X.B.5 Tests for Divergence and Convergence.  ( in  Progress)

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© 2000 M. Flashman

In this section we'll consider some elementary and direct tests for determining the question of convergence for an infinite series. These tests will be applied to examples of power series in an attempt to determine the intervals on which the power series may converge called intervals of convergence.

Theorem X.B.1 (The Divergence Test).  Suppose that `sum_{k=0}^{oo}c_k`  converges to S.
Then `lim_{k->oo}c_k = 0`.
 Proof: Let `S_n = c_0 + c_1 + c_2 + ... + c_k + ... + c_n = sum_{k=0}^{k=n}c_k`  so that `S_n =S_{n-1} + c_n`. 
By the hypothesis we have that as `n->oo`,  `S_n-> `S and  also `S _{n-1}->S` as well. But  `c_n  = S_n - S_{n-1}`,  so as `n->oo` ,

`c_n = S_n - S_{n-1}-> S - S =  0`.   EOP


  1. This result is called the divergence test because its use is primarily to show that an infinite series does not converge, i.e., that it diverges.
  2. Once again, don't forget the harmonic series. It shows that the fact that `c_k ->0` is only a necessary consequence of the convergence of a series, but is not sufficient by itself as a condition to guarantee that a series converges.
  3. There is a test that applies to alternating series which is almost a converse for this test. Review the previous discussion on the alternating harmonic series and see if you can state and justify such a result.
  4. The logic of necessary and sufficient conditions is sometimes confusing in the way that these conditions are expressed. One traditional form is to say, "If condition A is true, then condition B is true" or more simply: "If A then B."  B is decribed as the "necessary" condition, while A is described as the "sufficient" condition. Knowing that  condition A is  true is sufficient to infer that condition B is also true.  Condition B is a necessary  condition - without it being true, condition A cannot be true. This gives us the equivalent contrapositve statement  relating conditions A and B, namely, " If  condition B is not true , then condition A is also not true" or more simply "If  not  B, then not A."
    Try thinking these connections through with the following choices for A and B.
    A:  Ben earned a grade of A in the calculus course.  B: Ben passed the final examination in the calculus course.
    A: Michael earned a grade of A in all his high school courses.   B: Michael earned a grade of A in his sophomore High school English course.
    A: Jennifer  earned a doctorate degree in Sociology.     B: Jennifer passed the qualifying exminations  for  the doctorate degree in Sociology.
    A: The function `f`  is differentiable at `x=a`.  B: The function `f` is continuous at `x=a`.

Example X.B.7.  (a)   `sum_{k=0}^{oo} k/{5k + 100}`  diverges.
(b) If  `| x | >= 1` then `sum_{k=0}^{oo} x^k`  diverges.

Discussion.  (a) Apply the theorem to `c_k =  k/{5k + 100}`. As `k->oo`, `c_k->1/5 > 0`.  Therefore the series cannot converge.
(b) Since `| x |  >= 1`, the sequences `| x ^k| ` and `x^k` do  not approach `0`. [The term `|x^k|` is always at least `1`.]  Hence the series will not converge for those values of `x`.

The next test gives a very useful general test for convergence. Its greatest use for us will be to find more mechanical tests for convergence.

Theorem X.B.2 (The Foundation Test for Positive Series. "Bounded if and only if convergent").
Suppose that `c_k > 0`  for all `k`, then `S_n`  converges if and only if there some number `B`, for which 

`S_ n = sum_{k=0}^{k=n} c_k  < B`  for any `n`.

Proof: Notice first that `S_n = S_{n-1}+ c_n`   so  `S_n >S_{n-1}`  for all `n`. Thus the sequence `\{S_n\}` is an increasing sequence and from our previous discussion on sequences, the sequence converges if and only if there some number `B`, `S_n < B`  for any `n`.

We can apply this foundational result about positive series to justify three useful tests for convergence.

Theorem X.B.4 (The Comparison Tests for Positive Series). Suppose that `c_k  > d_k > 0`  for all `k`.

(i) If  `sum_{k=0}^{oo} c_k` converges then  `sum_{k=0}^{oo} d_k`  converges.
    (ii) If  `sum_{k=0}^{oo} d_k` diverges then `sum_{k=0}^{oo} c_k`  diverges.
Proof: First notice that since `c_k > d_k > 0`  for all `k` ,
`T_ n = d_0 + d_1 + d_2 + ... + d_ k + ... + d_ n < S_n = c_0 + c_1 + c_2 + ... + c_k + ... + c_n` for all `n`.
(i) Suppose   `sum_{k=0}^{oo} c_k`  converges. Then by X.B.3, there is some number `B` where `B>S_ n >T_ n` for all `n`. Now applying X.B.3 to the sequence` T_n` completes the argument.
(ii) Suppose   `sum_{k=0}^{oo} d_k` diverges . Then by part (i)  `sum_{k=0}^{oo} c_k` can not  converge. Therefore  `sum_{k=0}^{oo} c_k` diverges.

Example X.B.8: Consider the series (i) `sum_{k=1}^{oo}  (.5)^k` , (ii) `sum_{k=1}^{oo}{(.5)^k}/{k+1}`,
(iii) `sum_{k=1}^{oo}1/k` , and (iv) `sum_{k=1}^{oo} {3k}/{k^2+1}`.

For (i) and (ii) it should be clear that  `{(.5)^k}/{k+1}< (.5)^k`  for `k=1,2,3, ...`, and `sum_{k=1}^{oo} (.5)^k`   is a geometric series that converges. So by the comparison test, `sum_{k=1}^{oo}{(.5)^k}/{k+1}` converges also .
On the other hand  for `k= 1,2,3,...` , `1/k <{3k}/{k^2+1}`. [This is justified by observing that `k^2+1<3k^2`.]
Since `sum_{k=1}^{oo}1/k` diverges, we can apply X.B.3 to see that `sum_{k=1}^{oo} {3k}/{k^2+1}` diverges as well.

Theorem X.B.5 (The Integral Tests for Positive Series). Suppose that `c_k > 0`  for all `k` and for some decreasing continous function `f` , `f (k) = c_k`. Then the series `sum_{k=1}^{oo} c_k` converges if and only if  `int_{1}^{oo} f(x) dx` converges.
Proof: Consider the following inequalities based on the fact that `f(k)=c_k` and  `f` is a continuous decreasing function.

 `c_2 < int _1^2 f (x) dx < c_1`
  `c_3 < int_2^3 f (x) dx < c_2`
  `c_4 < int_3^4 f (x) dx < c_3`
`c_n< int_{n-1}^n f (x) dx <c_{n-1}`
So `sum_{k=1}^{k=n} c_k =c_1 +c_2 +... + c_n < c_1 + int_1^2 f (x)dx + ...+ int_{n-1}^n f (x)dx = c_1 + int_1^n f (x)dx` (*)
` int_1^n f (x)dx  = int_1^2 f (x)dx+ ... +int_{n-1}^n f (x)dx <c_1 +c_2 +... + c_{n-1}= sum_{k=1}^{k=n-1} c_k`. (**)

(i) Suppose `int_1^{oo}f (x) dx` converges to `I`. Then by (*) `sum_{k=0}^{k=n}c_k < I` for all `n` and the series converges.
(ii) Suppose  `int_1^{oo}f (x) dx` diverges.  `I_n = int_1^n f (x) dx >0` and is increasing for all `n`. Since the integral diverges, these integrals, `I_n` and the corresponding the partial sums `S_n =sum_{k=0}^{k=n}c_k` are not bounded. Therefore the series diverges.


Exanple X.B.9. Consider the series (i) `sum_{k=1}^{oo} 1/{k^2}` and (ii) `sum_{k=1}^{oo} {2k}/{k^2+1}`.
(i) The function `f(x) = 1/{x^2}` satisfies the hypothesis of the theorem. Thus we consider
`int _1^{oo} 1/{x^2} dx = lim_{B->oo}int_1^B 1/{x^2} dx = lim_{B->oo}- 1/x| _1^B =  lim_{B->oo}1 - 1/B = 1`.
Since the integral converges, we can apply the theorem to conclude that the series `sum_{k=1}^{oo} 1/{k^2}` also converges.
(ii) The function `f(x) = {2x}/{x^2+1}` satisfies the hypothesis of the theorem. Thus we consider
`int _1^{oo} {2x}/{x^2+1} dx = lim_{B->oo}int_1^B {2x}/{x^2+1} dx = lim_{B->oo}ln(x^2+1) | _1^B =  lim_{B->oo}ln(B^2+1)-ln(2) = oo`.
Since the integral diverges, we can apply the theorem to conclude that the series `sum_{k=1}^{oo} {2k}/{k^2+1}` also diverges.

This next theorem compares a positive series with a geometric series. In practice this is one of the most important tools we will develop for determining when a series converges or diverges.

Theorem X.B.6 (The Ratio Test for Positive Series).
  Suppose that `c_k>0`  for all `k` . Suppose that as  `k->oo`, the ratios  `{c_{k +1}}/{c_k}-> R`.
(a) If  `R < 1` then the series `sum_{k=0}^{oo}c_k` converges.
(b) If  `R > 1`  or does not exist, there is no limit then the series `sum_{k=0}^{oo}c_k`  diverges.
(c) If  `R = 1` then this test can give no information on the series convergence.


(b) We'll begin by considering the case when  `R>1`.
In this situation when `k`  is large, the ratios  `{c_{k +1}}/ {c_k}`  are close to the number `R`.  `R>1` means that there is a number `B >1` where when `k` is large  `{c_{k +1}}/{ c_k} > B >1`.
Choose a number `N` that is large enough so that the last inequality holds for all `k > N`.
Now since `c_N > 0` we can multiply  to obtain that `c_{N+1}>c_N * B >c_N > 0`.
In fact continuing this analysis for `c_{N+2}` we find that `c_{N+2} >c_[N+1} * B >c_N * B^2 > c_N > 0`.
In general you can see that for `k > N`,  `c_k > c_N>0`.
But this means that as `k->oo` ,  `c_k` cannot be approaching `0`. Therefore the series `sum_{k=0}^{oo}c_k` must diverge by the divergence test.

(a) Now to consider when `R < 1` we pursue a similar analysis. This time we see that there is some number `rho` with `rho < 1` where when `k` is large 
`0 <{c_{k +1}}/{c_k}< rho <1`.
Choose  `N`  large enough so that this last inequality is true for all `k > N`.
Then since  `{c_{N+1}}/{c_N}< rho <1`,   `c_{N+1}< c_N * rho <c_N` .
Similarly, `c_{N+2} < c_{N+1} * rho  < c_N * rho^2` .
Continuing we see that `c_{N+3}< c_{N+2} * rho < c_N * rho^3`.
In general we  have `c_{N+j}< c_{N+j-1}* rho< c_N * rho^j`.

Now we consider the geometric series `sum_{j=1}^{oo}c_N rho^j`.
This series converges to a number `L={c_N rho}/{1-rho}` because  `0 < rho < 1`. (At this point we are not concerned with the actual value of L, only that such a number exists!)
Let `B = L + sum_{k=0}^{N}c_k`.  You can now check that for all `n`,  `S_ n = sum_{k=0}^{k=n}c_k< B` . Thus by the comparison test, the series converges.


Example X.B.10.   Use the ratio test to determine the convergence of  `sum_{k=0}^{oo}{3^k}/{ k!}`.
Solution:Being systematic in applying the ratio test is often helpful in avoiding errors. Here is one way to keep your work organized in a step by step fashion:
Step 1: Write down `c_k`:`c_k = {3^k}/{ k!}`

Step 2:  Write down  `c_{k+1}`:. `c_{k+1}= {3^{k+1}}/ {(k+1)!}`

Step 3: Write down and simplify the ratio,`{c_{k +1}}/{c_k}`:
`{c_{k +1}}/{c_k}={{3^{k+1}}/ {(k+1)!}}/{{3^k}/{k!}}= {3^{k+1}*k!}/{3^k * (k+1)!}=3/ {k+1}`.

Step 4 : Think- find `R =lim_{k->oo}{c_{k +1}}/{c_k}`:
    Clearly as `k->oo`,   `{c_{k +1}}/{c_k}->0` so `R = 0`

Step 5:  Use the ratio test to draw a conclusion:  Since `R = 0` the series `sum_{k=0}^{oo}{3^k}/{ k!}` converges.

The Taylor polynomial of degree `n` for `e^x` evaluated at `3` is  `sum_{k=0}^{k=n}{3^k}/{ k!}`. It is an exercise at the end of this section to examine the error terms for these polynomials and thus show that `sum_{k=0}^{oo}{3^k}/{ k!}= e^3`.

Example X.B.11.
   The ratio test can show that the series `sum_{k=0}^{oo}{x^k}/{ k!}` converges for all positive values of x.
Following the work of the previous example we have
Step 1.`c_k = {x^k}/{ k!}`.
Step 2. `c_{k+1}= {x^{k+1}}/ {(k+1)!}`.
Step 3:  Write down and simplify the ratio:
`{c_{k +1}}/{c_k}={{x^{k+1}}/ {(k+1)!}}/{{x^k}/{k!}}= {x^{k+1}*k!}/{x^k * (k+1)!}=x/ {k+1}`.

Step 4: Think. Find R: Clearly as `k->oo`,   `{c_{k +1}}/{c_k}= x/ {k+1}->0`, so `R=0`.
Step 5. Use the ratio test: Since `R=0` for any `x>0`, the series `sum_{k=0}^{oo}{x^k}/{ k!}` converges for any positive ` x`.

Remark: The Ratio Test Meets the Divergence Test.
Combining the ratio test  with the divergence test can make a powerful tool for showing that a sequence converges to 0.
For instance, consider the last example in which the ratio test showed that `sum_{k=0}^{oo}{x^k}/{ k!}` converges for all positive values of x. Now by the divergence test we have that  `lim_{k->oo}{x^k}/{ k!}=0` for all positive values of x.
But `|t|>0` for any real number `t` so
`lim_{k->oo}{|t|^k}/{ k!}=0`.
And of course `|t|^k = |t^k|` so
`lim_{k->oo}{|t^k|}/{ k!}=0`. Therefore `lim_{k->oo}{t^k}/{ k!}=0` for any real number t.
This last statement is very useful- especially in showing that
`sum_{k=0}^{oo}{x^k}/{ k!} = e^x` for all `x`. [See exercise ***]

Example X.B.12.  Find all positive values of x for which the series `sum_{k=0}^{oo}{(2x)^k}/{ k+3}` converges.
Solution: Again being systematic in applying the ratio test is often helpful in avoiding errors. Here is the way we have been proceeding  in a step by step fashion:
Step 1: Write down `c_k`:` c_k = {(2x)^k}/{ k+3}`.
Step 2:  Write down  `c_{k+1}`:. `c_{k+1}= {(2x)^{k+1}}/ {(k+1)+3}`.
Step 3: Write down and simplify the ratio,`{c_{k +1}}/{c_k}`:
`{c_{k +1}}/{c_k}={{(2x)^{k+1}}/{(k+4)}}/{{(2x)^k}/{k+3}} = {(2x)^{k+1}*(k+3)}/{(2x)^k * (k+4)}=2x {k+3}/ {k+4}`.
Step 4 : Think- find `R =lim_{k->oo}{c_{k +1}}/{c_k}`:
Clearly as `k->oo`,   `{c_{k +1}}/{c_k}=2x {k+3}/ {k+4}-> 2x`, so `R=2x`.
Step 5: Analyze when `R<1` and when `R>1`:  `R=2x <1` when  `0<x<1/2` and `R>1` when `x>1/2`.
Step 6:  Use the ratio test to draw a conclusion:  The series `sum_{k=0}^{oo}{(2x)^k}/{ k+3}` converges for all `x` where `0<x<1/2 ` and diverges for all `x` where `x>1/2`.
Note: The ratio test does not apply when `R=1`. So when `x=1/2` a different analysis must be applied to the series:
`sum_{k=0}^{oo}{(2 1/2)^k}/{ k+3}= sum_{k=0}^{oo}1/{ k+3}`  which diverges. [This series is harmonic or use the integal test.]