$lim_{ntoinfty} n^{x}(a_{1}a_{2}ldots a_{n})^{frac{1}{n}}=ae^x$
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 Nov 20 '18 at 22:45
gimusi
1
asked Nov 20 '18 at 22:29
math.trouble
496
add a comment |
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 Nov 20 '18 at 22:45
gimusi
1
asked Nov 20 '18 at 22:29
math.trouble
496
add a comment |
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 Nov 20 '18 at 22:45
gimusi
1
asked Nov 20 '18 at 22:29
math.trouble
496
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 Nov 20 '18 at 22:45
gimusi
1
asked Nov 20 '18 at 22:29
math.trouble
496
edited Nov 20 '18 at 22:45
gimusi
1
asked Nov 20 '18 at 22:29
math.trouble
496
edited Nov 20 '18 at 22:45
gimusi
1
edited Nov 20 '18 at 22:45
gimusi
1
edited Nov 20 '18 at 22:45
gimusi
1
1
asked Nov 20 '18 at 22:29
math.trouble
496
asked Nov 20 '18 at 22:29
math.trouble
496
asked Nov 20 '18 at 22:29
math.trouble
496
496
add a comment |
add a comment |
1 Answer
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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 Nov 20 '18 at 22:43
gimusi
1
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
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votes
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 Nov 20 '18 at 22:43
gimusi
1
add a comment |
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 Nov 20 '18 at 22:43
gimusi
1
add a comment |
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 Nov 20 '18 at 22:43
gimusi
1
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 Nov 20 '18 at 22:43
gimusi
1
answered Nov 20 '18 at 22:43
gimusi
1
answered Nov 20 '18 at 22:43
gimusi
1
answered Nov 20 '18 at 22:43
gimusi
1
1
add a comment |
add a comment |
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