Mixing
Does convergence by the ratio test for sequences implies that the sequence is monotonic?
0
$begingroup$
I wonder, Let's say $(a_n)_n$ is a sequence.
If $ frac{a_{n+1}}{a_n} to L $ and $ L<1$. Then $ a_n to 0 $.
But, does it mean that $a_n$ is monotonically decreasing? (For every $n$, $a_n > a_{n+1} $ ) ?
Thank you.
calculus sequences-and-series convergence ratio
edited Jan 13 at 9:35
Robert Z
97.4k1066137
asked Jan 13 at 9:30
Dvir PeretzDvir Peretz
657
$endgroup$
add a comment |
0
$begingroup$
I wonder, Let's say $(a_n)_n$ is a sequence.
If $ frac{a_{n+1}}{a_n} to L $ and $ L<1$. Then $ a_n to 0 $.
But, does it mean that $a_n$ is monotonically decreasing? (For every $n$, $a_n > a_{n+1} $ ) ?
Thank you.
calculus sequences-and-series convergence ratio
edited Jan 13 at 9:35
Robert Z
97.4k1066137
asked Jan 13 at 9:30
Dvir PeretzDvir Peretz
657
$endgroup$
2
$begingroup$
Yes, if $a_n>0$ then $frac{a_{n+1}}{a_n} to L$ with $L<1$ implies that eventually the sequence is monotonically decreasing.
$endgroup$
– Robert Z
Jan 13 at 9:33
$begingroup$
First comment: you seem to consider only positive sequences but you didn't state that $a_n>0$. Secondly, $a_n >a_{n+1}$ for every $n$ is obviously false. You can change the values of $a_1$ and $a_2$ any way you like without changing the value of $lim frac {a_{n+1}} {a_n}$.
$endgroup$
– Kavi Rama Murthy
Jan 13 at 12:22
add a comment |
0
0
0
$begingroup$
I wonder, Let's say $(a_n)_n$ is a sequence.
If $ frac{a_{n+1}}{a_n} to L $ and $ L<1$. Then $ a_n to 0 $.
But, does it mean that $a_n$ is monotonically decreasing? (For every $n$, $a_n > a_{n+1} $ ) ?
Thank you.
calculus sequences-and-series convergence ratio
edited Jan 13 at 9:35
Robert Z
97.4k1066137
asked Jan 13 at 9:30
Dvir PeretzDvir Peretz
657
$endgroup$
I wonder, Let's say $(a_n)_n$ is a sequence.
If $ frac{a_{n+1}}{a_n} to L $ and $ L<1$. Then $ a_n to 0 $.
But, does it mean that $a_n$ is monotonically decreasing? (For every $n$, $a_n > a_{n+1} $ ) ?
Thank you.
calculus sequences-and-series convergence ratio
calculus sequences-and-series convergence ratio
edited Jan 13 at 9:35
Robert Z
97.4k1066137
asked Jan 13 at 9:30
Dvir PeretzDvir Peretz
657
edited Jan 13 at 9:35
Robert Z
97.4k1066137
asked Jan 13 at 9:30
Dvir PeretzDvir Peretz
657
edited Jan 13 at 9:35
Robert Z
97.4k1066137
edited Jan 13 at 9:35
Robert Z
97.4k1066137
edited Jan 13 at 9:35
Robert Z
97.4k1066137
97.4k1066137
asked Jan 13 at 9:30
Dvir PeretzDvir Peretz
657
asked Jan 13 at 9:30
Dvir PeretzDvir Peretz
657
asked Jan 13 at 9:30
Dvir PeretzDvir Peretz
657
657
2
$begingroup$
Yes, if $a_n>0$ then $frac{a_{n+1}}{a_n} to L$ with $L<1$ implies that eventually the sequence is monotonically decreasing.
$endgroup$
– Robert Z
Jan 13 at 9:33
$begingroup$
First comment: you seem to consider only positive sequences but you didn't state that $a_n>0$. Secondly, $a_n >a_{n+1}$ for every $n$ is obviously false. You can change the values of $a_1$ and $a_2$ any way you like without changing the value of $lim frac {a_{n+1}} {a_n}$.
$endgroup$
– Kavi Rama Murthy
Jan 13 at 12:22
add a comment |
2
$begingroup$
Yes, if $a_n>0$ then $frac{a_{n+1}}{a_n} to L$ with $L<1$ implies that eventually the sequence is monotonically decreasing.
$endgroup$
– Robert Z
Jan 13 at 9:33
$begingroup$
First comment: you seem to consider only positive sequences but you didn't state that $a_n>0$. Secondly, $a_n >a_{n+1}$ for every $n$ is obviously false. You can change the values of $a_1$ and $a_2$ any way you like without changing the value of $lim frac {a_{n+1}} {a_n}$.
$endgroup$
– Kavi Rama Murthy
Jan 13 at 12:22
2
2
$begingroup$
Yes, if $a_n>0$ then $frac{a_{n+1}}{a_n} to L$ with $L<1$ implies that eventually the sequence is monotonically decreasing.
$endgroup$
– Robert Z
Jan 13 at 9:33
$begingroup$
Yes, if $a_n>0$ then $frac{a_{n+1}}{a_n} to L$ with $L<1$ implies that eventually the sequence is monotonically decreasing.
$endgroup$
– Robert Z
Jan 13 at 9:33
$begingroup$
First comment: you seem to consider only positive sequences but you didn't state that $a_n>0$. Secondly, $a_n >a_{n+1}$ for every $n$ is obviously false. You can change the values of $a_1$ and $a_2$ any way you like without changing the value of $lim frac {a_{n+1}} {a_n}$.
$endgroup$
– Kavi Rama Murthy
Jan 13 at 12:22
$begingroup$
First comment: you seem to consider only positive sequences but you didn't state that $a_n>0$. Secondly, $a_n >a_{n+1}$ for every $n$ is obviously false. You can change the values of $a_1$ and $a_2$ any way you like without changing the value of $lim frac {a_{n+1}} {a_n}$.
$endgroup$
– Kavi Rama Murthy
Jan 13 at 12:22
add a comment |
0
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2
$begingroup$
Yes, if $a_n>0$ then $frac{a_{n+1}}{a_n} to L$ with $L<1$ implies that eventually the sequence is monotonically decreasing.
$endgroup$
– Robert Z
Jan 13 at 9:33
$begingroup$
First comment: you seem to consider only positive sequences but you didn't state that $a_n>0$. Secondly, $a_n >a_{n+1}$ for every $n$ is obviously false. You can change the values of $a_1$ and $a_2$ any way you like without changing the value of $lim frac {a_{n+1}} {a_n}$.
$endgroup$
– Kavi Rama Murthy
Jan 13 at 12:22