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course Mth 174
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?
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Your solution: Well, we know that 1/3^n will be a shrinking value as n grows from 1 to infinity. We also know that 1/3^n + 1 is greater than 1/3^n, so if our original is less than a convergent function, it also must be convergent.
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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).
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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? ****
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Your solution:
A sub. n = 1/(2n) !, a sub. (n+1) = 1/(2n+2)!
We will be using the test [a sub. (n+1) ]/a sub. n
So, given our 1st statement, we will now have (1/(2n+2)!) / (1/(2n)!) which equals (2n)! / (2n+2)!)
Expanded, we will get (2n * 2n-1 * 2n-2……*1) / 2n * 2n+1 * 2n +2 * 2n-1 * 2n-2….*1)
Dividing like terms leaves us with 1/(2n+1)(2n+2)
Now, we can easily see that as n->infinity, our series converges since the denominator will grow infinitely. This means that the expression is approaching zero, which means the radius of convergence is infinite.
confidence rating #$&*: fairly
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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. *&*&
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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?
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Your solution:
1-.09+.01-.001+.0001=.9090909090909
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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... .
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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) * x^3 + …?
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Your solution:
The general term is the coefficient of x^n
The factor of this is p( p-1) …..(p-n+1) which can be rewritten as p!/(p-n)!
The nth term would therefore be p!/(p-n)!/n! * x^2
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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) ! )
**** 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 + …?
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.
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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?
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Your solution:
The radius of convergence is:
px+(p(p-1))/2! * x^2 + p(p-1)(p-2)x^3
asub.n=p!/(n!*(p-n)!
asub.(n+1)=p!/(n+1)! * (p-(n+1))!
Asub.(n+1)/asub.n = p!/n! * (p-n)!/p!/(n+1)! * (p-(n+1))!
This becomes (p-n)/(n+1)
To see whats going on we will change this to p/n-1/1/n+1 and take the limit as n->infinity
This gives us a final answer of 1.
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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. *&*&
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&#Good responses. Let me know if you have questions. &#
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