1. Chapter 8
Infinite Series

2. Section 8.2
Convergence Tests

3. (Comparison Test) Theorem 8.2.1 Proof:
In this section we derive several tests that are useful in determining convergence.
(Comparison Test)
Let  an and  bn be infinite series of nonnegative terms. That is, an  0 and bn  0 for all n. Then
If  an converges and 0  bn  an for all n, then  bn converges.
If  an = +  and 0  an  bn for all n, then  bn = + .
Theorem 8.2.1
Proof:
Since bn  0 for all n, the sequence (tn) of partial sums of  bn is an increasing sequence.
In part (a) this sequence is bounded above by the sum of the series  an , so (tn) converges by the monotone convergence theorem (4.3.3).
Thus  bn converges.
In part (b) the sequence (tn) must be unbounded, for otherwise  an would have to converge.
But then lim tn = +  by Theorem 4.3.8, so that  bn = + . 

4. Example 8.2.2 Definition 8.2.4 Consider the series .
For all n  we have
In Example we saw that the series converges,
so must also converge.
Definition 8.2.4
If  | an | converges, then the series  an is said to converge absolutely (or to be absolutely convergent).
If  an converges but  | an | diverges, then  an is said to converge conditionally (or be conditionally convergent).

5. Theorem 8.2.5
If a series converges absolutely, then it converges.
Proof:
Suppose that  an is absolutely convergent, so that  | an | converges.
By the Cauchy criterion for series (Theorem 8.1.6), given any  > 0, there exists a natural number N such that n  m  N implies that | |  am | + … + |  an | | < .
But then
by the triangle inequality, so  an also converges. 
When the terms of a series are nonnegative, convergence and absolute convergence are really the same thing. Thus the comparison test can be viewed as a test for absolute convergence.
If we have a series  bn with some negative terms, the corresponding series | bn | will have only nonnegative terms, and we can use the comparison test on it.

6. If it happens that 0  |  bn  |  an for all n and if  an converges, then we can conclude that | bn | converges. That is,  bn converges absolutely.
In fact, changing the first few terms in a series will affect the value of the sum of the series, but it will not change whether or not the series is convergent.
So, given a nonnegative convergent series  an and a second series  bn, to conclude that  | bn | is convergent it suffices to show that 0  | bn |  an for all n greater than some N.
By using the comparison test and the geometric series, we can derive two more useful tests for convergence.

7. (Ratio Test) Theorem 8.2.7 Proof:
Let  an be a series of nonzero terms.
If lim sup | an + 1/an | < 1, then the series converges absolutely.
If lim inf | an + 1/an | > 1, then the series diverges.
Otherwise, lim inf | an + 1/an |  1  lim sup | an + 1/an |, and the test gives no information about convergence or divergence.
Theorem 8.2.7
Proof:
(a) Let lim sup | an + 1/an | = L.
If L < 1, then choose r so that L < r < 1.
By Theorem (a) there exists N  such that n  N implies that | an + 1/an |  r.
(If there were infinitely many terms | an + 1/an | greater than r, then a subsequence would converge to something greater than or equal to r, a contradiction to r being greater than the lim sup | an + 1/an |.)
That is, n  N implies | an + 1 |  r |an |.
So, | aN + 1 |  r |aN |, and | aN + 2 |  r |aN + 1 |  r2|aN |, etc.
It follows easily by induction that | aN + k |  rk |aN | for all k  .
Since 0 < r < 1, the geometric series   rk is convergent.
Thus  | aN | rk is also convergent,
and  an is absolutely convergent by the comparison test.

8. Theorem 8.2.7 (Ratio Test) Proof:
Let  an be a series of nonzero terms.
If lim sup | an + 1/an | < 1, then the series converges absolutely.
If lim inf | an + 1/an | > 1, then the series diverges.
Otherwise, lim inf | an + 1/an |  1  lim sup | an + 1/an |, and the test gives no information about convergence or divergence.
Proof:
(b) If lim inf | an + 1/an | > 1, then it follows that | an + 1| > | an | for all n sufficiently large.
Thus the sequence (an) cannot converge to zero, and by Theorem the series  an must diverge.
(c) This follows from the observation that the series  1/n2 converges and the harmonic series  1/n diverges.
In both series we have lim | an + 1/an | = 1. 

9. (Root Test) Theorem 8.2.8 Proof:
Given a series  an, let  = lim sup | an |1/n.
If  < 1, then the series converges absolutely.
If  > 1, then the series diverges.
Otherwise,  = 1, and the test gives no information about convergence or divergence.
Theorem 8.2.8
Proof:
(a) If  < 1, choose r so that  < r < 1.
By Theorem (a) we have
|  an |1/n  r for all n greater than some N.
That is, |  an |  r n for all n > N.
Since 0 < r < 1, the geometric series  r n is convergent,
so  an is absolutely convergent by the comparison test.
(b) If  > 1, then |  an |1/n  1 for infinitely many indices n.
That is, |  an |  1 for infinitely many terms.
Thus the sequence (an) cannot converge to zero and, by Theorem 8.1.5, the series  an must diverge.

10. Theorem 8.2.8 (Root Test) Proof:
Given a series  an, let  = lim sup | an |1/n.
If  < 1, then the series converges absolutely.
If  > 1, then the series diverges.
Otherwise,  = 1, and the test gives no information about convergence or divergence.
Proof:
(c) follows from considering the convergent series  1/n2 and the divergent series  1/n.
In Example we showed that lim n1/n = 1.
Thus,
Similarly, lim |1/n|1/n = 1, so the root test yields  = 1 for both series.
Thus when  = 1 we can draw no conclusion about the convergence or divergence of a given series. 
Note: This
is very useful in applying the root test to other sequences.

11. (Integral Test) Theorem 8.2.13 Proof:
Let f be a continuous function defined on [0, ), and suppose that f is positive and decreasing. That is, if x1 < x2, then f (x1)  f (x2) > 0. Then the series  f (n) converges iff
exists as a real number.
Theorem
Proof:
Let an = f (n) and bn =
Since f is decreasing, given any n  we have
f (n + 1)  f (x)  f (n) for all x  [n, n + 1].
Since the length of each subinterval is 1, it follows that
Geometrically, this means that the area under the curve y = f (x) from x = n to x = n + 1 is between f (n + 1) and f (n).
Thus 0 < an +1  bn  an for each n.
By the comparison test applied twice,  an converges iff  bn converges.
y = f (x)
But the partial sums of  bn are the integrals ,
so  bn converges precisely when
exists as a real number. 
Area is
f (n + 1)
Area is
f (n) n
n + 1

12. So, the p-series  1/n p converges if p > 1 and diverges if p  1.
Any series of the form  1/n p , where p  , is called a p-series.
Example
It is easy to see that the ratio and root tests both fail to determine convergence.
But, we can use the integral test.
For p  1, we have
The limit of this as n   will be finite if p > 1 and infinite if p < 1.
Thus by the integral test  1/n p converges if p > 1 and diverges if p < 1.
When p = 1, we get the harmonic series, which is divergent.
So, the p-series  1/n p converges if p > 1 and diverges if p  1.
This can be useful when applying the comparison test.

13. (Alternating Series Test)
If the terms in a series alternate between positive and negative values, the series is called an alternating series.
Our final test gives us a simple criterion for determining the convergence of an alternating series.
Theorem
(Alternating Series Test)
If (an) is a decreasing sequence of positive numbers and lim an = 0, then the series  (1)n +1an converges.
Proof:
The idea of the proof is to show that the sequence (sn) of partial sums converges by considering two subsequences:
the subsequence (s2n) coming from the sum of an even number of terms, and
the subsequence (s2n+1) coming from the sum of an odd number of terms.
In the text there are several examples of applying the various convergence tests to specific series.