Finding radius of convergence for $frac{x^n}{nsqrt{n}12^n}$
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I need help finding the radius of convergence for $frac{x^n}{nsqrt{n}12^n}$
I tried both the root and ratio tests but neither one is getting me anywhere.
Trying the root test I reduced to $lim limits_{n to infty}left(frac{x}{n^{1/2}n^{1/4}12}right)$ which is either incorrect or not helpful.
I need to find radius of convergence, as well as interval of convergence, then determine absolute and conditional convergence.
calculus sequences-and-series
edited Dec 6 at 10:37
user376343
2,7782822
asked Dec 6 at 1:33
SolidSnackDrive
1828
add a comment |
up vote
0
down vote
favorite
I need help finding the radius of convergence for $frac{x^n}{nsqrt{n}12^n}$
I tried both the root and ratio tests but neither one is getting me anywhere.
Trying the root test I reduced to $lim limits_{n to infty}left(frac{x}{n^{1/2}n^{1/4}12}right)$ which is either incorrect or not helpful.
I need to find radius of convergence, as well as interval of convergence, then determine absolute and conditional convergence.
calculus sequences-and-series
edited Dec 6 at 10:37
user376343
2,7782822
asked Dec 6 at 1:33
SolidSnackDrive
1828
Just, btw, the expression in the root test is $(x/12) (n^{- 3/2})^{1/n} = (x/12) n^{-1.5/n},$ not $(x/12) n^{-3/2}.$
– stochasticboy321
Dec 6 at 1:42
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I need help finding the radius of convergence for $frac{x^n}{nsqrt{n}12^n}$
I tried both the root and ratio tests but neither one is getting me anywhere.
Trying the root test I reduced to $lim limits_{n to infty}left(frac{x}{n^{1/2}n^{1/4}12}right)$ which is either incorrect or not helpful.
I need to find radius of convergence, as well as interval of convergence, then determine absolute and conditional convergence.
calculus sequences-and-series
edited Dec 6 at 10:37
user376343
2,7782822
asked Dec 6 at 1:33
SolidSnackDrive
1828
I need help finding the radius of convergence for $frac{x^n}{nsqrt{n}12^n}$
I tried both the root and ratio tests but neither one is getting me anywhere.
Trying the root test I reduced to $lim limits_{n to infty}left(frac{x}{n^{1/2}n^{1/4}12}right)$ which is either incorrect or not helpful.
I need to find radius of convergence, as well as interval of convergence, then determine absolute and conditional convergence.
calculus sequences-and-series
calculus sequences-and-series
edited Dec 6 at 10:37
user376343
2,7782822
asked Dec 6 at 1:33
SolidSnackDrive
1828
edited Dec 6 at 10:37
user376343
2,7782822
asked Dec 6 at 1:33
SolidSnackDrive
1828
edited Dec 6 at 10:37
user376343
2,7782822
edited Dec 6 at 10:37
user376343
2,7782822
edited Dec 6 at 10:37
user376343
2,7782822
2,7782822
asked Dec 6 at 1:33
SolidSnackDrive
1828
asked Dec 6 at 1:33
SolidSnackDrive
1828
asked Dec 6 at 1:33
SolidSnackDrive
1828
1828
Just, btw, the expression in the root test is $(x/12) (n^{- 3/2})^{1/n} = (x/12) n^{-1.5/n},$ not $(x/12) n^{-3/2}.$
– stochasticboy321
Dec 6 at 1:42
add a comment |
Just, btw, the expression in the root test is $(x/12) (n^{- 3/2})^{1/n} = (x/12) n^{-1.5/n},$ not $(x/12) n^{-3/2}.$
– stochasticboy321
Dec 6 at 1:42
Just, btw, the expression in the root test is $(x/12) (n^{- 3/2})^{1/n} = (x/12) n^{-1.5/n},$ not $(x/12) n^{-3/2}.$
– stochasticboy321
Dec 6 at 1:42
Just, btw, the expression in the root test is $(x/12) (n^{- 3/2})^{1/n} = (x/12) n^{-1.5/n},$ not $(x/12) n^{-3/2}.$
– stochasticboy321
Dec 6 at 1:42
add a comment |
2 Answers
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2
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For ratio test:
$$lim_{n to infty}frac{x^{n+1}}{(n+1)sqrt{n+1}cdot 12^{n+1}}frac{nsqrt{n} cdot 12^n}{x^n}= lim_{n to infty}frac{x}{12}lim_{n to infty}frac{n}{n+1}lim_{n to infty}sqrt{frac{n}{n+1}}$$
Can you evaluate the last two limits?
answered Dec 6 at 1:40
Siong Thye Goh
98.2k1463116
Second one goes to $1$, third one also $1$?
– SolidSnackDrive
Dec 6 at 1:50
yes, $frac{n}{n+1}= frac{1}{1+frac1n}$.
– Siong Thye Goh
Dec 6 at 1:51
add a comment |
up vote
1
down vote
The ratio test should work. The ratio of the $1/(n sqrt(n))$ part goes to $1$ and the rest gives $|x/12|$.
answered Dec 6 at 1:40
coffeemath
2,2731413
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
For ratio test:
$$lim_{n to infty}frac{x^{n+1}}{(n+1)sqrt{n+1}cdot 12^{n+1}}frac{nsqrt{n} cdot 12^n}{x^n}= lim_{n to infty}frac{x}{12}lim_{n to infty}frac{n}{n+1}lim_{n to infty}sqrt{frac{n}{n+1}}$$
Can you evaluate the last two limits?
answered Dec 6 at 1:40
Siong Thye Goh
98.2k1463116
Second one goes to $1$, third one also $1$?
– SolidSnackDrive
Dec 6 at 1:50
yes, $frac{n}{n+1}= frac{1}{1+frac1n}$.
– Siong Thye Goh
Dec 6 at 1:51
add a comment |
up vote
2
down vote
For ratio test:
$$lim_{n to infty}frac{x^{n+1}}{(n+1)sqrt{n+1}cdot 12^{n+1}}frac{nsqrt{n} cdot 12^n}{x^n}= lim_{n to infty}frac{x}{12}lim_{n to infty}frac{n}{n+1}lim_{n to infty}sqrt{frac{n}{n+1}}$$
Can you evaluate the last two limits?
answered Dec 6 at 1:40
Siong Thye Goh
98.2k1463116
Second one goes to $1$, third one also $1$?
– SolidSnackDrive
Dec 6 at 1:50
yes, $frac{n}{n+1}= frac{1}{1+frac1n}$.
– Siong Thye Goh
Dec 6 at 1:51
add a comment |
up vote
2
down vote
up vote
2
down vote
For ratio test:
$$lim_{n to infty}frac{x^{n+1}}{(n+1)sqrt{n+1}cdot 12^{n+1}}frac{nsqrt{n} cdot 12^n}{x^n}= lim_{n to infty}frac{x}{12}lim_{n to infty}frac{n}{n+1}lim_{n to infty}sqrt{frac{n}{n+1}}$$
Can you evaluate the last two limits?
answered Dec 6 at 1:40
Siong Thye Goh
98.2k1463116
For ratio test:
$$lim_{n to infty}frac{x^{n+1}}{(n+1)sqrt{n+1}cdot 12^{n+1}}frac{nsqrt{n} cdot 12^n}{x^n}= lim_{n to infty}frac{x}{12}lim_{n to infty}frac{n}{n+1}lim_{n to infty}sqrt{frac{n}{n+1}}$$
Can you evaluate the last two limits?
answered Dec 6 at 1:40
Siong Thye Goh
98.2k1463116
answered Dec 6 at 1:40
Siong Thye Goh
98.2k1463116
answered Dec 6 at 1:40
Siong Thye Goh
98.2k1463116
answered Dec 6 at 1:40
Siong Thye Goh
98.2k1463116
98.2k1463116
Second one goes to $1$, third one also $1$?
– SolidSnackDrive
Dec 6 at 1:50
yes, $frac{n}{n+1}= frac{1}{1+frac1n}$.
– Siong Thye Goh
Dec 6 at 1:51
add a comment |
Second one goes to $1$, third one also $1$?
– SolidSnackDrive
Dec 6 at 1:50
yes, $frac{n}{n+1}= frac{1}{1+frac1n}$.
– Siong Thye Goh
Dec 6 at 1:51
Second one goes to $1$, third one also $1$?
– SolidSnackDrive
Dec 6 at 1:50
Second one goes to $1$, third one also $1$?
– SolidSnackDrive
Dec 6 at 1:50
yes, $frac{n}{n+1}= frac{1}{1+frac1n}$.
– Siong Thye Goh
Dec 6 at 1:51
yes, $frac{n}{n+1}= frac{1}{1+frac1n}$.
– Siong Thye Goh
Dec 6 at 1:51
add a comment |
up vote
1
down vote
The ratio test should work. The ratio of the $1/(n sqrt(n))$ part goes to $1$ and the rest gives $|x/12|$.
answered Dec 6 at 1:40
coffeemath
2,2731413
add a comment |
up vote
1
down vote
The ratio test should work. The ratio of the $1/(n sqrt(n))$ part goes to $1$ and the rest gives $|x/12|$.
answered Dec 6 at 1:40
coffeemath
2,2731413
add a comment |
up vote
1
down vote
up vote
1
down vote
The ratio test should work. The ratio of the $1/(n sqrt(n))$ part goes to $1$ and the rest gives $|x/12|$.
answered Dec 6 at 1:40
coffeemath
2,2731413
The ratio test should work. The ratio of the $1/(n sqrt(n))$ part goes to $1$ and the rest gives $|x/12|$.
answered Dec 6 at 1:40
coffeemath
2,2731413
answered Dec 6 at 1:40
coffeemath
2,2731413
answered Dec 6 at 1:40
coffeemath
2,2731413
answered Dec 6 at 1:40
coffeemath
2,2731413
2,2731413
add a comment |
add a comment |
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Just, btw, the expression in the root test is $(x/12) (n^{- 3/2})^{1/n} = (x/12) n^{-1.5/n},$ not $(x/12) n^{-3/2}.$
– stochasticboy321
Dec 6 at 1:42