If $sum_{n=1}^{infty} a_n$ is a convergent series of positive terms then $sum_{n=1}^{infty}...
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If $sum_{n=1}^{infty} a_n$ is a convergent series of positive terms then $sum_{n=1}^{infty} frac{a_n^{1/n}}{n^{4/5}}$ coverges or diverges$?$
Using comparison test
$lim_{n to infty} frac{(a_n)^{1/n}}{n^{4/5} a_n}$
And I am getting nothing.
I know that any such convergent series of positive terms should be of the form $sum_{n=1}^{infty} (1/n^{a})$ where $a>1$. I tried to solve the problem by replacing $a_n$ with $1/n^{a}$, used comparison test, but found no result.
How to proceed?
sequences-and-series
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add a comment |
$begingroup$
If $sum_{n=1}^{infty} a_n$ is a convergent series of positive terms then $sum_{n=1}^{infty} frac{a_n^{1/n}}{n^{4/5}}$ coverges or diverges$?$
Using comparison test
$lim_{n to infty} frac{(a_n)^{1/n}}{n^{4/5} a_n}$
And I am getting nothing.
I know that any such convergent series of positive terms should be of the form $sum_{n=1}^{infty} (1/n^{a})$ where $a>1$. I tried to solve the problem by replacing $a_n$ with $1/n^{a}$, used comparison test, but found no result.
How to proceed?
sequences-and-series
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it can diverge. $a_n := frac{1}{n^2} implies a_n^{1/n} to 1$
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– mathworker21
Dec 30 '18 at 13:36
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I tried with this example but since $n^{4/5}$ is in denominator so $n^{th}$ term would tend to zero (in comparison test). So we can't say anything about convergence.
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– Mathsaddict
Dec 30 '18 at 13:42
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no. the sum $sum_n frac{1}{n^{4/5}}$ diverges
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– mathworker21
Dec 30 '18 at 13:43
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Can we directly find the behaviour of the series $sum_{n} frac{1}{n^{2/n} n^{4/5}}$ by applying limit on $n^{2/n}$ in all terms$?$
$endgroup$
– Mathsaddict
Dec 30 '18 at 13:56
add a comment |
$begingroup$
If $sum_{n=1}^{infty} a_n$ is a convergent series of positive terms then $sum_{n=1}^{infty} frac{a_n^{1/n}}{n^{4/5}}$ coverges or diverges$?$
Using comparison test
$lim_{n to infty} frac{(a_n)^{1/n}}{n^{4/5} a_n}$
And I am getting nothing.
I know that any such convergent series of positive terms should be of the form $sum_{n=1}^{infty} (1/n^{a})$ where $a>1$. I tried to solve the problem by replacing $a_n$ with $1/n^{a}$, used comparison test, but found no result.
How to proceed?
sequences-and-series
$endgroup$
If $sum_{n=1}^{infty} a_n$ is a convergent series of positive terms then $sum_{n=1}^{infty} frac{a_n^{1/n}}{n^{4/5}}$ coverges or diverges$?$
Using comparison test
$lim_{n to infty} frac{(a_n)^{1/n}}{n^{4/5} a_n}$
And I am getting nothing.
I know that any such convergent series of positive terms should be of the form $sum_{n=1}^{infty} (1/n^{a})$ where $a>1$. I tried to solve the problem by replacing $a_n$ with $1/n^{a}$, used comparison test, but found no result.
How to proceed?
sequences-and-series
sequences-and-series
asked Dec 30 '18 at 13:31
MathsaddictMathsaddict
3669
3669
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it can diverge. $a_n := frac{1}{n^2} implies a_n^{1/n} to 1$
$endgroup$
– mathworker21
Dec 30 '18 at 13:36
$begingroup$
I tried with this example but since $n^{4/5}$ is in denominator so $n^{th}$ term would tend to zero (in comparison test). So we can't say anything about convergence.
$endgroup$
– Mathsaddict
Dec 30 '18 at 13:42
$begingroup$
no. the sum $sum_n frac{1}{n^{4/5}}$ diverges
$endgroup$
– mathworker21
Dec 30 '18 at 13:43
$begingroup$
Can we directly find the behaviour of the series $sum_{n} frac{1}{n^{2/n} n^{4/5}}$ by applying limit on $n^{2/n}$ in all terms$?$
$endgroup$
– Mathsaddict
Dec 30 '18 at 13:56
add a comment |
$begingroup$
it can diverge. $a_n := frac{1}{n^2} implies a_n^{1/n} to 1$
$endgroup$
– mathworker21
Dec 30 '18 at 13:36
$begingroup$
I tried with this example but since $n^{4/5}$ is in denominator so $n^{th}$ term would tend to zero (in comparison test). So we can't say anything about convergence.
$endgroup$
– Mathsaddict
Dec 30 '18 at 13:42
$begingroup$
no. the sum $sum_n frac{1}{n^{4/5}}$ diverges
$endgroup$
– mathworker21
Dec 30 '18 at 13:43
$begingroup$
Can we directly find the behaviour of the series $sum_{n} frac{1}{n^{2/n} n^{4/5}}$ by applying limit on $n^{2/n}$ in all terms$?$
$endgroup$
– Mathsaddict
Dec 30 '18 at 13:56
$begingroup$
it can diverge. $a_n := frac{1}{n^2} implies a_n^{1/n} to 1$
$endgroup$
– mathworker21
Dec 30 '18 at 13:36
$begingroup$
it can diverge. $a_n := frac{1}{n^2} implies a_n^{1/n} to 1$
$endgroup$
– mathworker21
Dec 30 '18 at 13:36
$begingroup$
I tried with this example but since $n^{4/5}$ is in denominator so $n^{th}$ term would tend to zero (in comparison test). So we can't say anything about convergence.
$endgroup$
– Mathsaddict
Dec 30 '18 at 13:42
$begingroup$
I tried with this example but since $n^{4/5}$ is in denominator so $n^{th}$ term would tend to zero (in comparison test). So we can't say anything about convergence.
$endgroup$
– Mathsaddict
Dec 30 '18 at 13:42
$begingroup$
no. the sum $sum_n frac{1}{n^{4/5}}$ diverges
$endgroup$
– mathworker21
Dec 30 '18 at 13:43
$begingroup$
no. the sum $sum_n frac{1}{n^{4/5}}$ diverges
$endgroup$
– mathworker21
Dec 30 '18 at 13:43
$begingroup$
Can we directly find the behaviour of the series $sum_{n} frac{1}{n^{2/n} n^{4/5}}$ by applying limit on $n^{2/n}$ in all terms$?$
$endgroup$
– Mathsaddict
Dec 30 '18 at 13:56
$begingroup$
Can we directly find the behaviour of the series $sum_{n} frac{1}{n^{2/n} n^{4/5}}$ by applying limit on $n^{2/n}$ in all terms$?$
$endgroup$
– Mathsaddict
Dec 30 '18 at 13:56
add a comment |
1 Answer
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$begingroup$
You can have both situations.
1) If $a_n = q^n$, with $0<q<1$, then $sum_n a_n$ is convergent but $sum_n frac{a_n^{1/n}}{n^{4/5}} = sum_n frac{q}{n^{4/5}}$ diverges to $+infty$.
2) If $a_n = n^{-n}$, then $sum_n a_n$ is convergent and $sum_n frac{a_n^{1/n}}{n^{4/5}} = sum_n frac{1}{n^{1+4/5}}$ is convergent.
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add a comment |
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$begingroup$
You can have both situations.
1) If $a_n = q^n$, with $0<q<1$, then $sum_n a_n$ is convergent but $sum_n frac{a_n^{1/n}}{n^{4/5}} = sum_n frac{q}{n^{4/5}}$ diverges to $+infty$.
2) If $a_n = n^{-n}$, then $sum_n a_n$ is convergent and $sum_n frac{a_n^{1/n}}{n^{4/5}} = sum_n frac{1}{n^{1+4/5}}$ is convergent.
$endgroup$
add a comment |
$begingroup$
You can have both situations.
1) If $a_n = q^n$, with $0<q<1$, then $sum_n a_n$ is convergent but $sum_n frac{a_n^{1/n}}{n^{4/5}} = sum_n frac{q}{n^{4/5}}$ diverges to $+infty$.
2) If $a_n = n^{-n}$, then $sum_n a_n$ is convergent and $sum_n frac{a_n^{1/n}}{n^{4/5}} = sum_n frac{1}{n^{1+4/5}}$ is convergent.
$endgroup$
add a comment |
$begingroup$
You can have both situations.
1) If $a_n = q^n$, with $0<q<1$, then $sum_n a_n$ is convergent but $sum_n frac{a_n^{1/n}}{n^{4/5}} = sum_n frac{q}{n^{4/5}}$ diverges to $+infty$.
2) If $a_n = n^{-n}$, then $sum_n a_n$ is convergent and $sum_n frac{a_n^{1/n}}{n^{4/5}} = sum_n frac{1}{n^{1+4/5}}$ is convergent.
$endgroup$
You can have both situations.
1) If $a_n = q^n$, with $0<q<1$, then $sum_n a_n$ is convergent but $sum_n frac{a_n^{1/n}}{n^{4/5}} = sum_n frac{q}{n^{4/5}}$ diverges to $+infty$.
2) If $a_n = n^{-n}$, then $sum_n a_n$ is convergent and $sum_n frac{a_n^{1/n}}{n^{4/5}} = sum_n frac{1}{n^{1+4/5}}$ is convergent.
answered Dec 30 '18 at 13:42
RigelRigel
11.4k11320
11.4k11320
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$begingroup$
it can diverge. $a_n := frac{1}{n^2} implies a_n^{1/n} to 1$
$endgroup$
– mathworker21
Dec 30 '18 at 13:36
$begingroup$
I tried with this example but since $n^{4/5}$ is in denominator so $n^{th}$ term would tend to zero (in comparison test). So we can't say anything about convergence.
$endgroup$
– Mathsaddict
Dec 30 '18 at 13:42
$begingroup$
no. the sum $sum_n frac{1}{n^{4/5}}$ diverges
$endgroup$
– mathworker21
Dec 30 '18 at 13:43
$begingroup$
Can we directly find the behaviour of the series $sum_{n} frac{1}{n^{2/n} n^{4/5}}$ by applying limit on $n^{2/n}$ in all terms$?$
$endgroup$
– Mathsaddict
Dec 30 '18 at 13:56