Strong and weak Ratio test?
$begingroup$
Is this a valid test for convergence of $sum_{n=1}^infty a_n$ where $a_n$ are all positive?
Define:
$rho_n=a_n/a_{n+1}$
The series converges if $rho_n>1$ for all n>N
The series diverges if $rho_n le 1$ for all n>N
where $N$ is some positive integer. Note this is not the same as the usual ratio test which states that the series converges if $lim_{n to infty}rho_n>1$ and diverges if $lim_{n to infty}rho_n<1$ with no conclusion for 1.
I ask this because Kummer's test has been stated as: $rho_n=D_n a_n/a_{n+1}-D_{n+1}$ where $D_n$ is a positive term series, with convergence for $rho_n>0$ and divergence for $rho_n le 0$ and $D_n$ divergent, for some $n>N$. It has also been stated in the limit form where $lim_{n to infty}rho_n<0$ and $D_n$ divergent for divergence. Substituting $D_n=1$ into Kummers test gives the above statement (no limits), along with the usual ratio test involving limits.
sequences-and-series convergence
asked Dec 2 '18 at 4:25
Paul R.Paul R.
213
$endgroup$
add a comment |
$begingroup$
Is this a valid test for convergence of $sum_{n=1}^infty a_n$ where $a_n$ are all positive?
Define:
$rho_n=a_n/a_{n+1}$
The series converges if $rho_n>1$ for all n>N
The series diverges if $rho_n le 1$ for all n>N
where $N$ is some positive integer. Note this is not the same as the usual ratio test which states that the series converges if $lim_{n to infty}rho_n>1$ and diverges if $lim_{n to infty}rho_n<1$ with no conclusion for 1.
I ask this because Kummer's test has been stated as: $rho_n=D_n a_n/a_{n+1}-D_{n+1}$ where $D_n$ is a positive term series, with convergence for $rho_n>0$ and divergence for $rho_n le 0$ and $D_n$ divergent, for some $n>N$. It has also been stated in the limit form where $lim_{n to infty}rho_n<0$ and $D_n$ divergent for divergence. Substituting $D_n=1$ into Kummers test gives the above statement (no limits), along with the usual ratio test involving limits.
sequences-and-series convergence
asked Dec 2 '18 at 4:25
Paul R.Paul R.
213
$endgroup$
add a comment |
$begingroup$
Is this a valid test for convergence of $sum_{n=1}^infty a_n$ where $a_n$ are all positive?
Define:
$rho_n=a_n/a_{n+1}$
The series converges if $rho_n>1$ for all n>N
The series diverges if $rho_n le 1$ for all n>N
where $N$ is some positive integer. Note this is not the same as the usual ratio test which states that the series converges if $lim_{n to infty}rho_n>1$ and diverges if $lim_{n to infty}rho_n<1$ with no conclusion for 1.
I ask this because Kummer's test has been stated as: $rho_n=D_n a_n/a_{n+1}-D_{n+1}$ where $D_n$ is a positive term series, with convergence for $rho_n>0$ and divergence for $rho_n le 0$ and $D_n$ divergent, for some $n>N$. It has also been stated in the limit form where $lim_{n to infty}rho_n<0$ and $D_n$ divergent for divergence. Substituting $D_n=1$ into Kummers test gives the above statement (no limits), along with the usual ratio test involving limits.
sequences-and-series convergence
asked Dec 2 '18 at 4:25
Paul R.Paul R.
213
$endgroup$
Is this a valid test for convergence of $sum_{n=1}^infty a_n$ where $a_n$ are all positive?
Define:
$rho_n=a_n/a_{n+1}$
The series converges if $rho_n>1$ for all n>N
The series diverges if $rho_n le 1$ for all n>N
where $N$ is some positive integer. Note this is not the same as the usual ratio test which states that the series converges if $lim_{n to infty}rho_n>1$ and diverges if $lim_{n to infty}rho_n<1$ with no conclusion for 1.
I ask this because Kummer's test has been stated as: $rho_n=D_n a_n/a_{n+1}-D_{n+1}$ where $D_n$ is a positive term series, with convergence for $rho_n>0$ and divergence for $rho_n le 0$ and $D_n$ divergent, for some $n>N$. It has also been stated in the limit form where $lim_{n to infty}rho_n<0$ and $D_n$ divergent for divergence. Substituting $D_n=1$ into Kummers test gives the above statement (no limits), along with the usual ratio test involving limits.
sequences-and-series convergence
sequences-and-series convergence
asked Dec 2 '18 at 4:25
Paul R.Paul R.
213
asked Dec 2 '18 at 4:25
Paul R.Paul R.
213
asked Dec 2 '18 at 4:25
Paul R.Paul R.
213
asked Dec 2 '18 at 4:25
Paul R.Paul R.
213
asked Dec 2 '18 at 4:25
Paul R.Paul R.
213
213
add a comment |
add a comment |
2 Answers
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$begingroup$
No. For example, your test predicts that
$$ sum_{n geq 1} frac{1}{n^2} $$
diverges.
answered Dec 2 '18 at 4:51
davidlowryduda♦davidlowryduda
74.4k7118252
$endgroup$
add a comment |
$begingroup$
Ok, thanks, now I see the problem. Kummer's $rho_n$ is $D_n a_n/a_{n+1}-D_{n+1}$ and for convergence, there must be a $c>0$ such that $rho_n ge c$ which is NOT the same as $rho_n > 0$, which I mistakenly supposed. If $a_n=1/n^2$, then $rho_n=(1+1/n)^2$ which, although it is greater than zero for all $nge 1$, there is no $c>0$ that it is greater than or equal to for any $n$.
answered Dec 2 '18 at 5:05
Paul R.Paul R.
213
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
No. For example, your test predicts that
$$ sum_{n geq 1} frac{1}{n^2} $$
diverges.
answered Dec 2 '18 at 4:51
davidlowryduda♦davidlowryduda
74.4k7118252
$endgroup$
add a comment |
$begingroup$
No. For example, your test predicts that
$$ sum_{n geq 1} frac{1}{n^2} $$
diverges.
answered Dec 2 '18 at 4:51
davidlowryduda♦davidlowryduda
74.4k7118252
$endgroup$
add a comment |
$begingroup$
No. For example, your test predicts that
$$ sum_{n geq 1} frac{1}{n^2} $$
diverges.
answered Dec 2 '18 at 4:51
davidlowryduda♦davidlowryduda
74.4k7118252
$endgroup$
No. For example, your test predicts that
$$ sum_{n geq 1} frac{1}{n^2} $$
diverges.
answered Dec 2 '18 at 4:51
davidlowryduda♦davidlowryduda
74.4k7118252
answered Dec 2 '18 at 4:51
davidlowryduda♦davidlowryduda
74.4k7118252
answered Dec 2 '18 at 4:51
davidlowryduda♦davidlowryduda
74.4k7118252
answered Dec 2 '18 at 4:51
davidlowryduda♦davidlowryduda
74.4k7118252
74.4k7118252
add a comment |
add a comment |
$begingroup$
Ok, thanks, now I see the problem. Kummer's $rho_n$ is $D_n a_n/a_{n+1}-D_{n+1}$ and for convergence, there must be a $c>0$ such that $rho_n ge c$ which is NOT the same as $rho_n > 0$, which I mistakenly supposed. If $a_n=1/n^2$, then $rho_n=(1+1/n)^2$ which, although it is greater than zero for all $nge 1$, there is no $c>0$ that it is greater than or equal to for any $n$.
answered Dec 2 '18 at 5:05
Paul R.Paul R.
213
$endgroup$
add a comment |
$begingroup$
Ok, thanks, now I see the problem. Kummer's $rho_n$ is $D_n a_n/a_{n+1}-D_{n+1}$ and for convergence, there must be a $c>0$ such that $rho_n ge c$ which is NOT the same as $rho_n > 0$, which I mistakenly supposed. If $a_n=1/n^2$, then $rho_n=(1+1/n)^2$ which, although it is greater than zero for all $nge 1$, there is no $c>0$ that it is greater than or equal to for any $n$.
answered Dec 2 '18 at 5:05
Paul R.Paul R.
213
$endgroup$
add a comment |
$begingroup$
Ok, thanks, now I see the problem. Kummer's $rho_n$ is $D_n a_n/a_{n+1}-D_{n+1}$ and for convergence, there must be a $c>0$ such that $rho_n ge c$ which is NOT the same as $rho_n > 0$, which I mistakenly supposed. If $a_n=1/n^2$, then $rho_n=(1+1/n)^2$ which, although it is greater than zero for all $nge 1$, there is no $c>0$ that it is greater than or equal to for any $n$.
answered Dec 2 '18 at 5:05
Paul R.Paul R.
213
$endgroup$
Ok, thanks, now I see the problem. Kummer's $rho_n$ is $D_n a_n/a_{n+1}-D_{n+1}$ and for convergence, there must be a $c>0$ such that $rho_n ge c$ which is NOT the same as $rho_n > 0$, which I mistakenly supposed. If $a_n=1/n^2$, then $rho_n=(1+1/n)^2$ which, although it is greater than zero for all $nge 1$, there is no $c>0$ that it is greater than or equal to for any $n$.
answered Dec 2 '18 at 5:05
Paul R.Paul R.
213
answered Dec 2 '18 at 5:05
Paul R.Paul R.
213
answered Dec 2 '18 at 5:05
Paul R.Paul R.
213
answered Dec 2 '18 at 5:05
Paul R.Paul R.
213
213
add a comment |
add a comment |
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