When to use the root test. Is this not a good situation to use it?
$begingroup$
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
$endgroup$
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
$endgroup$
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
$endgroup$
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 Apr 10 at 2:43
Jwan622Jwan622
2,39211632
2,39211632
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2 Answers
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$begingroup$
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
$.
$endgroup$
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.
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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
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
$lim_{n to infty} frac{ln(n)}{n}
=0$.
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
$.
$endgroup$
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
$.
$endgroup$
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
$.
$endgroup$
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 Apr 10 at 3:32
marty cohenmarty cohen
75.6k549130
75.6k549130
add a comment |
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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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 Apr 10 at 3:00
MelodyMelody
1,28012
1,28012
add a comment |
add a comment |
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