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When to use the root test. Is this not a good situation to use it?

2

$begingroup$

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

share|cite|improve this question

asked 4 hours ago

Jwan622Jwan622

2,38011632

$endgroup$

add a comment | 

2

$begingroup$

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

share|cite|improve this question

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:

$$sum_n=1^infty fracx^nn^44^n$$

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

share|cite|improve this question

asked 4 hours ago

Jwan622Jwan622

2,38011632

$endgroup$

share|cite|improve this question

asked 4 hours ago

Jwan622Jwan622

2,38011632

share|cite|improve this question

asked 4 hours ago

Jwan622Jwan622

2,38011632

asked 4 hours ago

Jwan622Jwan622

2,38011632

asked 4 hours ago

Jwan622Jwan622

2,38011632

2 Answers
2

active

oldest

votes

4

$begingroup$

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.

share|cite|improve this answer

answered 3 hours ago

MelodyMelody

1,07412

$endgroup$

add a comment | 

2

$begingroup$

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\
&lt 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
$.

share|cite|improve this answer

answered 3 hours ago

marty cohenmarty cohen

75.2k549130

$endgroup$

add a comment | 

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4

$begingroup$

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.

share|cite|improve this answer

answered 3 hours ago

MelodyMelody

1,07412

$endgroup$

add a comment | 

4

$begingroup$

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.

share|cite|improve this answer

answered 3 hours ago

MelodyMelody

1,07412

$endgroup$

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.

share|cite|improve this answer

answered 3 hours ago

MelodyMelody

1,07412

$endgroup$

share|cite|improve this answer

answered 3 hours ago

MelodyMelody

1,07412

answered 3 hours ago

MelodyMelody

1,07412

answered 3 hours ago

MelodyMelody

1,07412

2

$begingroup$

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\
&lt 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
$.

share|cite|improve this answer

answered 3 hours ago

marty cohenmarty cohen

75.2k549130

$endgroup$

add a comment | 

2

$begingroup$

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\
&lt 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
$.

share|cite|improve this answer

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
$lim_n to infty fracln(n)n
=0$.

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\
&lt 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
$.

share|cite|improve this answer

answered 3 hours ago

marty cohenmarty cohen

75.2k549130

$endgroup$

share|cite|improve this answer

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