral becomes , diverges 
5) 5(telescoping series missing first 2 terms) series converges to 5[1 – (1/2 + 1/6)] = 5/3  
6) Geometric series a = 7, r = 7/4 > 1 diverges. 
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