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Calculate $lim_n rightarrow inftyn^x (a_1.a_2…a_n)^frac1n$?
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Suppose that $a_n$ is a sequence such that $displaystylelim_n rightarrowinfty n^xa_n=a $ for some real $x$. Calculate
$$lim_n rightarrow inftyn^x (a_1.a_2......a_n)^frac1n$$
My attempts : i take $a_1=a_2 =.......=a_n = a$
after that $lim_n rightarrow infty$ $n^x (a_1.a_2......a_n)^frac1n= infty . a = infty$
Is it correct ?? or not
pliz help me,
any HINTS/SOLUTion.....
calculus real-analysis
edited Jul 20 at 11:25
Nosrati
19.5k41544
asked Jul 20 at 11:18
stupid
55819
 |Â
show 3 more comments
up vote
5
down vote
favorite
Suppose that $a_n$ is a sequence such that $displaystylelim_n rightarrowinfty n^xa_n=a $ for some real $x$. Calculate
$$lim_n rightarrow inftyn^x (a_1.a_2......a_n)^frac1n$$
My attempts : i take $a_1=a_2 =.......=a_n = a$
after that $lim_n rightarrow infty$ $n^x (a_1.a_2......a_n)^frac1n= infty . a = infty$
Is it correct ?? or not
pliz help me,
any HINTS/SOLUTion.....
calculus real-analysis
edited Jul 20 at 11:25
Nosrati
19.5k41544
asked Jul 20 at 11:18
stupid
55819
Or this one: math.stackexchange.com/q/352935/42969.
– Martin R
Jul 20 at 11:29
Well, the first error that you make is in taking $a_i = a$. Since, then $lim n^x a_n = a lim n^x $ which equals $0$ if $x<0$, $a$ if $x=0$ and $textsign(a)infty$ if $x>0$.
– Stan Tendijck
Jul 20 at 11:30
1
No, there will be an extra factor. Note $$n^x,left(prod_i=1^n,a_iright)^frac1n=fracn^xleft(prod_i=1^n,i^xright)^frac1n,left(prod_i=1^n,x_iright)^frac1n=left(fracn^nn!right)^fracxn,left(prod_i=1^n,x_iright)^frac1n,.$$
– Batominovski
Jul 20 at 11:44
1
@MartinR But please don't delete both links. They are very useful to this problem.
– Batominovski
Jul 20 at 11:47
1
@JoséCarlosSantos This is not a duplicate. Please read the comments.
– Batominovski
Jul 20 at 13:54
 |Â
show 3 more comments
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Suppose that $a_n$ is a sequence such that $displaystylelim_n rightarrowinfty n^xa_n=a $ for some real $x$. Calculate
$$lim_n rightarrow inftyn^x (a_1.a_2......a_n)^frac1n$$
My attempts : i take $a_1=a_2 =.......=a_n = a$
after that $lim_n rightarrow infty$ $n^x (a_1.a_2......a_n)^frac1n= infty . a = infty$
Is it correct ?? or not
pliz help me,
any HINTS/SOLUTion.....
calculus real-analysis
edited Jul 20 at 11:25
Nosrati
19.5k41544
asked Jul 20 at 11:18
stupid
55819
Suppose that $a_n$ is a sequence such that $displaystylelim_n rightarrowinfty n^xa_n=a $ for some real $x$. Calculate
$$lim_n rightarrow inftyn^x (a_1.a_2......a_n)^frac1n$$
My attempts : i take $a_1=a_2 =.......=a_n = a$
after that $lim_n rightarrow infty$ $n^x (a_1.a_2......a_n)^frac1n= infty . a = infty$
Is it correct ?? or not
pliz help me,
any HINTS/SOLUTion.....
calculus real-analysis
edited Jul 20 at 11:25
Nosrati
19.5k41544
asked Jul 20 at 11:18
stupid
55819
edited Jul 20 at 11:25
Nosrati
19.5k41544
edited Jul 20 at 11:25
Nosrati
19.5k41544
edited Jul 20 at 11:25
Nosrati
19.5k41544
19.5k41544
asked Jul 20 at 11:18
stupid
55819
asked Jul 20 at 11:18
stupid
55819
asked Jul 20 at 11:18
stupid
55819
55819
Or this one: math.stackexchange.com/q/352935/42969.
– Martin R
Jul 20 at 11:29
Well, the first error that you make is in taking $a_i = a$. Since, then $lim n^x a_n = a lim n^x $ which equals $0$ if $x<0$, $a$ if $x=0$ and $textsign(a)infty$ if $x>0$.
– Stan Tendijck
Jul 20 at 11:30
1
No, there will be an extra factor. Note $$n^x,left(prod_i=1^n,a_iright)^frac1n=fracn^xleft(prod_i=1^n,i^xright)^frac1n,left(prod_i=1^n,x_iright)^frac1n=left(fracn^nn!right)^fracxn,left(prod_i=1^n,x_iright)^frac1n,.$$
– Batominovski
Jul 20 at 11:44
1
@MartinR But please don't delete both links. They are very useful to this problem.
– Batominovski
Jul 20 at 11:47
1
@JoséCarlosSantos This is not a duplicate. Please read the comments.
– Batominovski
Jul 20 at 13:54
 |Â
show 3 more comments
Or this one: math.stackexchange.com/q/352935/42969.
– Martin R
Jul 20 at 11:29
Well, the first error that you make is in taking $a_i = a$. Since, then $lim n^x a_n = a lim n^x $ which equals $0$ if $x<0$, $a$ if $x=0$ and $textsign(a)infty$ if $x>0$.
– Stan Tendijck
Jul 20 at 11:30
1
No, there will be an extra factor. Note $$n^x,left(prod_i=1^n,a_iright)^frac1n=fracn^xleft(prod_i=1^n,i^xright)^frac1n,left(prod_i=1^n,x_iright)^frac1n=left(fracn^nn!right)^fracxn,left(prod_i=1^n,x_iright)^frac1n,.$$
– Batominovski
Jul 20 at 11:44
1
@MartinR But please don't delete both links. They are very useful to this problem.
– Batominovski
Jul 20 at 11:47
1
@JoséCarlosSantos This is not a duplicate. Please read the comments.
– Batominovski
Jul 20 at 13:54
Or this one: math.stackexchange.com/q/352935/42969.
– Martin R
Jul 20 at 11:29
Or this one: math.stackexchange.com/q/352935/42969.
– Martin R
Jul 20 at 11:29
Well, the first error that you make is in taking $a_i = a$. Since, then $lim n^x a_n = a lim n^x $ which equals $0$ if $x<0$, $a$ if $x=0$ and $textsign(a)infty$ if $x>0$.
– Stan Tendijck
Jul 20 at 11:30
Well, the first error that you make is in taking $a_i = a$. Since, then $lim n^x a_n = a lim n^x $ which equals $0$ if $x<0$, $a$ if $x=0$ and $textsign(a)infty$ if $x>0$.
– Stan Tendijck
Jul 20 at 11:30
1
1
No, there will be an extra factor. Note $$n^x,left(prod_i=1^n,a_iright)^frac1n=fracn^xleft(prod_i=1^n,i^xright)^frac1n,left(prod_i=1^n,x_iright)^frac1n=left(fracn^nn!right)^fracxn,left(prod_i=1^n,x_iright)^frac1n,.$$
– Batominovski
Jul 20 at 11:44
No, there will be an extra factor. Note $$n^x,left(prod_i=1^n,a_iright)^frac1n=fracn^xleft(prod_i=1^n,i^xright)^frac1n,left(prod_i=1^n,x_iright)^frac1n=left(fracn^nn!right)^fracxn,left(prod_i=1^n,x_iright)^frac1n,.$$
– Batominovski
Jul 20 at 11:44
1
1
@MartinR But please don't delete both links. They are very useful to this problem.
– Batominovski
Jul 20 at 11:47
@MartinR But please don't delete both links. They are very useful to this problem.
– Batominovski
Jul 20 at 11:47
1
1
@JoséCarlosSantos This is not a duplicate. Please read the comments.
– Batominovski
Jul 20 at 13:54
@JoséCarlosSantos This is not a duplicate. Please read the comments.
– Batominovski
Jul 20 at 13:54
 |Â
show 3 more comments
3 Answers
3
active
oldest
votes
up vote
4
down vote
accepted
Assume that $a_n> 0$ for every positive integer $n$; otherwise the limit may not exist. Set $z_n:=n^x,a_n$ like Martin R recommended. Thus,
$$n^x,left(prod_i=1^n,a_iright)^frac1n=fracn^xleft(prodlimits_i=1^n,i^xright)^frac1n,left(prodlimits_i=1^n,z_iright)^frac1n=left(fracn^nn!right)^fracxn,left(prod_i=1^n,z_iright)^frac1n,.$$
Now, since $displaystylelim_ntoinfty,z_n=a$, we have $displaystylelim_ntoinfty,left(prod_i=1^n,z_iright)^frac1n=a$ (see Martin R's link in the comments above). Furthermore, Stirling's approximation $n!approx sqrt2pi nleft(fracntexteright)^n$ implies that
$$lim_ntoinfty,left(fracn^nn!right)^fracxn=lim_ntoinfty,left(fractexte^nsqrt2pi nright)^fracxn=exp(x),.$$ Consequently, $$lim_ntoinfty,n^x,left(prod_i=1^n,a_iright)^frac1n=a,exp(x),.$$
answered Jul 20 at 11:53
Batominovski
23.2k22777
add a comment |Â
up vote
2
down vote
Since
$a_n approx an^-x$,
$beginarray\
n^x (prod_k=1^na_k)^frac1n
&approx n^x left(prod_k=1^n(ak^-x)right)^frac1n\
&= n^x left(a^nn!^-xright)^frac1n\
&= n^x aleft(n!^1/nright)^-x\
&approx n^x aleft(n/eright)^-x\
&= ae^x\
endarray
$
answered Jul 24 at 19:06
marty cohen
69.2k446122
add a comment |Â
up vote
1
down vote
Consider $b_n=n^nx a_1 a_2dots a_n$ and then $b_n+1/b_n=(1+n^-1)^nx(n+1)^xa_n+1to e^xa$ and hence $b_n^1/nto ae^x$.
answered Jul 25 at 0:42
Paramanand Singh
45.2k553142
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Assume that $a_n> 0$ for every positive integer $n$; otherwise the limit may not exist. Set $z_n:=n^x,a_n$ like Martin R recommended. Thus,
$$n^x,left(prod_i=1^n,a_iright)^frac1n=fracn^xleft(prodlimits_i=1^n,i^xright)^frac1n,left(prodlimits_i=1^n,z_iright)^frac1n=left(fracn^nn!right)^fracxn,left(prod_i=1^n,z_iright)^frac1n,.$$
Now, since $displaystylelim_ntoinfty,z_n=a$, we have $displaystylelim_ntoinfty,left(prod_i=1^n,z_iright)^frac1n=a$ (see Martin R's link in the comments above). Furthermore, Stirling's approximation $n!approx sqrt2pi nleft(fracntexteright)^n$ implies that
$$lim_ntoinfty,left(fracn^nn!right)^fracxn=lim_ntoinfty,left(fractexte^nsqrt2pi nright)^fracxn=exp(x),.$$ Consequently, $$lim_ntoinfty,n^x,left(prod_i=1^n,a_iright)^frac1n=a,exp(x),.$$
answered Jul 20 at 11:53
Batominovski
23.2k22777
add a comment |Â
up vote
4
down vote
accepted
Assume that $a_n> 0$ for every positive integer $n$; otherwise the limit may not exist. Set $z_n:=n^x,a_n$ like Martin R recommended. Thus,
$$n^x,left(prod_i=1^n,a_iright)^frac1n=fracn^xleft(prodlimits_i=1^n,i^xright)^frac1n,left(prodlimits_i=1^n,z_iright)^frac1n=left(fracn^nn!right)^fracxn,left(prod_i=1^n,z_iright)^frac1n,.$$
Now, since $displaystylelim_ntoinfty,z_n=a$, we have $displaystylelim_ntoinfty,left(prod_i=1^n,z_iright)^frac1n=a$ (see Martin R's link in the comments above). Furthermore, Stirling's approximation $n!approx sqrt2pi nleft(fracntexteright)^n$ implies that
$$lim_ntoinfty,left(fracn^nn!right)^fracxn=lim_ntoinfty,left(fractexte^nsqrt2pi nright)^fracxn=exp(x),.$$ Consequently, $$lim_ntoinfty,n^x,left(prod_i=1^n,a_iright)^frac1n=a,exp(x),.$$
answered Jul 20 at 11:53
Batominovski
23.2k22777
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Assume that $a_n> 0$ for every positive integer $n$; otherwise the limit may not exist. Set $z_n:=n^x,a_n$ like Martin R recommended. Thus,
$$n^x,left(prod_i=1^n,a_iright)^frac1n=fracn^xleft(prodlimits_i=1^n,i^xright)^frac1n,left(prodlimits_i=1^n,z_iright)^frac1n=left(fracn^nn!right)^fracxn,left(prod_i=1^n,z_iright)^frac1n,.$$
Now, since $displaystylelim_ntoinfty,z_n=a$, we have $displaystylelim_ntoinfty,left(prod_i=1^n,z_iright)^frac1n=a$ (see Martin R's link in the comments above). Furthermore, Stirling's approximation $n!approx sqrt2pi nleft(fracntexteright)^n$ implies that
$$lim_ntoinfty,left(fracn^nn!right)^fracxn=lim_ntoinfty,left(fractexte^nsqrt2pi nright)^fracxn=exp(x),.$$ Consequently, $$lim_ntoinfty,n^x,left(prod_i=1^n,a_iright)^frac1n=a,exp(x),.$$
answered Jul 20 at 11:53
Batominovski
23.2k22777
Assume that $a_n> 0$ for every positive integer $n$; otherwise the limit may not exist. Set $z_n:=n^x,a_n$ like Martin R recommended. Thus,
$$n^x,left(prod_i=1^n,a_iright)^frac1n=fracn^xleft(prodlimits_i=1^n,i^xright)^frac1n,left(prodlimits_i=1^n,z_iright)^frac1n=left(fracn^nn!right)^fracxn,left(prod_i=1^n,z_iright)^frac1n,.$$
Now, since $displaystylelim_ntoinfty,z_n=a$, we have $displaystylelim_ntoinfty,left(prod_i=1^n,z_iright)^frac1n=a$ (see Martin R's link in the comments above). Furthermore, Stirling's approximation $n!approx sqrt2pi nleft(fracntexteright)^n$ implies that
$$lim_ntoinfty,left(fracn^nn!right)^fracxn=lim_ntoinfty,left(fractexte^nsqrt2pi nright)^fracxn=exp(x),.$$ Consequently, $$lim_ntoinfty,n^x,left(prod_i=1^n,a_iright)^frac1n=a,exp(x),.$$
answered Jul 20 at 11:53
Batominovski
23.2k22777
edited Jul 20 at 12:11
answered Jul 20 at 11:53
Batominovski
23.2k22777
answered Jul 20 at 11:53
Batominovski
23.2k22777
answered Jul 20 at 11:53
Batominovski
23.2k22777
23.2k22777
add a comment |Â
add a comment |Â
up vote
2
down vote
Since
$a_n approx an^-x$,
$beginarray\
n^x (prod_k=1^na_k)^frac1n
&approx n^x left(prod_k=1^n(ak^-x)right)^frac1n\
&= n^x left(a^nn!^-xright)^frac1n\
&= n^x aleft(n!^1/nright)^-x\
&approx n^x aleft(n/eright)^-x\
&= ae^x\
endarray
$
answered Jul 24 at 19:06
marty cohen
69.2k446122
add a comment |Â
up vote
2
down vote
Since
$a_n approx an^-x$,
$beginarray\
n^x (prod_k=1^na_k)^frac1n
&approx n^x left(prod_k=1^n(ak^-x)right)^frac1n\
&= n^x left(a^nn!^-xright)^frac1n\
&= n^x aleft(n!^1/nright)^-x\
&approx n^x aleft(n/eright)^-x\
&= ae^x\
endarray
$
answered Jul 24 at 19:06
marty cohen
69.2k446122
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Since
$a_n approx an^-x$,
$beginarray\
n^x (prod_k=1^na_k)^frac1n
&approx n^x left(prod_k=1^n(ak^-x)right)^frac1n\
&= n^x left(a^nn!^-xright)^frac1n\
&= n^x aleft(n!^1/nright)^-x\
&approx n^x aleft(n/eright)^-x\
&= ae^x\
endarray
$
answered Jul 24 at 19:06
marty cohen
69.2k446122
Since
$a_n approx an^-x$,
$beginarray\
n^x (prod_k=1^na_k)^frac1n
&approx n^x left(prod_k=1^n(ak^-x)right)^frac1n\
&= n^x left(a^nn!^-xright)^frac1n\
&= n^x aleft(n!^1/nright)^-x\
&approx n^x aleft(n/eright)^-x\
&= ae^x\
endarray
$
answered Jul 24 at 19:06
marty cohen
69.2k446122
answered Jul 24 at 19:06
marty cohen
69.2k446122
answered Jul 24 at 19:06
marty cohen
69.2k446122
answered Jul 24 at 19:06
marty cohen
69.2k446122
69.2k446122
add a comment |Â
add a comment |Â
up vote
1
down vote
Consider $b_n=n^nx a_1 a_2dots a_n$ and then $b_n+1/b_n=(1+n^-1)^nx(n+1)^xa_n+1to e^xa$ and hence $b_n^1/nto ae^x$.
answered Jul 25 at 0:42
Paramanand Singh
45.2k553142
add a comment |Â
up vote
1
down vote
Consider $b_n=n^nx a_1 a_2dots a_n$ and then $b_n+1/b_n=(1+n^-1)^nx(n+1)^xa_n+1to e^xa$ and hence $b_n^1/nto ae^x$.
answered Jul 25 at 0:42
Paramanand Singh
45.2k553142
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Consider $b_n=n^nx a_1 a_2dots a_n$ and then $b_n+1/b_n=(1+n^-1)^nx(n+1)^xa_n+1to e^xa$ and hence $b_n^1/nto ae^x$.
answered Jul 25 at 0:42
Paramanand Singh
45.2k553142
Consider $b_n=n^nx a_1 a_2dots a_n$ and then $b_n+1/b_n=(1+n^-1)^nx(n+1)^xa_n+1to e^xa$ and hence $b_n^1/nto ae^x$.
answered Jul 25 at 0:42
Paramanand Singh
45.2k553142
answered Jul 25 at 0:42
Paramanand Singh
45.2k553142
answered Jul 25 at 0:42
Paramanand Singh
45.2k553142
answered Jul 25 at 0:42
Paramanand Singh
45.2k553142
45.2k553142
add a comment |Â
add a comment |Â
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Or this one: math.stackexchange.com/q/352935/42969.
– Martin R
Jul 20 at 11:29
Well, the first error that you make is in taking $a_i = a$. Since, then $lim n^x a_n = a lim n^x $ which equals $0$ if $x<0$, $a$ if $x=0$ and $textsign(a)infty$ if $x>0$.
– Stan Tendijck
Jul 20 at 11:30
1
No, there will be an extra factor. Note $$n^x,left(prod_i=1^n,a_iright)^frac1n=fracn^xleft(prod_i=1^n,i^xright)^frac1n,left(prod_i=1^n,x_iright)^frac1n=left(fracn^nn!right)^fracxn,left(prod_i=1^n,x_iright)^frac1n,.$$
– Batominovski
Jul 20 at 11:44
1
@MartinR But please don't delete both links. They are very useful to this problem.
– Batominovski
Jul 20 at 11:47
1
@JoséCarlosSantos This is not a duplicate. Please read the comments.
– Batominovski
Jul 20 at 13:54