Mixing
How to prove that this infinite sum converges?
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$a_1,...,a_k$ are real numbers all bigger then $1$. Define the following sequence of numbers:
$$b_n=frac{1}{sqrt[n]{a_1^{n^2}+a_2^{n^2}+...+a_k^{n^2}}}$$ Show that $sum_{n=1}^{infty} b_n$ converges. I tried all the convergence tests that I know but none of them worked, and I have no other clue how to attack it. Can anyone please help?
sequences-and-series limits
asked Jan 21 at 17:18
OmerOmer
3619
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add a comment |
$begingroup$
$a_1,...,a_k$ are real numbers all bigger then $1$. Define the following sequence of numbers:
$$b_n=frac{1}{sqrt[n]{a_1^{n^2}+a_2^{n^2}+...+a_k^{n^2}}}$$ Show that $sum_{n=1}^{infty} b_n$ converges. I tried all the convergence tests that I know but none of them worked, and I have no other clue how to attack it. Can anyone please help?
sequences-and-series limits
asked Jan 21 at 17:18
OmerOmer
3619
$endgroup$
$begingroup$
What tests did you try? Please show your work from these tries.
$endgroup$
– jordan_glen
Jan 21 at 17:19
add a comment |
$begingroup$
$a_1,...,a_k$ are real numbers all bigger then $1$. Define the following sequence of numbers:
$$b_n=frac{1}{sqrt[n]{a_1^{n^2}+a_2^{n^2}+...+a_k^{n^2}}}$$ Show that $sum_{n=1}^{infty} b_n$ converges. I tried all the convergence tests that I know but none of them worked, and I have no other clue how to attack it. Can anyone please help?
sequences-and-series limits
asked Jan 21 at 17:18
OmerOmer
3619
$endgroup$
$a_1,...,a_k$ are real numbers all bigger then $1$. Define the following sequence of numbers:
$$b_n=frac{1}{sqrt[n]{a_1^{n^2}+a_2^{n^2}+...+a_k^{n^2}}}$$ Show that $sum_{n=1}^{infty} b_n$ converges. I tried all the convergence tests that I know but none of them worked, and I have no other clue how to attack it. Can anyone please help?
sequences-and-series limits
sequences-and-series limits
asked Jan 21 at 17:18
OmerOmer
3619
asked Jan 21 at 17:18
OmerOmer
3619
asked Jan 21 at 17:18
OmerOmer
3619
asked Jan 21 at 17:18
OmerOmer
3619
asked Jan 21 at 17:18
OmerOmer
3619
3619
$begingroup$
What tests did you try? Please show your work from these tries.
$endgroup$
– jordan_glen
Jan 21 at 17:19
add a comment |
$begingroup$
What tests did you try? Please show your work from these tries.
$endgroup$
– jordan_glen
Jan 21 at 17:19
$begingroup$
What tests did you try? Please show your work from these tries.
$endgroup$
– jordan_glen
Jan 21 at 17:19
$begingroup$
What tests did you try? Please show your work from these tries.
$endgroup$
– jordan_glen
Jan 21 at 17:19
add a comment |
1 Answer
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Since $0le b_nle a_1^{-n}$, convergence follows from comparison with a geometric series.
answered Jan 21 at 17:21
J.G.J.G.
28.9k22845
$endgroup$
1
$begingroup$
wow, I'm an idiot. It is so simple. Thank you!
$endgroup$
– Omer
Jan 21 at 17:22
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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$begingroup$
Since $0le b_nle a_1^{-n}$, convergence follows from comparison with a geometric series.
answered Jan 21 at 17:21
J.G.J.G.
28.9k22845
$endgroup$
1
$begingroup$
wow, I'm an idiot. It is so simple. Thank you!
$endgroup$
– Omer
Jan 21 at 17:22
add a comment |
$begingroup$
Since $0le b_nle a_1^{-n}$, convergence follows from comparison with a geometric series.
answered Jan 21 at 17:21
J.G.J.G.
28.9k22845
$endgroup$
1
$begingroup$
wow, I'm an idiot. It is so simple. Thank you!
$endgroup$
– Omer
Jan 21 at 17:22
add a comment |
$begingroup$
Since $0le b_nle a_1^{-n}$, convergence follows from comparison with a geometric series.
answered Jan 21 at 17:21
J.G.J.G.
28.9k22845
$endgroup$
Since $0le b_nle a_1^{-n}$, convergence follows from comparison with a geometric series.
answered Jan 21 at 17:21
J.G.J.G.
28.9k22845
answered Jan 21 at 17:21
J.G.J.G.
28.9k22845
answered Jan 21 at 17:21
J.G.J.G.
28.9k22845
answered Jan 21 at 17:21
J.G.J.G.
28.9k22845
28.9k22845
1
$begingroup$
wow, I'm an idiot. It is so simple. Thank you!
$endgroup$
– Omer
Jan 21 at 17:22
add a comment |
1
$begingroup$
wow, I'm an idiot. It is so simple. Thank you!
$endgroup$
– Omer
Jan 21 at 17:22
1
1
$begingroup$
wow, I'm an idiot. It is so simple. Thank you!
$endgroup$
– Omer
Jan 21 at 17:22
$begingroup$
wow, I'm an idiot. It is so simple. Thank you!
$endgroup$
– Omer
Jan 21 at 17:22
add a comment |
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What tests did you try? Please show your work from these tries.
$endgroup$
– jordan_glen
Jan 21 at 17:19