#$&*
course mth 174
5/26 9:50 pm
Asst # 13
*********************************************
Question:
`q **** query problem 9.4.8 (4th edition 9.4.4 3d edition 9.3.6). Using a comparison test determine whether the series sum(1/(3^n+1),n,1,infinity) converges. **** With what known series did you compare this series, and how did you show that the comparison was valid?
......!!!!!!!!...................................
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
We will compare the series sum of 1/(3^n +1) to the series sum 1/3^n. we need to know if 1/3^n converges or diverges. We can use a ration test to determine if the series converges or diverges so we have
1/(3^(n+1)) * 3^n/1 which is 3^n / 3^(n+1) which would give you 1/3 which is less than 1 so 1/3^n converges. We know that our original 1/(3^(n+1)) is less than our convergent comparison so the original also converges.
confidence rating #$&*: 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:
*********************************************
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:
We begin with the series 1/(2n!) and we are asked to use the ratio test.
We have 1/(2(n+1)!)* 2n!/1
1/(2n +2)! * 2n! /1
(2n) * (2n-1) * (2n -2) …..1 / (2n+2) *(2n+1)*(2n)*(2n-1)*(2n-2) ….1
Cancelling like terms we have
1/(2n+2)*(2n-1)
We know that this limit is 0 as values of x get larger and larger approaching infinity.
So since 0 is less than 1, the series converges and the radius of convergence will be infinitely large.
confidence rating #$&*: 3
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
For 9.3.12
*&*& The ratio test takes the limit as n -> infinity of a(n+1) / a(n). If the limit is less than 1 then the series converges in much the same way as a geometric series sum(r^n), with r equal to the limiting ratio.
In this case a(n+1) = 1 / (2n + 2)! and a(n) = 1 / (2n) ! so
a(n+1) / a(n)
= 1 / (2n+2) ! / [ 1 / (2n) ! ]
= (2n) ! / (2n + 2) !
= [ 2n * (2n-1) * (2n-2) * (2n - 3) * ... * 1 ] / [ (2n + 2) * (2n + 1) * (2n) * (2n - 1) * ... * 1 ]
= 1 / [ (2n+2) ( 2n+1) ].
As n -> infinity this result approaches zero. Thus the series converges for all values of n, and the radius of convergence is infinite. *&*&
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
ok
------------------------------------------------
Self-critique Rating:3
*********************************************
Question:
Query problem 9.4.52 (3d edition 9.4.40) (was 9.2.24) partial sums of 1-.1+.01-.001 ... to what does the series converge?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
This would be an alternating series because of the alternating signs, we need to figure out what the series is and we see that if an is .1^n this will give us the values of the series when n=0, 1, 2, 3. So we need to find out what the limit of .1^n is. We can see that as values of n get bigger and bigger approaching infinity, the series limit is 0. Since 0 is less than 1, the series converges.
confidence rating #$&*: 2
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
We have:
an = 10^(-n)
and
an+1 = 10^(-(n+1))
So, since 0 < an+1 < an, this series converges.
*&*& 0 < an+1 < an is not the appropriate test. For example if a(n) = (-1)^n * (.5 + 10^-n) we have the series .5 - .51 + .501 - .5001 etc. and 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):
### I think I am ok with my answer but I feel like I should have included a negative sign for .1^n in order to get the alternating signs or is this not necessary????
------------------------------------------------
Self-critique Rating:3
@&
You have the right idea. All you need to do is say that this is an alternating series sum(a_n) for which | a_n | approaches zero.
You said most, but not quite all, of this.
*@
*********************************************
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 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?????
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) ! )
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
#### 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?????
@&
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 Rating:3
*********************************************
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:
We have the series x / 3 + 2 x^2 / 5 + 3 x^2 / 7 + … And we want to find the radius of convergence. The radius of convergence is simply the reciprocal of the limit that is found for the series using the ratio test. So we begin by using the ratio test which has us take a(n+1) and divide it by a(n)
So for the first term we have x/3 using the ratio test we get (x+1)/3 * 3/x
This gives us (3x + 3)/(3x) which is 3/3 or 1
To confirm we didn’t make and error we use the ratio test on the 2nd term of the series getting
2(x+1)^2 / 5 * 5/(2x^2) we then have 2x^2 + 2X +1 / 2x^2 * 5/5 so we have 2/2 or 1 as our limit
Now that we know our limit is 1, the radius of convergence is the reciprocal of 1 which is 1.
Radius of convergence is 1.
confidence rating #$&*: 3
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
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 got the same answer but it appears that I have worked a different problem than the given solution. Please let me know if I am wrong in my solution or need to correct anything?????
@&
Good, but you can't base your ultimate conclusion on a pattern you have established only for the first few terms. You need to find the formula for the nth term, and use the general ratio a(n+1) / a(n).
The given solution use the expression n / (2 n + 1) for the nth term, but does not detail how to reason out this term.
Here's the reasoning:
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).
*@
------------------------------------------------
Self-critique Rating:3
*********************************************
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)!
So we have p!/(n+1)!(p-n+1)! * n!(p-n)! / p!
Now I will expand after cancelling the p! on numerator and denominator and get
n*(n-1)* (p-n)* (p-n-1) / (n+1) * (n-1) * (p-n+1) * (p-n-1)
n * (p-n) / (n+1) * (p-n+1) now we need to see what the limits are for n and p
for n we get limit approaching 1 and for p we also get limit approaching 1.
?????as I take limits of n for example 100*p-100 / 100+1*(p-100+1) is it correct to
Say 100 p -10,000 / 101p -9999 can I cancel out the p’s or not????????
Since the limits approach 1 the radius of convergence is 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. *&*&
.........................................
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
###I am still having trouble figuring out how to obtain the a(n) term as indicated in the previous problem, otherwise I understand the ratio test, I am not sure if I cancelled out like terms correctly when using the ratio test, but I did get 1 as the radius of convergence. Please let me know if I am incorrectly cancelling out terms???? Also when you have 2 variables in an expression and are trying to find the limits of each variable, I am unsure how to proceed??????
@&
Note the need to group the denominator:
a(n)= p!/ ( n!(p-n)! )
a(n+1) = p!/ ( (n+1)!(p-( n+1)) ! )
Note how the p - (n + 1) ! contrasts with your (p - n + 1) !.
Thus
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 Rating:
3
"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
3
"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
#*&!
&#Your work looks good. See my notes. Let me know if you have any questions. &#