Mixing
Limit of sum of sequences at infinity
Clash Royale CLAN TAG#URR8PPP
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Given two sequences $$a_n=int_0^1 (1-x^2)^n dx$$ and $$b_n=int_0^1 (1-x^3)^n dx$$ ,($nin N$) then find the value of $$L=lim_nto infty (10 sqrt [n]a_n +5sqrt [n]b_n)$$
My try:
$$a_n=int_0^1 (1-x^2)^n dx= frac 12 Bleft( n+1, 1/2right)=frac 12cdot frac Gamma(n+1)Gamma(1/2)Gammaleft(n +frac 32right)$$
Similarly $$b_n=frac 13cdot frac Gamma(n+1)Gamma(1/3)Gammaleft(n +frac 43right)$$
Now let's consider $$J=lim_nto infty (b_n)^1/n =lim_nto infty left(frac 13right)^1/ncdot left(frac Gamma(n+1)Gamma(1/3)Gammaleft(n +frac 43right)right) ^1/n$$ $$=lim_ntoinfty left(frac 13right)^1/ncdot left(lim_nto infty left(frac Gamma(n+1)Gamma(1/3)Gammaleft(n +frac 43right)right) ^1/nright)$$ $$=1cdot lim_nto infty left(frac Gamma(n+1)Gamma(1/3)Gammaleft(n +frac 43right)right) ^1/n$$ $$=exp left( lim_ntoinfty frac ln(Gamma(n+1)) +ln(Gamma(1/3))-ln(Gamma(n +4/3))nright)$$
Now by L'Hospital we have $$exp left( lim_nto infty frac ln(Gamma(n+1)) +ln(Gamma(1/3))-ln(Gamma(n +4/3))nright) =expleft( lim_nto infty left[psi(n+1) -psi(n+ 4/3)right] right)$$
But for large $x$, $psi(x)sim ln (x)$ Hence $$exp left( lim_nto infty left[psi(n+1) -psi(n+ 4/3)right]right)=exp left( lim_nto infty left[ln(n+1) -ln(n+ 4/3)right]right)=1$$
Similarly $$lim_nto infty (a_n)^1/n= 1$$
Hence giving me the answer as $15$.
I want to know whether I am correct or not. And if I am correct I want to know if there is any other simple and elementary method to solve the question.
Edit:
By some other method I dont mean that just converting the Gamma function to factorial representation and then using Stirling's approximation
calculus sequences-and-series limits gamma-function digamma-function
asked Sep 7 at 6:05
Manthanein
6,3831439
add a comment |Â
up vote
3
down vote
favorite
Given two sequences $$a_n=int_0^1 (1-x^2)^n dx$$ and $$b_n=int_0^1 (1-x^3)^n dx$$ ,($nin N$) then find the value of $$L=lim_nto infty (10 sqrt [n]a_n +5sqrt [n]b_n)$$
My try:
$$a_n=int_0^1 (1-x^2)^n dx= frac 12 Bleft( n+1, 1/2right)=frac 12cdot frac Gamma(n+1)Gamma(1/2)Gammaleft(n +frac 32right)$$
Similarly $$b_n=frac 13cdot frac Gamma(n+1)Gamma(1/3)Gammaleft(n +frac 43right)$$
Now let's consider $$J=lim_nto infty (b_n)^1/n =lim_nto infty left(frac 13right)^1/ncdot left(frac Gamma(n+1)Gamma(1/3)Gammaleft(n +frac 43right)right) ^1/n$$ $$=lim_ntoinfty left(frac 13right)^1/ncdot left(lim_nto infty left(frac Gamma(n+1)Gamma(1/3)Gammaleft(n +frac 43right)right) ^1/nright)$$ $$=1cdot lim_nto infty left(frac Gamma(n+1)Gamma(1/3)Gammaleft(n +frac 43right)right) ^1/n$$ $$=exp left( lim_ntoinfty frac ln(Gamma(n+1)) +ln(Gamma(1/3))-ln(Gamma(n +4/3))nright)$$
Now by L'Hospital we have $$exp left( lim_nto infty frac ln(Gamma(n+1)) +ln(Gamma(1/3))-ln(Gamma(n +4/3))nright) =expleft( lim_nto infty left[psi(n+1) -psi(n+ 4/3)right] right)$$
But for large $x$, $psi(x)sim ln (x)$ Hence $$exp left( lim_nto infty left[psi(n+1) -psi(n+ 4/3)right]right)=exp left( lim_nto infty left[ln(n+1) -ln(n+ 4/3)right]right)=1$$
Similarly $$lim_nto infty (a_n)^1/n= 1$$
Hence giving me the answer as $15$.
I want to know whether I am correct or not. And if I am correct I want to know if there is any other simple and elementary method to solve the question.
Edit:
By some other method I dont mean that just converting the Gamma function to factorial representation and then using Stirling's approximation
calculus sequences-and-series limits gamma-function digamma-function
asked Sep 7 at 6:05
Manthanein
6,3831439
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Given two sequences $$a_n=int_0^1 (1-x^2)^n dx$$ and $$b_n=int_0^1 (1-x^3)^n dx$$ ,($nin N$) then find the value of $$L=lim_nto infty (10 sqrt [n]a_n +5sqrt [n]b_n)$$
My try:
$$a_n=int_0^1 (1-x^2)^n dx= frac 12 Bleft( n+1, 1/2right)=frac 12cdot frac Gamma(n+1)Gamma(1/2)Gammaleft(n +frac 32right)$$
Similarly $$b_n=frac 13cdot frac Gamma(n+1)Gamma(1/3)Gammaleft(n +frac 43right)$$
Now let's consider $$J=lim_nto infty (b_n)^1/n =lim_nto infty left(frac 13right)^1/ncdot left(frac Gamma(n+1)Gamma(1/3)Gammaleft(n +frac 43right)right) ^1/n$$ $$=lim_ntoinfty left(frac 13right)^1/ncdot left(lim_nto infty left(frac Gamma(n+1)Gamma(1/3)Gammaleft(n +frac 43right)right) ^1/nright)$$ $$=1cdot lim_nto infty left(frac Gamma(n+1)Gamma(1/3)Gammaleft(n +frac 43right)right) ^1/n$$ $$=exp left( lim_ntoinfty frac ln(Gamma(n+1)) +ln(Gamma(1/3))-ln(Gamma(n +4/3))nright)$$
Now by L'Hospital we have $$exp left( lim_nto infty frac ln(Gamma(n+1)) +ln(Gamma(1/3))-ln(Gamma(n +4/3))nright) =expleft( lim_nto infty left[psi(n+1) -psi(n+ 4/3)right] right)$$
But for large $x$, $psi(x)sim ln (x)$ Hence $$exp left( lim_nto infty left[psi(n+1) -psi(n+ 4/3)right]right)=exp left( lim_nto infty left[ln(n+1) -ln(n+ 4/3)right]right)=1$$
Similarly $$lim_nto infty (a_n)^1/n= 1$$
Hence giving me the answer as $15$.
I want to know whether I am correct or not. And if I am correct I want to know if there is any other simple and elementary method to solve the question.
Edit:
By some other method I dont mean that just converting the Gamma function to factorial representation and then using Stirling's approximation
calculus sequences-and-series limits gamma-function digamma-function
asked Sep 7 at 6:05
Manthanein
6,3831439
Given two sequences $$a_n=int_0^1 (1-x^2)^n dx$$ and $$b_n=int_0^1 (1-x^3)^n dx$$ ,($nin N$) then find the value of $$L=lim_nto infty (10 sqrt [n]a_n +5sqrt [n]b_n)$$
My try:
$$a_n=int_0^1 (1-x^2)^n dx= frac 12 Bleft( n+1, 1/2right)=frac 12cdot frac Gamma(n+1)Gamma(1/2)Gammaleft(n +frac 32right)$$
Similarly $$b_n=frac 13cdot frac Gamma(n+1)Gamma(1/3)Gammaleft(n +frac 43right)$$
Now let's consider $$J=lim_nto infty (b_n)^1/n =lim_nto infty left(frac 13right)^1/ncdot left(frac Gamma(n+1)Gamma(1/3)Gammaleft(n +frac 43right)right) ^1/n$$ $$=lim_ntoinfty left(frac 13right)^1/ncdot left(lim_nto infty left(frac Gamma(n+1)Gamma(1/3)Gammaleft(n +frac 43right)right) ^1/nright)$$ $$=1cdot lim_nto infty left(frac Gamma(n+1)Gamma(1/3)Gammaleft(n +frac 43right)right) ^1/n$$ $$=exp left( lim_ntoinfty frac ln(Gamma(n+1)) +ln(Gamma(1/3))-ln(Gamma(n +4/3))nright)$$
Now by L'Hospital we have $$exp left( lim_nto infty frac ln(Gamma(n+1)) +ln(Gamma(1/3))-ln(Gamma(n +4/3))nright) =expleft( lim_nto infty left[psi(n+1) -psi(n+ 4/3)right] right)$$
But for large $x$, $psi(x)sim ln (x)$ Hence $$exp left( lim_nto infty left[psi(n+1) -psi(n+ 4/3)right]right)=exp left( lim_nto infty left[ln(n+1) -ln(n+ 4/3)right]right)=1$$
Similarly $$lim_nto infty (a_n)^1/n= 1$$
Hence giving me the answer as $15$.
I want to know whether I am correct or not. And if I am correct I want to know if there is any other simple and elementary method to solve the question.
Edit:
By some other method I dont mean that just converting the Gamma function to factorial representation and then using Stirling's approximation
calculus sequences-and-series limits gamma-function digamma-function
asked Sep 7 at 6:05
Manthanein
6,3831439
edited Sep 8 at 13:43
asked Sep 7 at 6:05
Manthanein
6,3831439
asked Sep 7 at 6:05
Manthanein
6,3831439
asked Sep 7 at 6:05
Manthanein
6,3831439
6,3831439
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
7
down vote
accepted
A much simpler approach. Let $k$ be a positive integer. Then for $xin [0,1]$,
$$1-xleq 1-x^kleq 1$$
and therefore
$$frac1n+1=int_0^1 (1-x)^n dxleq a_n(k):=int_0^1 (1-x^k)^n dxleq 1.$$
Hence, as $ntoinfty$,
$$1leftarrowfrac1sqrt[n]n+1lesqrt[n]a_n(k)leq sqrt[n]1=1$$
and by the Squeeze Theorem it follows that $sqrt[n]a_n(k)rightarrow 1$. In your case, we need the limit for $k=2$ and $k=3$.
edited Sep 7 at 7:33
amsmath
2,406114
answered Sep 7 at 6:23
Robert Z
85.5k1055123
add a comment |Â
up vote
4
down vote
There is a very simple method of evaluating $L$. Fort any bounded measurable function $f$ on $[0,1]$ we have $lim (int_0^1 |f|^p)^1/p to |f|_infty$ as $ p to infty$. Hence $a_n^1/n to max (1-x^2): 0leq x leq 1=1$. Similarly $b_n^1/n to 1$. So $L=10+5=15$. Actually, th result I have quoted is easy to prove in this case. Note that $int_0^1 (1-x^2)^n leq 1$. Also $int_0^1 (1-x^2)^n geq int_0^t (1-x^2)^n geq t(1-t^2)^n$ for each $t$. Take n-th root and conclude that $lim inf a_n^1/n geq 1-t^2$ for anty $t>0$. Let $t to 0$.
answered Sep 7 at 6:25
Kavi Rama Murthy
25.2k31335
Where does this result come from? Any references might help me a bit
– Manthanein
Sep 7 at 6:27
This is a 'popular' exercise from Rudin's book and you can get it easily from MSE.
– Kavi Rama Murthy
Sep 7 at 6:28
Some MSE links please?????
– Manthanein
Sep 7 at 6:29
@Manthanein I have included a proof now.
– Kavi Rama Murthy
Sep 7 at 6:36
Does this result have it's origin in the space theory or measure theory or something related. If yes then I am really sorry sir because I don't have quite an idea about those chapters. I am just in the 12th standard as of now. But I will try to understand the proofs so that I could use this result in near future.
– Manthanein
Sep 7 at 6:45
 |Â
show 2 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
A much simpler approach. Let $k$ be a positive integer. Then for $xin [0,1]$,
$$1-xleq 1-x^kleq 1$$
and therefore
$$frac1n+1=int_0^1 (1-x)^n dxleq a_n(k):=int_0^1 (1-x^k)^n dxleq 1.$$
Hence, as $ntoinfty$,
$$1leftarrowfrac1sqrt[n]n+1lesqrt[n]a_n(k)leq sqrt[n]1=1$$
and by the Squeeze Theorem it follows that $sqrt[n]a_n(k)rightarrow 1$. In your case, we need the limit for $k=2$ and $k=3$.
edited Sep 7 at 7:33
amsmath
2,406114
answered Sep 7 at 6:23
Robert Z
85.5k1055123
add a comment |Â
up vote
7
down vote
accepted
A much simpler approach. Let $k$ be a positive integer. Then for $xin [0,1]$,
$$1-xleq 1-x^kleq 1$$
and therefore
$$frac1n+1=int_0^1 (1-x)^n dxleq a_n(k):=int_0^1 (1-x^k)^n dxleq 1.$$
Hence, as $ntoinfty$,
$$1leftarrowfrac1sqrt[n]n+1lesqrt[n]a_n(k)leq sqrt[n]1=1$$
and by the Squeeze Theorem it follows that $sqrt[n]a_n(k)rightarrow 1$. In your case, we need the limit for $k=2$ and $k=3$.
edited Sep 7 at 7:33
amsmath
2,406114
answered Sep 7 at 6:23
Robert Z
85.5k1055123
add a comment |Â
up vote
7
down vote
accepted
up vote
7
down vote
accepted
A much simpler approach. Let $k$ be a positive integer. Then for $xin [0,1]$,
$$1-xleq 1-x^kleq 1$$
and therefore
$$frac1n+1=int_0^1 (1-x)^n dxleq a_n(k):=int_0^1 (1-x^k)^n dxleq 1.$$
Hence, as $ntoinfty$,
$$1leftarrowfrac1sqrt[n]n+1lesqrt[n]a_n(k)leq sqrt[n]1=1$$
and by the Squeeze Theorem it follows that $sqrt[n]a_n(k)rightarrow 1$. In your case, we need the limit for $k=2$ and $k=3$.
edited Sep 7 at 7:33
amsmath
2,406114
answered Sep 7 at 6:23
Robert Z
85.5k1055123
A much simpler approach. Let $k$ be a positive integer. Then for $xin [0,1]$,
$$1-xleq 1-x^kleq 1$$
and therefore
$$frac1n+1=int_0^1 (1-x)^n dxleq a_n(k):=int_0^1 (1-x^k)^n dxleq 1.$$
Hence, as $ntoinfty$,
$$1leftarrowfrac1sqrt[n]n+1lesqrt[n]a_n(k)leq sqrt[n]1=1$$
and by the Squeeze Theorem it follows that $sqrt[n]a_n(k)rightarrow 1$. In your case, we need the limit for $k=2$ and $k=3$.
edited Sep 7 at 7:33
amsmath
2,406114
answered Sep 7 at 6:23
Robert Z
85.5k1055123
edited Sep 7 at 7:33
amsmath
2,406114
edited Sep 7 at 7:33
amsmath
2,406114
edited Sep 7 at 7:33
amsmath
2,406114
2,406114
answered Sep 7 at 6:23
Robert Z
85.5k1055123
answered Sep 7 at 6:23
Robert Z
85.5k1055123
answered Sep 7 at 6:23
Robert Z
85.5k1055123
85.5k1055123
add a comment |Â
add a comment |Â
up vote
4
down vote
There is a very simple method of evaluating $L$. Fort any bounded measurable function $f$ on $[0,1]$ we have $lim (int_0^1 |f|^p)^1/p to |f|_infty$ as $ p to infty$. Hence $a_n^1/n to max (1-x^2): 0leq x leq 1=1$. Similarly $b_n^1/n to 1$. So $L=10+5=15$. Actually, th result I have quoted is easy to prove in this case. Note that $int_0^1 (1-x^2)^n leq 1$. Also $int_0^1 (1-x^2)^n geq int_0^t (1-x^2)^n geq t(1-t^2)^n$ for each $t$. Take n-th root and conclude that $lim inf a_n^1/n geq 1-t^2$ for anty $t>0$. Let $t to 0$.
answered Sep 7 at 6:25
Kavi Rama Murthy
25.2k31335
Where does this result come from? Any references might help me a bit
– Manthanein
Sep 7 at 6:27
This is a 'popular' exercise from Rudin's book and you can get it easily from MSE.
– Kavi Rama Murthy
Sep 7 at 6:28
Some MSE links please?????
– Manthanein
Sep 7 at 6:29
@Manthanein I have included a proof now.
– Kavi Rama Murthy
Sep 7 at 6:36
Does this result have it's origin in the space theory or measure theory or something related. If yes then I am really sorry sir because I don't have quite an idea about those chapters. I am just in the 12th standard as of now. But I will try to understand the proofs so that I could use this result in near future.
– Manthanein
Sep 7 at 6:45
 |Â
show 2 more comments
up vote
4
down vote
There is a very simple method of evaluating $L$. Fort any bounded measurable function $f$ on $[0,1]$ we have $lim (int_0^1 |f|^p)^1/p to |f|_infty$ as $ p to infty$. Hence $a_n^1/n to max (1-x^2): 0leq x leq 1=1$. Similarly $b_n^1/n to 1$. So $L=10+5=15$. Actually, th result I have quoted is easy to prove in this case. Note that $int_0^1 (1-x^2)^n leq 1$. Also $int_0^1 (1-x^2)^n geq int_0^t (1-x^2)^n geq t(1-t^2)^n$ for each $t$. Take n-th root and conclude that $lim inf a_n^1/n geq 1-t^2$ for anty $t>0$. Let $t to 0$.
answered Sep 7 at 6:25
Kavi Rama Murthy
25.2k31335
Where does this result come from? Any references might help me a bit
– Manthanein
Sep 7 at 6:27
This is a 'popular' exercise from Rudin's book and you can get it easily from MSE.
– Kavi Rama Murthy
Sep 7 at 6:28
Some MSE links please?????
– Manthanein
Sep 7 at 6:29
@Manthanein I have included a proof now.
– Kavi Rama Murthy
Sep 7 at 6:36
Does this result have it's origin in the space theory or measure theory or something related. If yes then I am really sorry sir because I don't have quite an idea about those chapters. I am just in the 12th standard as of now. But I will try to understand the proofs so that I could use this result in near future.
– Manthanein
Sep 7 at 6:45
 |Â
show 2 more comments
up vote
4
down vote
up vote
4
down vote
There is a very simple method of evaluating $L$. Fort any bounded measurable function $f$ on $[0,1]$ we have $lim (int_0^1 |f|^p)^1/p to |f|_infty$ as $ p to infty$. Hence $a_n^1/n to max (1-x^2): 0leq x leq 1=1$. Similarly $b_n^1/n to 1$. So $L=10+5=15$. Actually, th result I have quoted is easy to prove in this case. Note that $int_0^1 (1-x^2)^n leq 1$. Also $int_0^1 (1-x^2)^n geq int_0^t (1-x^2)^n geq t(1-t^2)^n$ for each $t$. Take n-th root and conclude that $lim inf a_n^1/n geq 1-t^2$ for anty $t>0$. Let $t to 0$.
answered Sep 7 at 6:25
Kavi Rama Murthy
25.2k31335
There is a very simple method of evaluating $L$. Fort any bounded measurable function $f$ on $[0,1]$ we have $lim (int_0^1 |f|^p)^1/p to |f|_infty$ as $ p to infty$. Hence $a_n^1/n to max (1-x^2): 0leq x leq 1=1$. Similarly $b_n^1/n to 1$. So $L=10+5=15$. Actually, th result I have quoted is easy to prove in this case. Note that $int_0^1 (1-x^2)^n leq 1$. Also $int_0^1 (1-x^2)^n geq int_0^t (1-x^2)^n geq t(1-t^2)^n$ for each $t$. Take n-th root and conclude that $lim inf a_n^1/n geq 1-t^2$ for anty $t>0$. Let $t to 0$.
answered Sep 7 at 6:25
Kavi Rama Murthy
25.2k31335
edited Sep 7 at 7:29
answered Sep 7 at 6:25
Kavi Rama Murthy
25.2k31335
answered Sep 7 at 6:25
Kavi Rama Murthy
25.2k31335
answered Sep 7 at 6:25
Kavi Rama Murthy
25.2k31335
25.2k31335
Where does this result come from? Any references might help me a bit
– Manthanein
Sep 7 at 6:27
This is a 'popular' exercise from Rudin's book and you can get it easily from MSE.
– Kavi Rama Murthy
Sep 7 at 6:28
Some MSE links please?????
– Manthanein
Sep 7 at 6:29
@Manthanein I have included a proof now.
– Kavi Rama Murthy
Sep 7 at 6:36
Does this result have it's origin in the space theory or measure theory or something related. If yes then I am really sorry sir because I don't have quite an idea about those chapters. I am just in the 12th standard as of now. But I will try to understand the proofs so that I could use this result in near future.
– Manthanein
Sep 7 at 6:45
 |Â
show 2 more comments
Where does this result come from? Any references might help me a bit
– Manthanein
Sep 7 at 6:27
This is a 'popular' exercise from Rudin's book and you can get it easily from MSE.
– Kavi Rama Murthy
Sep 7 at 6:28
Some MSE links please?????
– Manthanein
Sep 7 at 6:29
@Manthanein I have included a proof now.
– Kavi Rama Murthy
Sep 7 at 6:36
Does this result have it's origin in the space theory or measure theory or something related. If yes then I am really sorry sir because I don't have quite an idea about those chapters. I am just in the 12th standard as of now. But I will try to understand the proofs so that I could use this result in near future.
– Manthanein
Sep 7 at 6:45
Where does this result come from? Any references might help me a bit
– Manthanein
Sep 7 at 6:27
Where does this result come from? Any references might help me a bit
– Manthanein
Sep 7 at 6:27
This is a 'popular' exercise from Rudin's book and you can get it easily from MSE.
– Kavi Rama Murthy
Sep 7 at 6:28
This is a 'popular' exercise from Rudin's book and you can get it easily from MSE.
– Kavi Rama Murthy
Sep 7 at 6:28
Some MSE links please?????
– Manthanein
Sep 7 at 6:29
Some MSE links please?????
– Manthanein
Sep 7 at 6:29
@Manthanein I have included a proof now.
– Kavi Rama Murthy
Sep 7 at 6:36
@Manthanein I have included a proof now.
– Kavi Rama Murthy
Sep 7 at 6:36
Does this result have it's origin in the space theory or measure theory or something related. If yes then I am really sorry sir because I don't have quite an idea about those chapters. I am just in the 12th standard as of now. But I will try to understand the proofs so that I could use this result in near future.
– Manthanein
Sep 7 at 6:45
Does this result have it's origin in the space theory or measure theory or something related. If yes then I am really sorry sir because I don't have quite an idea about those chapters. I am just in the 12th standard as of now. But I will try to understand the proofs so that I could use this result in near future.
– Manthanein
Sep 7 at 6:45
 |Â
show 2 more comments
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