When to use the root test. Is this not a good situation to use it? The 2019 Stack Overflow...
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When to use the root test. Is this not a good situation to use it?
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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 $sum_{n=1}^{infty}(-1)^nfrac{x^{2n}}{(2n)!}$ by root and ratio test are different?How would I use root/ratio test on $sum_{n=1}^inftyleft(frac{n}{n+1}right)^{n^2}$?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:
$$sum_{n=1}^{infty} frac{x^n}{n^44^n}$$
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:
$$biggr lbrack frac{a_{n+1}}{a_n} biggr rbrack = biggr lbrack frac{x^{n+1}}{(n+1)^4 * 4^{n+1}} * frac{n^4*4^n}{x^n} biggr rbrack = biggr lbrack frac{x*n^4}{(n+1)^4 * 4} biggr rbrack = frac{x}{4}$$
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 4 hours ago
Jwan622Jwan622
2,38011632
$endgroup$
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:
$$sum_{n=1}^{infty} frac{x^n}{n^44^n}$$
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:
$$biggr lbrack frac{a_{n+1}}{a_n} biggr rbrack = biggr lbrack frac{x^{n+1}}{(n+1)^4 * 4^{n+1}} * frac{n^4*4^n}{x^n} biggr rbrack = biggr lbrack frac{x*n^4}{(n+1)^4 * 4} biggr rbrack = frac{x}{4}$$
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 4 hours ago
Jwan622Jwan622
2,38011632
$endgroup$
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:
$$sum_{n=1}^{infty} frac{x^n}{n^44^n}$$
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:
$$biggr lbrack frac{a_{n+1}}{a_n} biggr rbrack = biggr lbrack frac{x^{n+1}}{(n+1)^4 * 4^{n+1}} * frac{n^4*4^n}{x^n} biggr rbrack = biggr lbrack frac{x*n^4}{(n+1)^4 * 4} biggr rbrack = frac{x}{4}$$
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 4 hours ago
Jwan622Jwan622
2,38011632
$endgroup$
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:
$$sum_{n=1}^{infty} frac{x^n}{n^44^n}$$
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:
$$biggr lbrack frac{a_{n+1}}{a_n} biggr rbrack = biggr lbrack frac{x^{n+1}}{(n+1)^4 * 4^{n+1}} * frac{n^4*4^n}{x^n} biggr rbrack = biggr lbrack frac{x*n^4}{(n+1)^4 * 4} biggr rbrack = frac{x}{4}$$
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 4 hours ago
Jwan622Jwan622
2,38011632
asked 4 hours ago
Jwan622Jwan622
2,38011632
asked 4 hours ago
Jwan622Jwan622
2,38011632
asked 4 hours ago
Jwan622Jwan622
2,38011632
asked 4 hours ago
Jwan622Jwan622
2,38011632
2,38011632
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It doesn't cause any problems, because $lim_{ntoinfty}sqrt[n]{n^4}=1.$ 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 3 hours ago
MelodyMelody
1,07412
$endgroup$
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
$lim_{n to infty} frac{ln(n)}{n}
=0$.
An easy,
but nonelementary proof of this is this:
$begin{array}\
ln(n)
&=int_1^n dfrac{dt}{t}\
&<int_1^n dfrac{dt}{t^{1/2}}\
&=2t^{1/2}|_1^n\
< 2sqrt{n}\
text{so}\
dfrac{ln(n)}{n}
&<dfrac{2}{sqrt{n}}\
end{array}
$
Therefore
$ (n^k)^{1/n}
=n^{k/n}
=e^{k ln(n)/n}
lt e^{2k/sqrt{n}}
to 1
$.
answered 3 hours ago
marty cohenmarty cohen
75.2k549130
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It doesn't cause any problems, because $lim_{ntoinfty}sqrt[n]{n^4}=1.$ 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 3 hours ago
MelodyMelody
1,07412
$endgroup$
add a comment |
$begingroup$
It doesn't cause any problems, because $lim_{ntoinfty}sqrt[n]{n^4}=1.$ 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 3 hours ago
MelodyMelody
1,07412
$endgroup$
add a comment |
$begingroup$
It doesn't cause any problems, because $lim_{ntoinfty}sqrt[n]{n^4}=1.$ 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 3 hours ago
MelodyMelody
1,07412
$endgroup$
It doesn't cause any problems, because $lim_{ntoinfty}sqrt[n]{n^4}=1.$ 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 3 hours ago
MelodyMelody
1,07412
answered 3 hours ago
MelodyMelody
1,07412
answered 3 hours ago
MelodyMelody
1,07412
answered 3 hours ago
MelodyMelody
1,07412
1,07412
add a comment |
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
$lim_{n to infty} frac{ln(n)}{n}
=0$.
An easy,
but nonelementary proof of this is this:
$begin{array}\
ln(n)
&=int_1^n dfrac{dt}{t}\
&<int_1^n dfrac{dt}{t^{1/2}}\
&=2t^{1/2}|_1^n\
< 2sqrt{n}\
text{so}\
dfrac{ln(n)}{n}
&<dfrac{2}{sqrt{n}}\
end{array}
$
Therefore
$ (n^k)^{1/n}
=n^{k/n}
=e^{k ln(n)/n}
lt e^{2k/sqrt{n}}
to 1
$.
answered 3 hours ago
marty cohenmarty cohen
75.2k549130
$endgroup$
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
$lim_{n to infty} frac{ln(n)}{n}
=0$.
An easy,
but nonelementary proof of this is this:
$begin{array}\
ln(n)
&=int_1^n dfrac{dt}{t}\
&<int_1^n dfrac{dt}{t^{1/2}}\
&=2t^{1/2}|_1^n\
< 2sqrt{n}\
text{so}\
dfrac{ln(n)}{n}
&<dfrac{2}{sqrt{n}}\
end{array}
$
Therefore
$ (n^k)^{1/n}
=n^{k/n}
=e^{k ln(n)/n}
lt e^{2k/sqrt{n}}
to 1
$.
answered 3 hours ago
marty cohenmarty cohen
75.2k549130
$endgroup$
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
$lim_{n to infty} frac{ln(n)}{n}
=0$.
An easy,
but nonelementary proof of this is this:
$begin{array}\
ln(n)
&=int_1^n dfrac{dt}{t}\
&<int_1^n dfrac{dt}{t^{1/2}}\
&=2t^{1/2}|_1^n\
< 2sqrt{n}\
text{so}\
dfrac{ln(n)}{n}
&<dfrac{2}{sqrt{n}}\
end{array}
$
Therefore
$ (n^k)^{1/n}
=n^{k/n}
=e^{k ln(n)/n}
lt e^{2k/sqrt{n}}
to 1
$.
answered 3 hours ago
marty cohenmarty cohen
75.2k549130
$endgroup$
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
$lim_{n to infty} frac{ln(n)}{n}
=0$.
An easy,
but nonelementary proof of this is this:
$begin{array}\
ln(n)
&=int_1^n dfrac{dt}{t}\
&<int_1^n dfrac{dt}{t^{1/2}}\
&=2t^{1/2}|_1^n\
< 2sqrt{n}\
text{so}\
dfrac{ln(n)}{n}
&<dfrac{2}{sqrt{n}}\
end{array}
$
Therefore
$ (n^k)^{1/n}
=n^{k/n}
=e^{k ln(n)/n}
lt e^{2k/sqrt{n}}
to 1
$.
answered 3 hours ago
marty cohenmarty cohen
75.2k549130
answered 3 hours ago
marty cohenmarty cohen
75.2k549130
answered 3 hours ago
marty cohenmarty cohen
75.2k549130
answered 3 hours ago
marty cohenmarty cohen
75.2k549130
75.2k549130
add a comment |
add a comment |
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