#$&*
course Mth 174
7/26 8
Asst # 13
*********************************************
Question:
`q **** query problem 9.4.8 (4th edition 9.4.4 3d edition 9.3.6). Using a comparison test determine whether the series sum(1/(3^n+1),n,1,infinity) converges. **** With what known series did you compare this series, and how did you show that the comparison was valid?
......!!!!!!!!...................................
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Comparison Test:
We can condense a(n) = 1/(3^n+1) to b(n)= 1/3^n. 1/3^n converges.
The comparison test says that if is b(n) converges a(n) converges.
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
3
.............................................
Given Solution:
GOOD STUDENT SOLUTION WITH COMMENT: For large values of n, this series is similar to 1 / 3^n. As n approaches infinity, 1 / 3^n approaches 0. A larger number in the denominator means that the value of the function will be smaller. So, (1 / 3^n) > (1 / (3^n + 1)). We know that since 1 / 3^n converges, so does 1 / (3^n + 1).
COMMENT: We know that 1 / (3^n) converges by the ratio test: The limit of a(n+1) / a(n) is (1 / 3^(n+1) ) / (1 / 3^n) = 1/3, so the series converges.
We can also determine this from an integral test. The integral of b^x, from x = 0 to infinity, converges whenever b is less than 1 (antiderivative is 1 / ln(b) * b^x; as x -> infinity the expression b^x approaches zero, as long as b < 1).
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
OK
------------------------------------------------
Self-critique Rating:
OK
*********************************************
Question:
**** Query 9.4.14 (4th edition 9.4.10) (was 9.3.12). Use the ratio test to determine whether the series the series sum( 1 / (2 n) ! ) converges or diverges.
The text did not ask the following question, but this is covered in an assigned section so you should be able to answer: What is the radius of convergence of the series and how did you use the ratio test to establish your result? ****
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
a(n) = the sum which is = 1/(2n)!
a(n+1) = 1/(2(n+1)! = 1/(2n + 2)!
We take the limit (| a(n+1) | / | a(n) | , n, infinity)
1/(2n + 2)! / (1/(2n)!) = [2n!/(2n+2)! = 2n * (2n-1) * (2n-2) * (2n-3)… * 2 * 1] / [(2n + 2) * (2n + 1) * (2n) * (2n - 1)…* 2 * 1]
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
3
.............................................
Given Solution:
For 9.3.12
*&*& The ratio test takes the limit as n -> infinity of a(n+1) / a(n). If the limit is less than 1 then the series converges in much the same way as a geometric series sum(r^n), with r equal to the limiting ratio.
In this case a(n+1) = 1 / (2n + 2)! and a(n) = 1 / (2n) ! so
a(n+1) / a(n)
= 1 / (2n+2) ! / [ 1 / (2n) ! ]
= (2n) ! / (2n + 2) !
= [ 2n * (2n-1) * (2n-2) * (2n - 3) * ... * 1 ] / [ (2n + 2) * (2n + 1) * (2n) * (2n - 1) * ... * 1 ]
= 1 / [ (2n+2) ( 2n+1) ].
As n -> infinity this result approaches zero. Thus the series converges for all values of n, and the radius of convergence is infinite. *&*&
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
OK
------------------------------------------------
Self-critique Rating:
OK
*********************************************
Question:
Query problem 9.4.52 (3d edition 9.4.40) (was 9.2.24) partial sums of 1-.1+.01-.001 ... to what does the series converge?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
This pattern has factor .1^n or 1/10^n
The signs are changing back and forth so we use the alternating series:
where a(n) = 1/10^n
a(n + 1) = 1/10^(n + 1)
So,
1/10^(n + 1) < 1/10^n , and lim (1/10^n, n, infinity) = 0 , thus the series converges.
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
3
.............................................
Given Solution:
STUDENT SOLUTION:
We have:
an = 10^(-n)
and
an+1 = 10^(-(n+1))
So, since 0 < an+1 < an, this series converges.
INSTRUCTOR RESPONSE
Right idea, but 0 < a(n+1) < a(n) is not quite the correct test. The correct test requires absolute values, and the limiting value of the nth term must be zero.
Consider a different series with a(n) = (-1)^n * (.5 + 10^-n), giving us .5 - .51 + .501 - .5001 etc.. The partial sumsare .5, -.01, 4.91, -.091, etc.. The partial sums jump back and forth by about .5 units and never approach a limit.
What you have is an alternating series where | a(n) | -> 0. This is the criterion for convergence of an alternating series. *&*&
This is an alternating series with | a(n) | = .1^n, for n = 0, 1, 2, ... .
Thus limit{n->infinity}(a(n)) = 0.
An alternating series for which | a(n) | -> 0 is convergent.
sum(1/n^.999) diverges and sum(1 / n^1.001) converges, but doing partial sums on your calculator will never reveal this. The calculator is very limited in determining convergence or divergence.
However there is a pattern to the partial sums, which are 1, .9, .91, .909, .9091, .90909, ... . It's easy enough to show that the pattern continues, so the convergent value is .9090909... .
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
OK
------------------------------------------------
Self-critique Rating:
OK
*********************************************
Question:
**** Query 9.5.6. (3d edition 9.4.24). What is your expression for the general term of the series p x + p(p-1) / 2! * x^2 + p(p-1)(p-2)/ 3! * x^3 + …?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
The general term is the coefficient to the x^n.
p ( p - 1 ) ( p - 2 ) * ... * ( p - (n-1) ) / n !
a(n) = p ! / (n ! * (p - n) !)
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
1
.............................................
Given Solution:
** The general term is the coefficient of x^n.
In this case you would have the factors of x^2, x^3 etc. go down to p-1, p-2 etc.; the last factor is p minus one greater than the exponent of x. So the factor of x^n would go down to p - n + 1.
This factor would therefore be p ( p - 1) ( p - 2) ... ( p - n + 1).
This expression can be written as p ! / (p-n) !. All the terms of p ! after (p - n + 1) will divide out, leaving you the desired expression p ( p - 1) ( p - 2) ... ( p - n + 1).
The nth term is therefore (p ! / (p - n) !) / n! * x^n, of the form a(n) x^2 with a(n) = p ! / (n ! * (p - n) ! )
STUDENT QUESTION:
I can follow that the general term would be x^n because with each term of the series, the x power is decreasing by 1.
When I start looking at the other parts of the series I see that in the denominator n is decreasing by 1. For the exponent, n is decreasing by 1. On the numerator, n is increasing by 1. I can see where p(p-n)! comes from but I am having a hard time following from this point to the point where we find the nth term?????
INSTRUCTOR RESPONSE
@& The term
p(p-1) / 2 ! * x^2
is the n = 2 term, since x is raised to the power 2.
The term
p(p-1)(p-2) / 3 ! * x^3
is the n = 3 term.
What is different about the coefficients of these terms, and what the difference have to do with the power of n?
The first this we would notice is that the denominators for n = 2 and n = 3 are respectively 2 ! and 3 !.
Then we notice that the numerators p ( p - 1) and p ( p - 1 ) ( p - 2) for n = 2 and n = 3 consist respectively of 2 and 3 terms, with a clear pattern to the terms.
So we expect that the coefficients will continue as
p ( p - 1 ) ( p - 2 ) ( p - 3 ) / 4 !
p ( p - 1 ) ( p - 2 ) ( p - 3 ) ( p - 4 ) / 5 !
etc..
The denominator is equal to n !, as previously observed. The last number subtracted from p is one less than n, so the nth term would be
p ( p - 1 ) ( p - 2 ) * ... * ( p - (n-1) ) / n !
which can be written
p ( p - 1 ) ( p - 2 ) * ... * ( p - n + 1) ) / n ! .
p ( p - 1 ) ( p - 2 ) * ... * ( p - n + 1) )
is the same as
p ( p - 1 ) ( p - 2 ) * ... * ( p - n + 1) * (p - n) * (p - n - 1) * ... * 3 * 2 * 1 / ( (p - n) * (p - n - 1) * ... * 3 * 2 * 1 )
with the entire denominator matching the part of the numerator beyond the factor p - n + 1.
The numerator is still p ! and the denominator is (p - n) ! so
p ( p - 1 ) ( p - 2 ) * ... * ( p - n + 1) )
= p ( p - 1 ) ( p - 2 ) * ... * ( p - n + 1) * (p - n) * (p - n - 1) * ... * 3 * 2 * 1 / ( (p - n) * (p - n - 1) * ... * 3 * 2 * 1 )
= p ! / (p - n)!.
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
I didn’t understand why you divided by (p - n) * (p - n - 1)…
in the following part of the solution:
p ( p - 1 ) ( p - 2 ) * ... * ( p - n + 1) * (p - n) * (p - n - 1) * ... * 3 * 2 * 1 / ( (p - n) * (p - n - 1) * ... * 3 * 2 * 1 )
------------------------------------------------
Self-critique Rating:
@&
Having divided by that factor, we are left with a simple series, and can more easily express the general term.
*@
OK
*********************************************
Question:
**** Query 9.5.18 (was 9.4.18). What is the radius of convergence of the series x / 3 + 2 x^2 / 5 + 3 x^2 / 7 + …?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
a(n) = n / (2 n + 1)
a(n+1) = (n+1) / ( 2(n + 1) + 1 ) = (n+1) / (2n + 3)
Take the limit as n -> infinity:
a(n + 1 ) / a(n) = (n/(2n+ 1)) / ((n+1) / (2n + 3)) = (2n^2+ 5n + 3) / (2n^2 + n) = 1
Radius = 1
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
2
.............................................
Given Solution:
For this sequence we have the following coefficients:
a(1) = 1 / 3
a(2) = 2 / 5
a(3) = 3 / 7
The numerators for n = 1, 2, 3 are respectively 1, 2, 3, so the numerator of the nth term will be n.
The denominators for n = 1, 2, 3 are 3, 5 and 7. When n changes by 1 the denominator changes by 2, so one terms of the denominator will be 2 * n. The values of 2 * n are respectively 2, 4 and 6, which don't quite match the denominators 3, 5, 7. However if we add 1, we do get a match. So the denominator of the nth coefficient is 2 n + 1.
The coefficient of the nth term is thus
a(n) = n / (2 n + 1).
To find the radius of convergence you first find the limit of the ratio | a(n+1) / a(n) | as n -> infinity. The radius of convergence is the reciprocal of this limit.
a(n+1) / a(n) = ( n / (2n+1) ) / (n+1 / (2(n+1) + 1) ) = (2n + 3) / (2n + 1) * n / (n+1).
(2n + 3) / (2n + 1) = ( 1 + 3 / (2n) ) / (1 + 1 / (2n) ), obtained by dividing both numerator and denominator by 2n. In this form we see that as n -> infinity, this expression approaches ( 1 + 0) / ( 1 + 0) = 1.
Similarly n / (n+1) = 1 / (1 + 1/n), which also approaches 1.
Thus (2n + 3) / (2n + 1) * n / (n+1) approaches 1 * 1 = 1, and the limit of a(n+1) / a(n) is therefore 1.
The radius of convergence is the reciprocal of this ratio, which is 1.
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
I have a hard time understanding why you took the reciprocal of a(n+1) / a(n) which is a(n) / a(n+1).
I also wasn’t sure how you got (2n + 3) / (2n + 1) * n / (n+1).
Isn’t ( n / (2n+1) ) / (n+1 / (2(n+1) + 1) ) = (n / (2n + 1)) * ((2n + 3) / (n + 1) ?
@&
It is, and if you multiply numerators and denominators you get
(2n + 3) / (2n + 1) * n / (n+1)
*@
@&
On one level, that's just what you do. As stated:
"you first find the limit of the ratio | a(n+1) / a(n) | as n -> infinity. The radius of convergence is the reciprocal of this limit."
Of course that's just a rule, and I want you to not only know the rule but understand why it is so.
I'm not going to give you a completely rigorous answer to that question, but consider the following, which should give you what you need to understand the rigorous answer given in your text and in class notes:
A geometric sequence 1 + r + r^2 + ... converges if and only if | r | < 1.
If the limiting value of | a(n+1) x^(n+1 / a(n) x^n | = | a(n+1) / a(n) | * x is less than 1, then the power series eventually acts pretty much like a geometric series with common ratio a(n+1) / a(n) * x, and it converges.
| a(n+1) / a(n) * x | < 1 implies that
| x | < | a(n) / a(n+1) |.
Thus the radius of convergence is the reciprocal of the limiting ratio of | a(n+1) / a(n) |.
*@
OK
*********************************************
Question:
**** Query 9.5.34 (4th edition 9.5.28 3d edition 9.4.24). What is the radius of convergence of the series p x + p(p-1) / 2! * x^2 + p(p-1)(p-2) * x^3 + … and how did you obtain your result?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
a(n) = p ! / (n ! * (p - n) ! )
a(n+1) = p ! / [ (n+1) ! * ( p - (n+1) )!
a(n+1) / a(n) = { p ! / [ (n+1) ! * ( p - (n+1) ) ! ] } / {p ! / (n ! * (p - n) ! ) }
= (n ! * ( p - n) !) / {(n+1)! * (p - n - 1) ! }
= (p - n) / (n + 1)
lim (p - n) / (n + 1) = 1
Radius = 1
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
2
.............................................
Given Solution:
*&*& As seen in 9.4.6 we have
a(n) = p ! / (n ! * (p - n) ! ) so
a(n+1) = p ! / [ (n+1) ! * ( p - (n+1) ) ! ] and
a(n+1) / a(n) = { p ! / [ (n+1) ! * ( p - (n+1) ) ! ] } / {p ! / (n ! * (p - n) ! ) }
= (n ! * ( p - n) !) / {(n+1)! * (p - n - 1) ! }
= (p - n) / (n + 1).
This expression can be written as
(p / n - 1) / (1/n + 1). As n -> infinity both p/n and 1/n approach zero so our limit is -1 / 1 = -1.
Thus the limiting value of | a(n+1) / a(n) | is 1 and the radius of convergence is 1/1 = 1. *&*&
ADDITIONAL DETAIL IN SIMPLIFYING THE RATIO:
a(n+1) / a(n) = p!/ ( (n+1)!(p-n-1)! ) * n ! ( p - n ) ! / p !
= n ! / (n + 1) ! * (p- n)! / (p - n 1 - 1) !
n ! matches (n+1) ! except for the factor n + 1 in its denominator, and (p - n)! matches (p - n - 1) ! except for the factor p - n in the numerator. So
a(n+1) / a(n) = (p - n) / (n + 1),
which is shown as in the given solution to have limit 1 as n -> infinity.
.........................................
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
How does this: a(n+1) / a(n) = p!/ ( (n+1)!(p-n-1)! ) * n ! ( p - n ) ! / p !
equal this? :
= n ! / (n + 1) ! * (p- n)! / (p - n 1 - 1) !
@&
n ! and (p - n) ! both occur in numerators.
(n+1) ! and (p - n - 1) both occur in denominators.
p ! occurs in both a numerator and a denominator.
(p - n 1 - 1) ! includes a typo; it should read (p - n - 1)!
*@
------------------------------------------------
Self-critique Rating:
@&
n ! and (p - n) ! both occur in numerators.
(n+1) ! and (p - n - 1) both occur in denominators.
p ! occurs in both a numerator and a denominator.
(p - n 1 - 1) ! includes a typo; it should read (p - n - 1)!
*@
OK"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
OK"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
#*&!
&#Good responses. See my notes and let me know if you have questions. &#