When to use the root test. Is this not a good situation to use it? The 2019 Stack Overflow Developer Survey Results Are InWhich test would be appropriate to use on this series to show convergence/divergence?Integral test vs root test vs ratio testHow to show convergence or divergence of a series when the ratio test is inconclusive?Root test with nested power function?Confused about using alternating test, ratio test, and root test (please help).Radius and interval of convergence of $sum_n=1^infty(-1)^nfracx^2n(2n)!$ by root and ratio test are different?How would I use root/ratio test on $sum_n=1^inftyleft(fracnn+1right)^n^2$?How would I know when to use what test for convergence?convergence of a sum fails with root testIntuition for Root Test.
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When to use the root test. Is this not a good situation to use it?
The 2019 Stack Overflow Developer Survey Results Are InWhich test would be appropriate to use on this series to show convergence/divergence?Integral test vs root test vs ratio testHow to show convergence or divergence of a series when the ratio test is inconclusive?Root test with nested power function?Confused about using alternating test, ratio test, and root test (please help).Radius and interval of convergence of $sum_n=1^infty(-1)^nfracx^2n(2n)!$ by root and ratio test are different?How would I use root/ratio test on $sum_n=1^inftyleft(fracnn+1right)^n^2$?How would I know when to use what test for convergence?convergence of a sum fails with root testIntuition for Root Test.
$begingroup$
I'm having trouble seeing when to use the root test. nth powers occur, but I think the ratio test is easier:
Here is the problem:
$$sum_n=1^infty fracx^nn^44^n$$
So the ratio test seems to work here, but can't the root test be used to? The problem is that the $n^4$ doesnt play well with the root test right?
Here is the beginning of my solution with the ratio test:
$$biggr lbrack fraca_n+1a_n biggr rbrack = biggr lbrack fracx^n+1(n+1)^4 * 4^n+1 * fracn^4*4^nx^n biggr rbrack = biggr lbrack fracx*n^4(n+1)^4 * 4 biggr rbrack = fracx4$$
So I don't think the explanation for when to use the root test is totally right right? I can't really use it here because the $n^4$ causes some problems with the root test right?
sequences-and-series
asked 2 days ago
Jwan622Jwan622
2,38711632
$endgroup$
add a comment |
$begingroup$
I'm having trouble seeing when to use the root test. nth powers occur, but I think the ratio test is easier:
Here is the problem:
$$sum_n=1^infty fracx^nn^44^n$$
So the ratio test seems to work here, but can't the root test be used to? The problem is that the $n^4$ doesnt play well with the root test right?
Here is the beginning of my solution with the ratio test:
$$biggr lbrack fraca_n+1a_n biggr rbrack = biggr lbrack fracx^n+1(n+1)^4 * 4^n+1 * fracn^4*4^nx^n biggr rbrack = biggr lbrack fracx*n^4(n+1)^4 * 4 biggr rbrack = fracx4$$
So I don't think the explanation for when to use the root test is totally right right? I can't really use it here because the $n^4$ causes some problems with the root test right?
sequences-and-series
asked 2 days ago
Jwan622Jwan622
2,38711632
$endgroup$
add a comment |
$begingroup$
I'm having trouble seeing when to use the root test. nth powers occur, but I think the ratio test is easier:
Here is the problem:
$$sum_n=1^infty fracx^nn^44^n$$
So the ratio test seems to work here, but can't the root test be used to? The problem is that the $n^4$ doesnt play well with the root test right?
Here is the beginning of my solution with the ratio test:
$$biggr lbrack fraca_n+1a_n biggr rbrack = biggr lbrack fracx^n+1(n+1)^4 * 4^n+1 * fracn^4*4^nx^n biggr rbrack = biggr lbrack fracx*n^4(n+1)^4 * 4 biggr rbrack = fracx4$$
So I don't think the explanation for when to use the root test is totally right right? I can't really use it here because the $n^4$ causes some problems with the root test right?
sequences-and-series
asked 2 days ago
Jwan622Jwan622
2,38711632
$endgroup$
I'm having trouble seeing when to use the root test. nth powers occur, but I think the ratio test is easier:
Here is the problem:
$$sum_n=1^infty fracx^nn^44^n$$
So the ratio test seems to work here, but can't the root test be used to? The problem is that the $n^4$ doesnt play well with the root test right?
Here is the beginning of my solution with the ratio test:
$$biggr lbrack fraca_n+1a_n biggr rbrack = biggr lbrack fracx^n+1(n+1)^4 * 4^n+1 * fracn^4*4^nx^n biggr rbrack = biggr lbrack fracx*n^4(n+1)^4 * 4 biggr rbrack = fracx4$$
So I don't think the explanation for when to use the root test is totally right right? I can't really use it here because the $n^4$ causes some problems with the root test right?
sequences-and-series
sequences-and-series
asked 2 days ago
Jwan622Jwan622
2,38711632
asked 2 days ago
Jwan622Jwan622
2,38711632
asked 2 days ago
Jwan622Jwan622
2,38711632
asked 2 days ago
Jwan622Jwan622
2,38711632
asked 2 days ago
Jwan622Jwan622
2,38711632
2,38711632
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
When doing a root test,
powers of $n$ can be ignored
because,
for any fixed $k$,
$lim_n to infty (n^k)^1/n
=1
$.
This is because
$ (n^k)^1/n
=n^k/n
=e^k ln(n)/n
$
and
$lim_n to infty fracln(n)n
=0$.
An easy,
but nonelementary proof of this is this:
$beginarray\
ln(n)
&=int_1^n dfracdtt\
&<int_1^n dfracdtt^1/2\
&=2t^1/2|_1^n\
< 2sqrtn\
textso\
dfracln(n)n
&<dfrac2sqrtn\
endarray
$
Therefore
$ (n^k)^1/n
=n^k/n
=e^k ln(n)/n
lt e^2k/sqrtn
to 1
$.
answered 2 days ago
marty cohenmarty cohen
75.3k549130
$endgroup$
add a comment |
$begingroup$
It doesn't cause any problems, because $lim_ntoinftysqrt[n]n^4=1.$ Actually, the root test is stronger than the ratio test. Sometimes the root test limit exists, but the ratio test limit does not. However, if they both exist, then they are equal. Which is why if one limit is $1$ you shouldn't try the other, even though the root test is stronger.
answered 2 days ago
MelodyMelody
1,09412
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
When doing a root test,
powers of $n$ can be ignored
because,
for any fixed $k$,
$lim_n to infty (n^k)^1/n
=1
$.
This is because
$ (n^k)^1/n
=n^k/n
=e^k ln(n)/n
$
and
$lim_n to infty fracln(n)n
=0$.
An easy,
but nonelementary proof of this is this:
$beginarray\
ln(n)
&=int_1^n dfracdtt\
&<int_1^n dfracdtt^1/2\
&=2t^1/2|_1^n\
< 2sqrtn\
textso\
dfracln(n)n
&<dfrac2sqrtn\
endarray
$
Therefore
$ (n^k)^1/n
=n^k/n
=e^k ln(n)/n
lt e^2k/sqrtn
to 1
$.
answered 2 days ago
marty cohenmarty cohen
75.3k549130
$endgroup$
add a comment |
$begingroup$
When doing a root test,
powers of $n$ can be ignored
because,
for any fixed $k$,
$lim_n to infty (n^k)^1/n
=1
$.
This is because
$ (n^k)^1/n
=n^k/n
=e^k ln(n)/n
$
and
$lim_n to infty fracln(n)n
=0$.
An easy,
but nonelementary proof of this is this:
$beginarray\
ln(n)
&=int_1^n dfracdtt\
&<int_1^n dfracdtt^1/2\
&=2t^1/2|_1^n\
< 2sqrtn\
textso\
dfracln(n)n
&<dfrac2sqrtn\
endarray
$
Therefore
$ (n^k)^1/n
=n^k/n
=e^k ln(n)/n
lt e^2k/sqrtn
to 1
$.
answered 2 days ago
marty cohenmarty cohen
75.3k549130
$endgroup$
add a comment |
$begingroup$
When doing a root test,
powers of $n$ can be ignored
because,
for any fixed $k$,
$lim_n to infty (n^k)^1/n
=1
$.
This is because
$ (n^k)^1/n
=n^k/n
=e^k ln(n)/n
$
and
$lim_n to infty fracln(n)n
=0$.
An easy,
but nonelementary proof of this is this:
$beginarray\
ln(n)
&=int_1^n dfracdtt\
&<int_1^n dfracdtt^1/2\
&=2t^1/2|_1^n\
< 2sqrtn\
textso\
dfracln(n)n
&<dfrac2sqrtn\
endarray
$
Therefore
$ (n^k)^1/n
=n^k/n
=e^k ln(n)/n
lt e^2k/sqrtn
to 1
$.
answered 2 days ago
marty cohenmarty cohen
75.3k549130
$endgroup$
When doing a root test,
powers of $n$ can be ignored
because,
for any fixed $k$,
$lim_n to infty (n^k)^1/n
=1
$.
This is because
$ (n^k)^1/n
=n^k/n
=e^k ln(n)/n
$
and
$lim_n to infty fracln(n)n
=0$.
An easy,
but nonelementary proof of this is this:
$beginarray\
ln(n)
&=int_1^n dfracdtt\
&<int_1^n dfracdtt^1/2\
&=2t^1/2|_1^n\
< 2sqrtn\
textso\
dfracln(n)n
&<dfrac2sqrtn\
endarray
$
Therefore
$ (n^k)^1/n
=n^k/n
=e^k ln(n)/n
lt e^2k/sqrtn
to 1
$.
answered 2 days ago
marty cohenmarty cohen
75.3k549130
answered 2 days ago
marty cohenmarty cohen
75.3k549130
answered 2 days ago
marty cohenmarty cohen
75.3k549130
answered 2 days ago
marty cohenmarty cohen
75.3k549130
75.3k549130
add a comment |
add a comment |
$begingroup$
It doesn't cause any problems, because $lim_ntoinftysqrt[n]n^4=1.$ Actually, the root test is stronger than the ratio test. Sometimes the root test limit exists, but the ratio test limit does not. However, if they both exist, then they are equal. Which is why if one limit is $1$ you shouldn't try the other, even though the root test is stronger.
answered 2 days ago
MelodyMelody
1,09412
$endgroup$
add a comment |
$begingroup$
It doesn't cause any problems, because $lim_ntoinftysqrt[n]n^4=1.$ Actually, the root test is stronger than the ratio test. Sometimes the root test limit exists, but the ratio test limit does not. However, if they both exist, then they are equal. Which is why if one limit is $1$ you shouldn't try the other, even though the root test is stronger.
answered 2 days ago
MelodyMelody
1,09412
$endgroup$
add a comment |
$begingroup$
It doesn't cause any problems, because $lim_ntoinftysqrt[n]n^4=1.$ Actually, the root test is stronger than the ratio test. Sometimes the root test limit exists, but the ratio test limit does not. However, if they both exist, then they are equal. Which is why if one limit is $1$ you shouldn't try the other, even though the root test is stronger.
answered 2 days ago
MelodyMelody
1,09412
$endgroup$
It doesn't cause any problems, because $lim_ntoinftysqrt[n]n^4=1.$ Actually, the root test is stronger than the ratio test. Sometimes the root test limit exists, but the ratio test limit does not. However, if they both exist, then they are equal. Which is why if one limit is $1$ you shouldn't try the other, even though the root test is stronger.
answered 2 days ago
MelodyMelody
1,09412
answered 2 days ago
MelodyMelody
1,09412
answered 2 days ago
MelodyMelody
1,09412
answered 2 days ago
MelodyMelody
1,09412
1,09412
add a comment |
add a comment |
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-sequences-and-series
