How do I determine the divergence/convergence of $sum_n frac{1}{log(log(n))}$?
$begingroup$
I am working through some problems in Durrett's probability book, and one of them involves a variant of the law of iterated logarithm.
I've managed to reduce the result to showing that
$$sum_n frac{1}{log log (n)}exp(-log log(n)) < infty$$
Using upperbounds for tail probabilities of standard normal. But, as I'm really not good with this stuff, I'm unsure how I am supposed to show this (if it's indeed true?).
Clearly without the exponential, the series would diverge since $log (n) leq n$. But, with the exponential it seems as though this could converge.
Could someone advise me how to complete this last step?
convergence
edited Dec 1 '18 at 6:15
Chinnapparaj R
5,4872928
asked Dec 1 '18 at 6:12
XiaomiXiaomi
1,057115
$endgroup$
add a comment |
$begingroup$
I am working through some problems in Durrett's probability book, and one of them involves a variant of the law of iterated logarithm.
I've managed to reduce the result to showing that
$$sum_n frac{1}{log log (n)}exp(-log log(n)) < infty$$
Using upperbounds for tail probabilities of standard normal. But, as I'm really not good with this stuff, I'm unsure how I am supposed to show this (if it's indeed true?).
Clearly without the exponential, the series would diverge since $log (n) leq n$. But, with the exponential it seems as though this could converge.
Could someone advise me how to complete this last step?
convergence
edited Dec 1 '18 at 6:15
Chinnapparaj R
5,4872928
asked Dec 1 '18 at 6:12
XiaomiXiaomi
1,057115
$endgroup$
add a comment |
$begingroup$
I am working through some problems in Durrett's probability book, and one of them involves a variant of the law of iterated logarithm.
I've managed to reduce the result to showing that
$$sum_n frac{1}{log log (n)}exp(-log log(n)) < infty$$
Using upperbounds for tail probabilities of standard normal. But, as I'm really not good with this stuff, I'm unsure how I am supposed to show this (if it's indeed true?).
Clearly without the exponential, the series would diverge since $log (n) leq n$. But, with the exponential it seems as though this could converge.
Could someone advise me how to complete this last step?
convergence
edited Dec 1 '18 at 6:15
Chinnapparaj R
5,4872928
asked Dec 1 '18 at 6:12
XiaomiXiaomi
1,057115
$endgroup$
I am working through some problems in Durrett's probability book, and one of them involves a variant of the law of iterated logarithm.
I've managed to reduce the result to showing that
$$sum_n frac{1}{log log (n)}exp(-log log(n)) < infty$$
Using upperbounds for tail probabilities of standard normal. But, as I'm really not good with this stuff, I'm unsure how I am supposed to show this (if it's indeed true?).
Clearly without the exponential, the series would diverge since $log (n) leq n$. But, with the exponential it seems as though this could converge.
Could someone advise me how to complete this last step?
convergence
convergence
edited Dec 1 '18 at 6:15
Chinnapparaj R
5,4872928
asked Dec 1 '18 at 6:12
XiaomiXiaomi
1,057115
edited Dec 1 '18 at 6:15
Chinnapparaj R
5,4872928
asked Dec 1 '18 at 6:12
XiaomiXiaomi
1,057115
edited Dec 1 '18 at 6:15
Chinnapparaj R
5,4872928
edited Dec 1 '18 at 6:15
Chinnapparaj R
5,4872928
edited Dec 1 '18 at 6:15
Chinnapparaj R
5,4872928
5,4872928
asked Dec 1 '18 at 6:12
XiaomiXiaomi
1,057115
asked Dec 1 '18 at 6:12
XiaomiXiaomi
1,057115
asked Dec 1 '18 at 6:12
XiaomiXiaomi
1,057115
1,057115
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
We can simplify
$$exp(- log log n) = frac{1}{e^{log log n}} = frac{1}{log n}$$
Since $(log log n) log n leq n$, this diverges.
answered Dec 1 '18 at 6:17
plattyplatty
3,370320
$endgroup$
add a comment |
$begingroup$
$sum_n frac{1}{n}$ is divergent. So is $sum_n frac{1}{log(log(n))}$
But I feel $sum_n frac{exp(-log(log(n))}{log(log(n))}$ might still be convergent, you just need to find another way to prove it.
answered Dec 1 '18 at 6:19
MoonKnightMoonKnight
1,369611
$endgroup$
add a comment |
$begingroup$
$$
frac 1{(log log n )exp(log log n)} = frac 1 {log log n cdot log n} geqslant frac 1{log ^2 n} geqslant frac 1{n^{1/2 times 2}} = frac 1n,
$$
so it still diverges.
answered Dec 1 '18 at 6:18
xbhxbh
6,1351522
$endgroup$
$begingroup$
Oh, ok. Thanks a lot. Do you have any advice on how to show $sum_n P(Z>sqrt{2log log n}(1+epsilon)) < infty$ for $Z sim N(0,1)$?
$endgroup$
– Xiaomi
Dec 1 '18 at 6:21
$begingroup$
@Xiaomi Sorry, I suck at probability theory.
$endgroup$
– xbh
Dec 1 '18 at 7:15
add a comment |
$begingroup$
Cauchy condensation shows
$$sum_n frac{1}{log(log(n))} sim sum_n frac{2^n}{log(log(2^n))} = sum_n frac{2^n}{log(nlog(2))} = sum_n frac{2^n}{log(n) + log(log(2))} $$
So, it is divergent.
answered Dec 1 '18 at 6:20
trancelocationtrancelocation
11.3k1724
$endgroup$
$begingroup$
Thanks a lot. Do you have any advice on how to show $sum_n P(Z>sqrt{2log log n}(1+epsilon)) < infty$ for $Z sim N(0,1)$? The common probability inequalities do not seem tight enough to show this..
$endgroup$
– Xiaomi
Dec 1 '18 at 6:24
$begingroup$
You may use the integral for $N(0,1)$ for each member of the series and estimate it: $$P(Z>sqrt{2log log n}(1+epsilon)) sim int_{sqrt{2log log n}(1+epsilon)}^{infty} e^{frac{-x^2}{2}}dx$$
$endgroup$
– trancelocation
Dec 1 '18 at 6:48
add a comment |
$begingroup$
Credits to user MoonKnight.
$log n <n ; $
And once more:
$log (log n) lt log n lt n.$
$dfrac {1}{n} lt dfrac{1}{log n} lt dfrac{1}{log (log n)}.$
Comparison test.
answered Dec 1 '18 at 7:15
Peter SzilasPeter Szilas
11.3k2822
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We can simplify
$$exp(- log log n) = frac{1}{e^{log log n}} = frac{1}{log n}$$
Since $(log log n) log n leq n$, this diverges.
answered Dec 1 '18 at 6:17
plattyplatty
3,370320
$endgroup$
add a comment |
$begingroup$
We can simplify
$$exp(- log log n) = frac{1}{e^{log log n}} = frac{1}{log n}$$
Since $(log log n) log n leq n$, this diverges.
answered Dec 1 '18 at 6:17
plattyplatty
3,370320
$endgroup$
add a comment |
$begingroup$
We can simplify
$$exp(- log log n) = frac{1}{e^{log log n}} = frac{1}{log n}$$
Since $(log log n) log n leq n$, this diverges.
answered Dec 1 '18 at 6:17
plattyplatty
3,370320
$endgroup$
We can simplify
$$exp(- log log n) = frac{1}{e^{log log n}} = frac{1}{log n}$$
Since $(log log n) log n leq n$, this diverges.
answered Dec 1 '18 at 6:17
plattyplatty
3,370320
answered Dec 1 '18 at 6:17
plattyplatty
3,370320
answered Dec 1 '18 at 6:17
plattyplatty
3,370320
answered Dec 1 '18 at 6:17
plattyplatty
3,370320
3,370320
add a comment |
add a comment |
$begingroup$
$sum_n frac{1}{n}$ is divergent. So is $sum_n frac{1}{log(log(n))}$
But I feel $sum_n frac{exp(-log(log(n))}{log(log(n))}$ might still be convergent, you just need to find another way to prove it.
answered Dec 1 '18 at 6:19
MoonKnightMoonKnight
1,369611
$endgroup$
add a comment |
$begingroup$
$sum_n frac{1}{n}$ is divergent. So is $sum_n frac{1}{log(log(n))}$
But I feel $sum_n frac{exp(-log(log(n))}{log(log(n))}$ might still be convergent, you just need to find another way to prove it.
answered Dec 1 '18 at 6:19
MoonKnightMoonKnight
1,369611
$endgroup$
add a comment |
$begingroup$
$sum_n frac{1}{n}$ is divergent. So is $sum_n frac{1}{log(log(n))}$
But I feel $sum_n frac{exp(-log(log(n))}{log(log(n))}$ might still be convergent, you just need to find another way to prove it.
answered Dec 1 '18 at 6:19
MoonKnightMoonKnight
1,369611
$endgroup$
$sum_n frac{1}{n}$ is divergent. So is $sum_n frac{1}{log(log(n))}$
But I feel $sum_n frac{exp(-log(log(n))}{log(log(n))}$ might still be convergent, you just need to find another way to prove it.
answered Dec 1 '18 at 6:19
MoonKnightMoonKnight
1,369611
answered Dec 1 '18 at 6:19
MoonKnightMoonKnight
1,369611
answered Dec 1 '18 at 6:19
MoonKnightMoonKnight
1,369611
answered Dec 1 '18 at 6:19
MoonKnightMoonKnight
1,369611
1,369611
add a comment |
add a comment |
$begingroup$
$$
frac 1{(log log n )exp(log log n)} = frac 1 {log log n cdot log n} geqslant frac 1{log ^2 n} geqslant frac 1{n^{1/2 times 2}} = frac 1n,
$$
so it still diverges.
answered Dec 1 '18 at 6:18
xbhxbh
6,1351522
$endgroup$
$begingroup$
Oh, ok. Thanks a lot. Do you have any advice on how to show $sum_n P(Z>sqrt{2log log n}(1+epsilon)) < infty$ for $Z sim N(0,1)$?
$endgroup$
– Xiaomi
Dec 1 '18 at 6:21
$begingroup$
@Xiaomi Sorry, I suck at probability theory.
$endgroup$
– xbh
Dec 1 '18 at 7:15
add a comment |
$begingroup$
$$
frac 1{(log log n )exp(log log n)} = frac 1 {log log n cdot log n} geqslant frac 1{log ^2 n} geqslant frac 1{n^{1/2 times 2}} = frac 1n,
$$
so it still diverges.
answered Dec 1 '18 at 6:18
xbhxbh
6,1351522
$endgroup$
$begingroup$
Oh, ok. Thanks a lot. Do you have any advice on how to show $sum_n P(Z>sqrt{2log log n}(1+epsilon)) < infty$ for $Z sim N(0,1)$?
$endgroup$
– Xiaomi
Dec 1 '18 at 6:21
$begingroup$
@Xiaomi Sorry, I suck at probability theory.
$endgroup$
– xbh
Dec 1 '18 at 7:15
add a comment |
$begingroup$
$$
frac 1{(log log n )exp(log log n)} = frac 1 {log log n cdot log n} geqslant frac 1{log ^2 n} geqslant frac 1{n^{1/2 times 2}} = frac 1n,
$$
so it still diverges.
answered Dec 1 '18 at 6:18
xbhxbh
6,1351522
$endgroup$
$$
frac 1{(log log n )exp(log log n)} = frac 1 {log log n cdot log n} geqslant frac 1{log ^2 n} geqslant frac 1{n^{1/2 times 2}} = frac 1n,
$$
so it still diverges.
answered Dec 1 '18 at 6:18
xbhxbh
6,1351522
answered Dec 1 '18 at 6:18
xbhxbh
6,1351522
answered Dec 1 '18 at 6:18
xbhxbh
6,1351522
answered Dec 1 '18 at 6:18
xbhxbh
6,1351522
6,1351522
$begingroup$
Oh, ok. Thanks a lot. Do you have any advice on how to show $sum_n P(Z>sqrt{2log log n}(1+epsilon)) < infty$ for $Z sim N(0,1)$?
$endgroup$
– Xiaomi
Dec 1 '18 at 6:21
$begingroup$
@Xiaomi Sorry, I suck at probability theory.
$endgroup$
– xbh
Dec 1 '18 at 7:15
add a comment |
$begingroup$
Oh, ok. Thanks a lot. Do you have any advice on how to show $sum_n P(Z>sqrt{2log log n}(1+epsilon)) < infty$ for $Z sim N(0,1)$?
$endgroup$
– Xiaomi
Dec 1 '18 at 6:21
$begingroup$
@Xiaomi Sorry, I suck at probability theory.
$endgroup$
– xbh
Dec 1 '18 at 7:15
$begingroup$
Oh, ok. Thanks a lot. Do you have any advice on how to show $sum_n P(Z>sqrt{2log log n}(1+epsilon)) < infty$ for $Z sim N(0,1)$?
$endgroup$
– Xiaomi
Dec 1 '18 at 6:21
$begingroup$
Oh, ok. Thanks a lot. Do you have any advice on how to show $sum_n P(Z>sqrt{2log log n}(1+epsilon)) < infty$ for $Z sim N(0,1)$?
$endgroup$
– Xiaomi
Dec 1 '18 at 6:21
$begingroup$
@Xiaomi Sorry, I suck at probability theory.
$endgroup$
– xbh
Dec 1 '18 at 7:15
$begingroup$
@Xiaomi Sorry, I suck at probability theory.
$endgroup$
– xbh
Dec 1 '18 at 7:15
add a comment |
$begingroup$
Cauchy condensation shows
$$sum_n frac{1}{log(log(n))} sim sum_n frac{2^n}{log(log(2^n))} = sum_n frac{2^n}{log(nlog(2))} = sum_n frac{2^n}{log(n) + log(log(2))} $$
So, it is divergent.
answered Dec 1 '18 at 6:20
trancelocationtrancelocation
11.3k1724
$endgroup$
$begingroup$
Thanks a lot. Do you have any advice on how to show $sum_n P(Z>sqrt{2log log n}(1+epsilon)) < infty$ for $Z sim N(0,1)$? The common probability inequalities do not seem tight enough to show this..
$endgroup$
– Xiaomi
Dec 1 '18 at 6:24
$begingroup$
You may use the integral for $N(0,1)$ for each member of the series and estimate it: $$P(Z>sqrt{2log log n}(1+epsilon)) sim int_{sqrt{2log log n}(1+epsilon)}^{infty} e^{frac{-x^2}{2}}dx$$
$endgroup$
– trancelocation
Dec 1 '18 at 6:48
add a comment |
$begingroup$
Cauchy condensation shows
$$sum_n frac{1}{log(log(n))} sim sum_n frac{2^n}{log(log(2^n))} = sum_n frac{2^n}{log(nlog(2))} = sum_n frac{2^n}{log(n) + log(log(2))} $$
So, it is divergent.
answered Dec 1 '18 at 6:20
trancelocationtrancelocation
11.3k1724
$endgroup$
$begingroup$
Thanks a lot. Do you have any advice on how to show $sum_n P(Z>sqrt{2log log n}(1+epsilon)) < infty$ for $Z sim N(0,1)$? The common probability inequalities do not seem tight enough to show this..
$endgroup$
– Xiaomi
Dec 1 '18 at 6:24
$begingroup$
You may use the integral for $N(0,1)$ for each member of the series and estimate it: $$P(Z>sqrt{2log log n}(1+epsilon)) sim int_{sqrt{2log log n}(1+epsilon)}^{infty} e^{frac{-x^2}{2}}dx$$
$endgroup$
– trancelocation
Dec 1 '18 at 6:48
add a comment |
$begingroup$
Cauchy condensation shows
$$sum_n frac{1}{log(log(n))} sim sum_n frac{2^n}{log(log(2^n))} = sum_n frac{2^n}{log(nlog(2))} = sum_n frac{2^n}{log(n) + log(log(2))} $$
So, it is divergent.
answered Dec 1 '18 at 6:20
trancelocationtrancelocation
11.3k1724
$endgroup$
Cauchy condensation shows
$$sum_n frac{1}{log(log(n))} sim sum_n frac{2^n}{log(log(2^n))} = sum_n frac{2^n}{log(nlog(2))} = sum_n frac{2^n}{log(n) + log(log(2))} $$
So, it is divergent.
answered Dec 1 '18 at 6:20
trancelocationtrancelocation
11.3k1724
answered Dec 1 '18 at 6:20
trancelocationtrancelocation
11.3k1724
answered Dec 1 '18 at 6:20
trancelocationtrancelocation
11.3k1724
answered Dec 1 '18 at 6:20
trancelocationtrancelocation
11.3k1724
11.3k1724
$begingroup$
Thanks a lot. Do you have any advice on how to show $sum_n P(Z>sqrt{2log log n}(1+epsilon)) < infty$ for $Z sim N(0,1)$? The common probability inequalities do not seem tight enough to show this..
$endgroup$
– Xiaomi
Dec 1 '18 at 6:24
$begingroup$
You may use the integral for $N(0,1)$ for each member of the series and estimate it: $$P(Z>sqrt{2log log n}(1+epsilon)) sim int_{sqrt{2log log n}(1+epsilon)}^{infty} e^{frac{-x^2}{2}}dx$$
$endgroup$
– trancelocation
Dec 1 '18 at 6:48
add a comment |
$begingroup$
Thanks a lot. Do you have any advice on how to show $sum_n P(Z>sqrt{2log log n}(1+epsilon)) < infty$ for $Z sim N(0,1)$? The common probability inequalities do not seem tight enough to show this..
$endgroup$
– Xiaomi
Dec 1 '18 at 6:24
$begingroup$
You may use the integral for $N(0,1)$ for each member of the series and estimate it: $$P(Z>sqrt{2log log n}(1+epsilon)) sim int_{sqrt{2log log n}(1+epsilon)}^{infty} e^{frac{-x^2}{2}}dx$$
$endgroup$
– trancelocation
Dec 1 '18 at 6:48
$begingroup$
Thanks a lot. Do you have any advice on how to show $sum_n P(Z>sqrt{2log log n}(1+epsilon)) < infty$ for $Z sim N(0,1)$? The common probability inequalities do not seem tight enough to show this..
$endgroup$
– Xiaomi
Dec 1 '18 at 6:24
$begingroup$
Thanks a lot. Do you have any advice on how to show $sum_n P(Z>sqrt{2log log n}(1+epsilon)) < infty$ for $Z sim N(0,1)$? The common probability inequalities do not seem tight enough to show this..
$endgroup$
– Xiaomi
Dec 1 '18 at 6:24
$begingroup$
You may use the integral for $N(0,1)$ for each member of the series and estimate it: $$P(Z>sqrt{2log log n}(1+epsilon)) sim int_{sqrt{2log log n}(1+epsilon)}^{infty} e^{frac{-x^2}{2}}dx$$
$endgroup$
– trancelocation
Dec 1 '18 at 6:48
$begingroup$
You may use the integral for $N(0,1)$ for each member of the series and estimate it: $$P(Z>sqrt{2log log n}(1+epsilon)) sim int_{sqrt{2log log n}(1+epsilon)}^{infty} e^{frac{-x^2}{2}}dx$$
$endgroup$
– trancelocation
Dec 1 '18 at 6:48
add a comment |
$begingroup$
Credits to user MoonKnight.
$log n <n ; $
And once more:
$log (log n) lt log n lt n.$
$dfrac {1}{n} lt dfrac{1}{log n} lt dfrac{1}{log (log n)}.$
Comparison test.
answered Dec 1 '18 at 7:15
Peter SzilasPeter Szilas
11.3k2822
$endgroup$
add a comment |
$begingroup$
Credits to user MoonKnight.
$log n <n ; $
And once more:
$log (log n) lt log n lt n.$
$dfrac {1}{n} lt dfrac{1}{log n} lt dfrac{1}{log (log n)}.$
Comparison test.
answered Dec 1 '18 at 7:15
Peter SzilasPeter Szilas
11.3k2822
$endgroup$
add a comment |
$begingroup$
Credits to user MoonKnight.
$log n <n ; $
And once more:
$log (log n) lt log n lt n.$
$dfrac {1}{n} lt dfrac{1}{log n} lt dfrac{1}{log (log n)}.$
Comparison test.
answered Dec 1 '18 at 7:15
Peter SzilasPeter Szilas
11.3k2822
$endgroup$
Credits to user MoonKnight.
$log n <n ; $
And once more:
$log (log n) lt log n lt n.$
$dfrac {1}{n} lt dfrac{1}{log n} lt dfrac{1}{log (log n)}.$
Comparison test.
answered Dec 1 '18 at 7:15
Peter SzilasPeter Szilas
11.3k2822
answered Dec 1 '18 at 7:15
Peter SzilasPeter Szilas
11.3k2822
answered Dec 1 '18 at 7:15
Peter SzilasPeter Szilas
11.3k2822
answered Dec 1 '18 at 7:15
Peter SzilasPeter Szilas
11.3k2822
11.3k2822
add a comment |
add a comment |
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