$lim_{ntoinfty} n^{x}(a_{1}a_{2}ldots a_{n})^{frac{1}{n}}=ae^x$
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Let $ (a_{n})$ be positive sequence, $a,x in R quad $ and $ lim_{ntoinfty} n^{x}a_{n}=a$.
Prove that $lim_{ntoinfty} n^{x}(a_{1}a_{2}ldots a_{n})^{frac{1}{n}}=ae^x$
I know that $lim_{ntoinfty} (a_{1}a_{2}ldots a_{n})^{frac{1}{n}}=lim_{ntoinfty} a_{n}$ but don't have idea how to use it
limits
edited yesterday
gimusi
86k74292
asked yesterday
math.trouble
455
add a comment |
up vote
2
down vote
favorite
Let $ (a_{n})$ be positive sequence, $a,x in R quad $ and $ lim_{ntoinfty} n^{x}a_{n}=a$.
Prove that $lim_{ntoinfty} n^{x}(a_{1}a_{2}ldots a_{n})^{frac{1}{n}}=ae^x$
I know that $lim_{ntoinfty} (a_{1}a_{2}ldots a_{n})^{frac{1}{n}}=lim_{ntoinfty} a_{n}$ but don't have idea how to use it
limits
edited yesterday
gimusi
86k74292
asked yesterday
math.trouble
455
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $ (a_{n})$ be positive sequence, $a,x in R quad $ and $ lim_{ntoinfty} n^{x}a_{n}=a$.
Prove that $lim_{ntoinfty} n^{x}(a_{1}a_{2}ldots a_{n})^{frac{1}{n}}=ae^x$
I know that $lim_{ntoinfty} (a_{1}a_{2}ldots a_{n})^{frac{1}{n}}=lim_{ntoinfty} a_{n}$ but don't have idea how to use it
limits
edited yesterday
gimusi
86k74292
asked yesterday
math.trouble
455
Let $ (a_{n})$ be positive sequence, $a,x in R quad $ and $ lim_{ntoinfty} n^{x}a_{n}=a$.
Prove that $lim_{ntoinfty} n^{x}(a_{1}a_{2}ldots a_{n})^{frac{1}{n}}=ae^x$
I know that $lim_{ntoinfty} (a_{1}a_{2}ldots a_{n})^{frac{1}{n}}=lim_{ntoinfty} a_{n}$ but don't have idea how to use it
limits
limits
edited yesterday
gimusi
86k74292
asked yesterday
math.trouble
455
edited yesterday
gimusi
86k74292
asked yesterday
math.trouble
455
edited yesterday
gimusi
86k74292
edited yesterday
gimusi
86k74292
edited yesterday
gimusi
86k74292
86k74292
asked yesterday
math.trouble
455
asked yesterday
math.trouble
455
asked yesterday
math.trouble
455
455
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
HINT
By ratio root criteria we have
$$frac{(n+1)^{x(n+1)}a_{1}a_{2}ldots a_{n+1}}{n^{xn}a_{1}a_{2}ldots a_{n}}=(n+1)^xa_{n+1}left(1+frac1nright)^{nx}$$
answered yesterday
gimusi
86k74292
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
HINT
By ratio root criteria we have
$$frac{(n+1)^{x(n+1)}a_{1}a_{2}ldots a_{n+1}}{n^{xn}a_{1}a_{2}ldots a_{n}}=(n+1)^xa_{n+1}left(1+frac1nright)^{nx}$$
answered yesterday
gimusi
86k74292
add a comment |
up vote
3
down vote
accepted
HINT
By ratio root criteria we have
$$frac{(n+1)^{x(n+1)}a_{1}a_{2}ldots a_{n+1}}{n^{xn}a_{1}a_{2}ldots a_{n}}=(n+1)^xa_{n+1}left(1+frac1nright)^{nx}$$
answered yesterday
gimusi
86k74292
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
HINT
By ratio root criteria we have
$$frac{(n+1)^{x(n+1)}a_{1}a_{2}ldots a_{n+1}}{n^{xn}a_{1}a_{2}ldots a_{n}}=(n+1)^xa_{n+1}left(1+frac1nright)^{nx}$$
answered yesterday
gimusi
86k74292
HINT
By ratio root criteria we have
$$frac{(n+1)^{x(n+1)}a_{1}a_{2}ldots a_{n+1}}{n^{xn}a_{1}a_{2}ldots a_{n}}=(n+1)^xa_{n+1}left(1+frac1nright)^{nx}$$
answered yesterday
gimusi
86k74292
answered yesterday
gimusi
86k74292
answered yesterday
gimusi
86k74292
answered yesterday
gimusi
86k74292
86k74292
add a comment |
add a comment |
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