Mixing
Value of $lim_n to infty left(frac(n+1)(n+2)(n+3)…(3n)n^2nright)^1/n$
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
I was asked to evaluate the following expression:
$lim_n to infty left(frac(n+1)(n+2)(n+3)...(3n)n^2nright)^1/n$
My first step was to assume that the limit existed, and set that value to $y$.
$ y = lim_n to infty left(frac(n+1)(n+2)(n+3)...(3n)n^2nright)^1/n$
And then, I took the natural logarithm of both sides of the equation. I obtained the expression:
$ ln y = lim_n to infty frac1n cdot left(ln(1+frac1n) + ln(1+frac2n) + ... + ln(1+frac2nn)right) $
This simplified to:
$ ln y = lim_n to infty frac1n cdot sum_k = 1^colorRed2n ln(1+frackn) $
I realize that this is similar to the form of a Riemann sum, which can then be manipulated to give the expression in the form of a definite integral. However, the part bolded in red, which is $ 2n$, throws me off. I have only seen Riemann sums be evaluated when the upper limit is $ n - k $, where $k$ is a constant.
Therefore, how would I go about evaluating this expression?
Thank you for all help in advance.
calculus limits
asked Jul 25 at 6:08
JustMath
264
add a comment |Â
up vote
3
down vote
favorite
I was asked to evaluate the following expression:
$lim_n to infty left(frac(n+1)(n+2)(n+3)...(3n)n^2nright)^1/n$
My first step was to assume that the limit existed, and set that value to $y$.
$ y = lim_n to infty left(frac(n+1)(n+2)(n+3)...(3n)n^2nright)^1/n$
And then, I took the natural logarithm of both sides of the equation. I obtained the expression:
$ ln y = lim_n to infty frac1n cdot left(ln(1+frac1n) + ln(1+frac2n) + ... + ln(1+frac2nn)right) $
This simplified to:
$ ln y = lim_n to infty frac1n cdot sum_k = 1^colorRed2n ln(1+frackn) $
I realize that this is similar to the form of a Riemann sum, which can then be manipulated to give the expression in the form of a definite integral. However, the part bolded in red, which is $ 2n$, throws me off. I have only seen Riemann sums be evaluated when the upper limit is $ n - k $, where $k$ is a constant.
Therefore, how would I go about evaluating this expression?
Thank you for all help in advance.
calculus limits
asked Jul 25 at 6:08
JustMath
264
Hint: Use Cesaro-Stolz
– Paramanand Singh
Jul 25 at 8:34
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I was asked to evaluate the following expression:
$lim_n to infty left(frac(n+1)(n+2)(n+3)...(3n)n^2nright)^1/n$
My first step was to assume that the limit existed, and set that value to $y$.
$ y = lim_n to infty left(frac(n+1)(n+2)(n+3)...(3n)n^2nright)^1/n$
And then, I took the natural logarithm of both sides of the equation. I obtained the expression:
$ ln y = lim_n to infty frac1n cdot left(ln(1+frac1n) + ln(1+frac2n) + ... + ln(1+frac2nn)right) $
This simplified to:
$ ln y = lim_n to infty frac1n cdot sum_k = 1^colorRed2n ln(1+frackn) $
I realize that this is similar to the form of a Riemann sum, which can then be manipulated to give the expression in the form of a definite integral. However, the part bolded in red, which is $ 2n$, throws me off. I have only seen Riemann sums be evaluated when the upper limit is $ n - k $, where $k$ is a constant.
Therefore, how would I go about evaluating this expression?
Thank you for all help in advance.
calculus limits
asked Jul 25 at 6:08
JustMath
264
I was asked to evaluate the following expression:
$lim_n to infty left(frac(n+1)(n+2)(n+3)...(3n)n^2nright)^1/n$
My first step was to assume that the limit existed, and set that value to $y$.
$ y = lim_n to infty left(frac(n+1)(n+2)(n+3)...(3n)n^2nright)^1/n$
And then, I took the natural logarithm of both sides of the equation. I obtained the expression:
$ ln y = lim_n to infty frac1n cdot left(ln(1+frac1n) + ln(1+frac2n) + ... + ln(1+frac2nn)right) $
This simplified to:
$ ln y = lim_n to infty frac1n cdot sum_k = 1^colorRed2n ln(1+frackn) $
I realize that this is similar to the form of a Riemann sum, which can then be manipulated to give the expression in the form of a definite integral. However, the part bolded in red, which is $ 2n$, throws me off. I have only seen Riemann sums be evaluated when the upper limit is $ n - k $, where $k$ is a constant.
Therefore, how would I go about evaluating this expression?
Thank you for all help in advance.
calculus limits
asked Jul 25 at 6:08
JustMath
264
edited Jul 25 at 6:14
asked Jul 25 at 6:08
JustMath
264
asked Jul 25 at 6:08
JustMath
264
asked Jul 25 at 6:08
JustMath
264
264
Hint: Use Cesaro-Stolz
– Paramanand Singh
Jul 25 at 8:34
add a comment |Â
Hint: Use Cesaro-Stolz
– Paramanand Singh
Jul 25 at 8:34
Hint: Use Cesaro-Stolz
– Paramanand Singh
Jul 25 at 8:34
Hint: Use Cesaro-Stolz
– Paramanand Singh
Jul 25 at 8:34
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
3
down vote
Consider
$$int_0^2f(x),dx$$
where
$$f(x)=ln(1+x).$$
Splitting $[1,2]$ into $2n$ intervals of length $1/n$ gives a Riemann
sum
$$frac1nsum_k=1^2nf(k/n)=frac1nsum_k=1^2nlnleft(
1+frac knright)$$
which is exactly yours.
Alternatively you could use Stirling's formula.
answered Jul 25 at 6:16
Lord Shark the Unknown
85k950111
add a comment |Â
up vote
1
down vote
We use the following result: if $(a_n)$ is a sequence such that $a_n >0$ for all $n$ and $(fraca_n+1a_n)$ is convergent, then $(a_n^1/n)$ is convergent and
$$lim_n to inftyfraca_n+1a_n=lim_n to inftya_n^1/n.$$
Now let $a_n:= frac(n+1)(n+2)(n+3)...(3n)n^2n$.
Some easy computations give
$$fraca_n+1a_n=frac(3n+1)(3n+2)(3n+3)(n+1)^3 cdot (1- frac1n+1)^2n to frac27e^2.$$
answered Jul 25 at 6:41
Fred
37.2k1237
add a comment |Â
up vote
0
down vote
$(2)^1/nrightarrow 1$, so the answer is not altered by adding an $n$ and dropping $2n$ from the middle of the numerator.
Rearrange the terms in the numerator to produce the following pairing:
$$((n)(3n))((n+1)(3n-1))((n+2)(3n-2))cdots ((2n-1)(2n+1))$$
This is $prodlimits_i=1^n (2n-i)(2n+i) = prodlimits_i=1^n (4n^2-i^2)$
After dividing by $n^2n$, you obtain $prodlimits_i=1^n (4-(i/n)^2)$.
So the limit you are looking for equals to $exp(limlimits_nrightarrow infty sumlimits_i=1^nfrac1nlog(4-left(fracinright)^2))$.
Clearly, as $n$ tends to infinity, the sum $sumlimits_i=1^nfrac1nlog(4-left(fracinright)^2))$ tends to the integral $intlimits_0^1 log(4-x^2) , dx = intlimits_0^1 log(2-x) , dx + intlimits_0^1 log(2+x) , dx$.
It is easy to compute both definit integrals.
But your calculation is also good. I that case, you obtain a definite integral on $[0,2]$.
answered Jul 25 at 6:40
A. Pongrácz
1,804116
add a comment |Â
up vote
0
down vote
Just a comment after Lord Shark the Unknown's answer.
When I see things looking like factorials, my first reaction is to think about Stirling's approximation of them.
$$a_n=left(fracprod_i=1^2n (n+i)n^2nright)^frac 1n=left(frac(3n)!n^2n,n!right)^frac 1n$$
$$log(a_n)=frac 1n left(log((3n)!)-log(n!)-2n log(n)right)$$ Now, Stirling's approximation
$$log(p!)=p (log (p)-1)+frac12 left(log (2 pi )+log
left(pright)right)+frac112
p+Oleft(frac1p^3right)$$ Apply it and simplify to get
$$log(a_n)=frac 1n left( (3log (3)-2)n+fraclog (3)2-frac118
n+Oleft(frac1n^3right) right)$$ $$log(a_n)=(3log (3)-2)+fraclog (3)2n-frac118
n^2+Oleft(frac1n^4right) $$ Now, using
$$a_n=e^log(a_n)=frac27 e^2 left(1+fraclog (3)2 nright)+Oleft(frac1n^2right) $$ which shows the limit and how it is approached.
answered Jul 25 at 7:21
Claude Leibovici
111k1055126
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Consider
$$int_0^2f(x),dx$$
where
$$f(x)=ln(1+x).$$
Splitting $[1,2]$ into $2n$ intervals of length $1/n$ gives a Riemann
sum
$$frac1nsum_k=1^2nf(k/n)=frac1nsum_k=1^2nlnleft(
1+frac knright)$$
which is exactly yours.
Alternatively you could use Stirling's formula.
answered Jul 25 at 6:16
Lord Shark the Unknown
85k950111
add a comment |Â
up vote
3
down vote
Consider
$$int_0^2f(x),dx$$
where
$$f(x)=ln(1+x).$$
Splitting $[1,2]$ into $2n$ intervals of length $1/n$ gives a Riemann
sum
$$frac1nsum_k=1^2nf(k/n)=frac1nsum_k=1^2nlnleft(
1+frac knright)$$
which is exactly yours.
Alternatively you could use Stirling's formula.
answered Jul 25 at 6:16
Lord Shark the Unknown
85k950111
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Consider
$$int_0^2f(x),dx$$
where
$$f(x)=ln(1+x).$$
Splitting $[1,2]$ into $2n$ intervals of length $1/n$ gives a Riemann
sum
$$frac1nsum_k=1^2nf(k/n)=frac1nsum_k=1^2nlnleft(
1+frac knright)$$
which is exactly yours.
Alternatively you could use Stirling's formula.
answered Jul 25 at 6:16
Lord Shark the Unknown
85k950111
Consider
$$int_0^2f(x),dx$$
where
$$f(x)=ln(1+x).$$
Splitting $[1,2]$ into $2n$ intervals of length $1/n$ gives a Riemann
sum
$$frac1nsum_k=1^2nf(k/n)=frac1nsum_k=1^2nlnleft(
1+frac knright)$$
which is exactly yours.
Alternatively you could use Stirling's formula.
answered Jul 25 at 6:16
Lord Shark the Unknown
85k950111
answered Jul 25 at 6:16
Lord Shark the Unknown
85k950111
answered Jul 25 at 6:16
Lord Shark the Unknown
85k950111
answered Jul 25 at 6:16
Lord Shark the Unknown
85k950111
85k950111
add a comment |Â
add a comment |Â
up vote
1
down vote
We use the following result: if $(a_n)$ is a sequence such that $a_n >0$ for all $n$ and $(fraca_n+1a_n)$ is convergent, then $(a_n^1/n)$ is convergent and
$$lim_n to inftyfraca_n+1a_n=lim_n to inftya_n^1/n.$$
Now let $a_n:= frac(n+1)(n+2)(n+3)...(3n)n^2n$.
Some easy computations give
$$fraca_n+1a_n=frac(3n+1)(3n+2)(3n+3)(n+1)^3 cdot (1- frac1n+1)^2n to frac27e^2.$$
answered Jul 25 at 6:41
Fred
37.2k1237
add a comment |Â
up vote
1
down vote
We use the following result: if $(a_n)$ is a sequence such that $a_n >0$ for all $n$ and $(fraca_n+1a_n)$ is convergent, then $(a_n^1/n)$ is convergent and
$$lim_n to inftyfraca_n+1a_n=lim_n to inftya_n^1/n.$$
Now let $a_n:= frac(n+1)(n+2)(n+3)...(3n)n^2n$.
Some easy computations give
$$fraca_n+1a_n=frac(3n+1)(3n+2)(3n+3)(n+1)^3 cdot (1- frac1n+1)^2n to frac27e^2.$$
answered Jul 25 at 6:41
Fred
37.2k1237
add a comment |Â
up vote
1
down vote
up vote
1
down vote
We use the following result: if $(a_n)$ is a sequence such that $a_n >0$ for all $n$ and $(fraca_n+1a_n)$ is convergent, then $(a_n^1/n)$ is convergent and
$$lim_n to inftyfraca_n+1a_n=lim_n to inftya_n^1/n.$$
Now let $a_n:= frac(n+1)(n+2)(n+3)...(3n)n^2n$.
Some easy computations give
$$fraca_n+1a_n=frac(3n+1)(3n+2)(3n+3)(n+1)^3 cdot (1- frac1n+1)^2n to frac27e^2.$$
answered Jul 25 at 6:41
Fred
37.2k1237
We use the following result: if $(a_n)$ is a sequence such that $a_n >0$ for all $n$ and $(fraca_n+1a_n)$ is convergent, then $(a_n^1/n)$ is convergent and
$$lim_n to inftyfraca_n+1a_n=lim_n to inftya_n^1/n.$$
Now let $a_n:= frac(n+1)(n+2)(n+3)...(3n)n^2n$.
Some easy computations give
$$fraca_n+1a_n=frac(3n+1)(3n+2)(3n+3)(n+1)^3 cdot (1- frac1n+1)^2n to frac27e^2.$$
answered Jul 25 at 6:41
Fred
37.2k1237
answered Jul 25 at 6:41
Fred
37.2k1237
answered Jul 25 at 6:41
Fred
37.2k1237
answered Jul 25 at 6:41
Fred
37.2k1237
37.2k1237
add a comment |Â
add a comment |Â
up vote
0
down vote
$(2)^1/nrightarrow 1$, so the answer is not altered by adding an $n$ and dropping $2n$ from the middle of the numerator.
Rearrange the terms in the numerator to produce the following pairing:
$$((n)(3n))((n+1)(3n-1))((n+2)(3n-2))cdots ((2n-1)(2n+1))$$
This is $prodlimits_i=1^n (2n-i)(2n+i) = prodlimits_i=1^n (4n^2-i^2)$
After dividing by $n^2n$, you obtain $prodlimits_i=1^n (4-(i/n)^2)$.
So the limit you are looking for equals to $exp(limlimits_nrightarrow infty sumlimits_i=1^nfrac1nlog(4-left(fracinright)^2))$.
Clearly, as $n$ tends to infinity, the sum $sumlimits_i=1^nfrac1nlog(4-left(fracinright)^2))$ tends to the integral $intlimits_0^1 log(4-x^2) , dx = intlimits_0^1 log(2-x) , dx + intlimits_0^1 log(2+x) , dx$.
It is easy to compute both definit integrals.
But your calculation is also good. I that case, you obtain a definite integral on $[0,2]$.
answered Jul 25 at 6:40
A. Pongrácz
1,804116
add a comment |Â
up vote
0
down vote
$(2)^1/nrightarrow 1$, so the answer is not altered by adding an $n$ and dropping $2n$ from the middle of the numerator.
Rearrange the terms in the numerator to produce the following pairing:
$$((n)(3n))((n+1)(3n-1))((n+2)(3n-2))cdots ((2n-1)(2n+1))$$
This is $prodlimits_i=1^n (2n-i)(2n+i) = prodlimits_i=1^n (4n^2-i^2)$
After dividing by $n^2n$, you obtain $prodlimits_i=1^n (4-(i/n)^2)$.
So the limit you are looking for equals to $exp(limlimits_nrightarrow infty sumlimits_i=1^nfrac1nlog(4-left(fracinright)^2))$.
Clearly, as $n$ tends to infinity, the sum $sumlimits_i=1^nfrac1nlog(4-left(fracinright)^2))$ tends to the integral $intlimits_0^1 log(4-x^2) , dx = intlimits_0^1 log(2-x) , dx + intlimits_0^1 log(2+x) , dx$.
It is easy to compute both definit integrals.
But your calculation is also good. I that case, you obtain a definite integral on $[0,2]$.
answered Jul 25 at 6:40
A. Pongrácz
1,804116
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$(2)^1/nrightarrow 1$, so the answer is not altered by adding an $n$ and dropping $2n$ from the middle of the numerator.
Rearrange the terms in the numerator to produce the following pairing:
$$((n)(3n))((n+1)(3n-1))((n+2)(3n-2))cdots ((2n-1)(2n+1))$$
This is $prodlimits_i=1^n (2n-i)(2n+i) = prodlimits_i=1^n (4n^2-i^2)$
After dividing by $n^2n$, you obtain $prodlimits_i=1^n (4-(i/n)^2)$.
So the limit you are looking for equals to $exp(limlimits_nrightarrow infty sumlimits_i=1^nfrac1nlog(4-left(fracinright)^2))$.
Clearly, as $n$ tends to infinity, the sum $sumlimits_i=1^nfrac1nlog(4-left(fracinright)^2))$ tends to the integral $intlimits_0^1 log(4-x^2) , dx = intlimits_0^1 log(2-x) , dx + intlimits_0^1 log(2+x) , dx$.
It is easy to compute both definit integrals.
But your calculation is also good. I that case, you obtain a definite integral on $[0,2]$.
answered Jul 25 at 6:40
A. Pongrácz
1,804116
$(2)^1/nrightarrow 1$, so the answer is not altered by adding an $n$ and dropping $2n$ from the middle of the numerator.
Rearrange the terms in the numerator to produce the following pairing:
$$((n)(3n))((n+1)(3n-1))((n+2)(3n-2))cdots ((2n-1)(2n+1))$$
This is $prodlimits_i=1^n (2n-i)(2n+i) = prodlimits_i=1^n (4n^2-i^2)$
After dividing by $n^2n$, you obtain $prodlimits_i=1^n (4-(i/n)^2)$.
So the limit you are looking for equals to $exp(limlimits_nrightarrow infty sumlimits_i=1^nfrac1nlog(4-left(fracinright)^2))$.
Clearly, as $n$ tends to infinity, the sum $sumlimits_i=1^nfrac1nlog(4-left(fracinright)^2))$ tends to the integral $intlimits_0^1 log(4-x^2) , dx = intlimits_0^1 log(2-x) , dx + intlimits_0^1 log(2+x) , dx$.
It is easy to compute both definit integrals.
But your calculation is also good. I that case, you obtain a definite integral on $[0,2]$.
answered Jul 25 at 6:40
A. Pongrácz
1,804116
answered Jul 25 at 6:40
A. Pongrácz
1,804116
answered Jul 25 at 6:40
A. Pongrácz
1,804116
answered Jul 25 at 6:40
A. Pongrácz
1,804116
1,804116
add a comment |Â
add a comment |Â
up vote
0
down vote
Just a comment after Lord Shark the Unknown's answer.
When I see things looking like factorials, my first reaction is to think about Stirling's approximation of them.
$$a_n=left(fracprod_i=1^2n (n+i)n^2nright)^frac 1n=left(frac(3n)!n^2n,n!right)^frac 1n$$
$$log(a_n)=frac 1n left(log((3n)!)-log(n!)-2n log(n)right)$$ Now, Stirling's approximation
$$log(p!)=p (log (p)-1)+frac12 left(log (2 pi )+log
left(pright)right)+frac112
p+Oleft(frac1p^3right)$$ Apply it and simplify to get
$$log(a_n)=frac 1n left( (3log (3)-2)n+fraclog (3)2-frac118
n+Oleft(frac1n^3right) right)$$ $$log(a_n)=(3log (3)-2)+fraclog (3)2n-frac118
n^2+Oleft(frac1n^4right) $$ Now, using
$$a_n=e^log(a_n)=frac27 e^2 left(1+fraclog (3)2 nright)+Oleft(frac1n^2right) $$ which shows the limit and how it is approached.
answered Jul 25 at 7:21
Claude Leibovici
111k1055126
add a comment |Â
up vote
0
down vote
Just a comment after Lord Shark the Unknown's answer.
When I see things looking like factorials, my first reaction is to think about Stirling's approximation of them.
$$a_n=left(fracprod_i=1^2n (n+i)n^2nright)^frac 1n=left(frac(3n)!n^2n,n!right)^frac 1n$$
$$log(a_n)=frac 1n left(log((3n)!)-log(n!)-2n log(n)right)$$ Now, Stirling's approximation
$$log(p!)=p (log (p)-1)+frac12 left(log (2 pi )+log
left(pright)right)+frac112
p+Oleft(frac1p^3right)$$ Apply it and simplify to get
$$log(a_n)=frac 1n left( (3log (3)-2)n+fraclog (3)2-frac118
n+Oleft(frac1n^3right) right)$$ $$log(a_n)=(3log (3)-2)+fraclog (3)2n-frac118
n^2+Oleft(frac1n^4right) $$ Now, using
$$a_n=e^log(a_n)=frac27 e^2 left(1+fraclog (3)2 nright)+Oleft(frac1n^2right) $$ which shows the limit and how it is approached.
answered Jul 25 at 7:21
Claude Leibovici
111k1055126
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Just a comment after Lord Shark the Unknown's answer.
When I see things looking like factorials, my first reaction is to think about Stirling's approximation of them.
$$a_n=left(fracprod_i=1^2n (n+i)n^2nright)^frac 1n=left(frac(3n)!n^2n,n!right)^frac 1n$$
$$log(a_n)=frac 1n left(log((3n)!)-log(n!)-2n log(n)right)$$ Now, Stirling's approximation
$$log(p!)=p (log (p)-1)+frac12 left(log (2 pi )+log
left(pright)right)+frac112
p+Oleft(frac1p^3right)$$ Apply it and simplify to get
$$log(a_n)=frac 1n left( (3log (3)-2)n+fraclog (3)2-frac118
n+Oleft(frac1n^3right) right)$$ $$log(a_n)=(3log (3)-2)+fraclog (3)2n-frac118
n^2+Oleft(frac1n^4right) $$ Now, using
$$a_n=e^log(a_n)=frac27 e^2 left(1+fraclog (3)2 nright)+Oleft(frac1n^2right) $$ which shows the limit and how it is approached.
answered Jul 25 at 7:21
Claude Leibovici
111k1055126
Just a comment after Lord Shark the Unknown's answer.
When I see things looking like factorials, my first reaction is to think about Stirling's approximation of them.
$$a_n=left(fracprod_i=1^2n (n+i)n^2nright)^frac 1n=left(frac(3n)!n^2n,n!right)^frac 1n$$
$$log(a_n)=frac 1n left(log((3n)!)-log(n!)-2n log(n)right)$$ Now, Stirling's approximation
$$log(p!)=p (log (p)-1)+frac12 left(log (2 pi )+log
left(pright)right)+frac112
p+Oleft(frac1p^3right)$$ Apply it and simplify to get
$$log(a_n)=frac 1n left( (3log (3)-2)n+fraclog (3)2-frac118
n+Oleft(frac1n^3right) right)$$ $$log(a_n)=(3log (3)-2)+fraclog (3)2n-frac118
n^2+Oleft(frac1n^4right) $$ Now, using
$$a_n=e^log(a_n)=frac27 e^2 left(1+fraclog (3)2 nright)+Oleft(frac1n^2right) $$ which shows the limit and how it is approached.
answered Jul 25 at 7:21
Claude Leibovici
111k1055126
answered Jul 25 at 7:21
Claude Leibovici
111k1055126
answered Jul 25 at 7:21
Claude Leibovici
111k1055126
answered Jul 25 at 7:21
Claude Leibovici
111k1055126
111k1055126
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2862076%2fvalue-of-lim-n-to-infty-left-fracn1n2n3-3nn2n-ri%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Hint: Use Cesaro-Stolz
– Paramanand Singh
Jul 25 at 8:34