Radius of Finite Convergence
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I am trying to solve a problem where I have to find the radius of finite convergence problem. I believe that I solved the problem correctly, receiving an answer of 1. However, I was informed that this was incorrect but I am not sure why.
My process involved using the ratio test and ultimately received:
(x+2) lim n/(n+1) which goes to 1
Any help would be greatly appreciated
calculus sequences-and-series convergence power-series divergent-series
asked Dec 3 at 3:38
cars cars
11
|
show 3 more comments
up vote
0
down vote
favorite
I am trying to solve a problem where I have to find the radius of finite convergence problem. I believe that I solved the problem correctly, receiving an answer of 1. However, I was informed that this was incorrect but I am not sure why.
My process involved using the ratio test and ultimately received:
(x+2) lim n/(n+1) which goes to 1
Any help would be greatly appreciated
calculus sequences-and-series convergence power-series divergent-series
asked Dec 3 at 3:38
cars cars
11
UPDATE: I believe that I may have figured out where I went wrong and should have received (n/(n+1))^n which goes to 1/e. If anyone sees this am I correct in my correction or am I still off?
– cars cars
Dec 3 at 3:45
The ratio is calculated wrong. You have missed $n^n/(n+1)^{n+1}$.
– xbh
Dec 3 at 3:45
yes, so does the +1 portion of n+1 cancel with the remaining n+1 in the numerator that originated from (n+1)! leading to the answer that I found in my update or is my assumption wrong?
– cars cars
Dec 3 at 3:48
Now you are on track.
– xbh
Dec 3 at 3:49
And then this will give me that the interval of convergence is from 1/e - 2 to -1/e - 2 correct? Thank you for your help
– cars cars
Dec 3 at 3:50
|
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to solve a problem where I have to find the radius of finite convergence problem. I believe that I solved the problem correctly, receiving an answer of 1. However, I was informed that this was incorrect but I am not sure why.
My process involved using the ratio test and ultimately received:
(x+2) lim n/(n+1) which goes to 1
Any help would be greatly appreciated
calculus sequences-and-series convergence power-series divergent-series
asked Dec 3 at 3:38
cars cars
11
I am trying to solve a problem where I have to find the radius of finite convergence problem. I believe that I solved the problem correctly, receiving an answer of 1. However, I was informed that this was incorrect but I am not sure why.
My process involved using the ratio test and ultimately received:
(x+2) lim n/(n+1) which goes to 1
Any help would be greatly appreciated
calculus sequences-and-series convergence power-series divergent-series
calculus sequences-and-series convergence power-series divergent-series
asked Dec 3 at 3:38
cars cars
11
asked Dec 3 at 3:38
cars cars
11
edited Dec 3 at 3:44
asked Dec 3 at 3:38
cars cars
11
asked Dec 3 at 3:38
cars cars
11
asked Dec 3 at 3:38
cars cars
11
11
UPDATE: I believe that I may have figured out where I went wrong and should have received (n/(n+1))^n which goes to 1/e. If anyone sees this am I correct in my correction or am I still off?
– cars cars
Dec 3 at 3:45
The ratio is calculated wrong. You have missed $n^n/(n+1)^{n+1}$.
– xbh
Dec 3 at 3:45
yes, so does the +1 portion of n+1 cancel with the remaining n+1 in the numerator that originated from (n+1)! leading to the answer that I found in my update or is my assumption wrong?
– cars cars
Dec 3 at 3:48
Now you are on track.
– xbh
Dec 3 at 3:49
And then this will give me that the interval of convergence is from 1/e - 2 to -1/e - 2 correct? Thank you for your help
– cars cars
Dec 3 at 3:50
|
show 3 more comments
UPDATE: I believe that I may have figured out where I went wrong and should have received (n/(n+1))^n which goes to 1/e. If anyone sees this am I correct in my correction or am I still off?
– cars cars
Dec 3 at 3:45
The ratio is calculated wrong. You have missed $n^n/(n+1)^{n+1}$.
– xbh
Dec 3 at 3:45
yes, so does the +1 portion of n+1 cancel with the remaining n+1 in the numerator that originated from (n+1)! leading to the answer that I found in my update or is my assumption wrong?
– cars cars
Dec 3 at 3:48
Now you are on track.
– xbh
Dec 3 at 3:49
And then this will give me that the interval of convergence is from 1/e - 2 to -1/e - 2 correct? Thank you for your help
– cars cars
Dec 3 at 3:50
UPDATE: I believe that I may have figured out where I went wrong and should have received (n/(n+1))^n which goes to 1/e. If anyone sees this am I correct in my correction or am I still off?
– cars cars
Dec 3 at 3:45
UPDATE: I believe that I may have figured out where I went wrong and should have received (n/(n+1))^n which goes to 1/e. If anyone sees this am I correct in my correction or am I still off?
– cars cars
Dec 3 at 3:45
The ratio is calculated wrong. You have missed $n^n/(n+1)^{n+1}$.
– xbh
Dec 3 at 3:45
The ratio is calculated wrong. You have missed $n^n/(n+1)^{n+1}$.
– xbh
Dec 3 at 3:45
yes, so does the +1 portion of n+1 cancel with the remaining n+1 in the numerator that originated from (n+1)! leading to the answer that I found in my update or is my assumption wrong?
– cars cars
Dec 3 at 3:48
yes, so does the +1 portion of n+1 cancel with the remaining n+1 in the numerator that originated from (n+1)! leading to the answer that I found in my update or is my assumption wrong?
– cars cars
Dec 3 at 3:48
Now you are on track.
– xbh
Dec 3 at 3:49
Now you are on track.
– xbh
Dec 3 at 3:49
And then this will give me that the interval of convergence is from 1/e - 2 to -1/e - 2 correct? Thank you for your help
– cars cars
Dec 3 at 3:50
And then this will give me that the interval of convergence is from 1/e - 2 to -1/e - 2 correct? Thank you for your help
– cars cars
Dec 3 at 3:50
|
show 3 more comments
1 Answer
1
active
oldest
votes
up vote
0
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If you use the Ratio Test, you have
$$
begin{split}
lim_{n to infty} left|dfrac{dfrac{(n+1)!(x+2)^{n+1}}{(n+1)^{n+1}}}{dfrac{n!(x+2)^n}{n^n}} right|&= lim_{n to infty} left|dfrac{(n+1)!(x+2)^{n+1}n^n}{n!(x+2)^n(n+1)^{n+1}} right| \
&= lim_{n to infty} left| dfrac{(n+1)(x+2)n^n}{(n+1)(n+1)^n} right| \
&= lim_{n to infty} left| dfrac{(x+2)n^n}{(n+1)^n} right| \
&= lim_{n to infty} left|(x+2) ; left( dfrac{n}{n+1}right)^n right| \
&= lim_{n to infty} left|(x+2) ; left( dfrac{n+1}{n}right)^{-n} right| \
&= lim_{n to infty} left|(x+2) ; left( 1+ dfrac{1}{n}right)^{-n} right| \
&= lim_{n to infty} left|(x+2) ; left(left( 1+ dfrac{1}{n}right)^{n}right)^{-1} right| \
&= |x+2| cdot e^{-1},
end{split}
$$
which is what you found. The trick with these harder Ratio Test problems is working slowly and being careful with the Algebra.
answered Dec 3 at 5:20
mathematics2x2life
8,03221738
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
If you use the Ratio Test, you have
$$
begin{split}
lim_{n to infty} left|dfrac{dfrac{(n+1)!(x+2)^{n+1}}{(n+1)^{n+1}}}{dfrac{n!(x+2)^n}{n^n}} right|&= lim_{n to infty} left|dfrac{(n+1)!(x+2)^{n+1}n^n}{n!(x+2)^n(n+1)^{n+1}} right| \
&= lim_{n to infty} left| dfrac{(n+1)(x+2)n^n}{(n+1)(n+1)^n} right| \
&= lim_{n to infty} left| dfrac{(x+2)n^n}{(n+1)^n} right| \
&= lim_{n to infty} left|(x+2) ; left( dfrac{n}{n+1}right)^n right| \
&= lim_{n to infty} left|(x+2) ; left( dfrac{n+1}{n}right)^{-n} right| \
&= lim_{n to infty} left|(x+2) ; left( 1+ dfrac{1}{n}right)^{-n} right| \
&= lim_{n to infty} left|(x+2) ; left(left( 1+ dfrac{1}{n}right)^{n}right)^{-1} right| \
&= |x+2| cdot e^{-1},
end{split}
$$
which is what you found. The trick with these harder Ratio Test problems is working slowly and being careful with the Algebra.
answered Dec 3 at 5:20
mathematics2x2life
8,03221738
add a comment |
up vote
0
down vote
If you use the Ratio Test, you have
$$
begin{split}
lim_{n to infty} left|dfrac{dfrac{(n+1)!(x+2)^{n+1}}{(n+1)^{n+1}}}{dfrac{n!(x+2)^n}{n^n}} right|&= lim_{n to infty} left|dfrac{(n+1)!(x+2)^{n+1}n^n}{n!(x+2)^n(n+1)^{n+1}} right| \
&= lim_{n to infty} left| dfrac{(n+1)(x+2)n^n}{(n+1)(n+1)^n} right| \
&= lim_{n to infty} left| dfrac{(x+2)n^n}{(n+1)^n} right| \
&= lim_{n to infty} left|(x+2) ; left( dfrac{n}{n+1}right)^n right| \
&= lim_{n to infty} left|(x+2) ; left( dfrac{n+1}{n}right)^{-n} right| \
&= lim_{n to infty} left|(x+2) ; left( 1+ dfrac{1}{n}right)^{-n} right| \
&= lim_{n to infty} left|(x+2) ; left(left( 1+ dfrac{1}{n}right)^{n}right)^{-1} right| \
&= |x+2| cdot e^{-1},
end{split}
$$
which is what you found. The trick with these harder Ratio Test problems is working slowly and being careful with the Algebra.
answered Dec 3 at 5:20
mathematics2x2life
8,03221738
add a comment |
up vote
0
down vote
up vote
0
down vote
If you use the Ratio Test, you have
$$
begin{split}
lim_{n to infty} left|dfrac{dfrac{(n+1)!(x+2)^{n+1}}{(n+1)^{n+1}}}{dfrac{n!(x+2)^n}{n^n}} right|&= lim_{n to infty} left|dfrac{(n+1)!(x+2)^{n+1}n^n}{n!(x+2)^n(n+1)^{n+1}} right| \
&= lim_{n to infty} left| dfrac{(n+1)(x+2)n^n}{(n+1)(n+1)^n} right| \
&= lim_{n to infty} left| dfrac{(x+2)n^n}{(n+1)^n} right| \
&= lim_{n to infty} left|(x+2) ; left( dfrac{n}{n+1}right)^n right| \
&= lim_{n to infty} left|(x+2) ; left( dfrac{n+1}{n}right)^{-n} right| \
&= lim_{n to infty} left|(x+2) ; left( 1+ dfrac{1}{n}right)^{-n} right| \
&= lim_{n to infty} left|(x+2) ; left(left( 1+ dfrac{1}{n}right)^{n}right)^{-1} right| \
&= |x+2| cdot e^{-1},
end{split}
$$
which is what you found. The trick with these harder Ratio Test problems is working slowly and being careful with the Algebra.
answered Dec 3 at 5:20
mathematics2x2life
8,03221738
If you use the Ratio Test, you have
$$
begin{split}
lim_{n to infty} left|dfrac{dfrac{(n+1)!(x+2)^{n+1}}{(n+1)^{n+1}}}{dfrac{n!(x+2)^n}{n^n}} right|&= lim_{n to infty} left|dfrac{(n+1)!(x+2)^{n+1}n^n}{n!(x+2)^n(n+1)^{n+1}} right| \
&= lim_{n to infty} left| dfrac{(n+1)(x+2)n^n}{(n+1)(n+1)^n} right| \
&= lim_{n to infty} left| dfrac{(x+2)n^n}{(n+1)^n} right| \
&= lim_{n to infty} left|(x+2) ; left( dfrac{n}{n+1}right)^n right| \
&= lim_{n to infty} left|(x+2) ; left( dfrac{n+1}{n}right)^{-n} right| \
&= lim_{n to infty} left|(x+2) ; left( 1+ dfrac{1}{n}right)^{-n} right| \
&= lim_{n to infty} left|(x+2) ; left(left( 1+ dfrac{1}{n}right)^{n}right)^{-1} right| \
&= |x+2| cdot e^{-1},
end{split}
$$
which is what you found. The trick with these harder Ratio Test problems is working slowly and being careful with the Algebra.
answered Dec 3 at 5:20
mathematics2x2life
8,03221738
answered Dec 3 at 5:20
mathematics2x2life
8,03221738
answered Dec 3 at 5:20
mathematics2x2life
8,03221738
answered Dec 3 at 5:20
mathematics2x2life
8,03221738
8,03221738
add a comment |
add a comment |
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UPDATE: I believe that I may have figured out where I went wrong and should have received (n/(n+1))^n which goes to 1/e. If anyone sees this am I correct in my correction or am I still off?
– cars cars
Dec 3 at 3:45
The ratio is calculated wrong. You have missed $n^n/(n+1)^{n+1}$.
– xbh
Dec 3 at 3:45
yes, so does the +1 portion of n+1 cancel with the remaining n+1 in the numerator that originated from (n+1)! leading to the answer that I found in my update or is my assumption wrong?
– cars cars
Dec 3 at 3:48
Now you are on track.
– xbh
Dec 3 at 3:49
And then this will give me that the interval of convergence is from 1/e - 2 to -1/e - 2 correct? Thank you for your help
– cars cars
Dec 3 at 3:50