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Show that $sum_n=1^infty frac(2n)!4^n(n!)^2$ is divergent
Clash Royale CLAN TAG#URR8PPP
up vote
2
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Given the series $S=sum_n=1^infty frac(2n)!4^n(n!)^2$ I'm trying to prove its divergent but with no luck. It doesn't tend to $infty$ as $n$ grows large and ratio test is inconclusive.
I have been trying to do a comparison test but all series less than $S$ that I was able to come up with were convergent. So I'm starting to suspect that the series might actually be convergent, any hints?
Thanks!
sequences-and-series
asked Jul 20 at 18:05
Shadowfirex
456
add a comment |Â
up vote
2
down vote
favorite
Given the series $S=sum_n=1^infty frac(2n)!4^n(n!)^2$ I'm trying to prove its divergent but with no luck. It doesn't tend to $infty$ as $n$ grows large and ratio test is inconclusive.
I have been trying to do a comparison test but all series less than $S$ that I was able to come up with were convergent. So I'm starting to suspect that the series might actually be convergent, any hints?
Thanks!
sequences-and-series
asked Jul 20 at 18:05
Shadowfirex
456
Hint: The series is divergent.
– Cornman
Jul 20 at 18:07
2
@Comman - the OP declares he wants to prove that the series is divergent. How does "the series is divergent" provide a hint?
– uniquesolution
Jul 20 at 18:08
1
Did you try Stirling's formula?
– uniquesolution
Jul 20 at 18:08
@Cornman: how is that a hint in this context? Care to elaborate?
– Clayton
Jul 20 at 18:10
2
@uniquesolution I do not know what you are reading, but I read "So I'm starting to suspect that the series might actually be convergent"
– Cornman
Jul 20 at 18:10
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Given the series $S=sum_n=1^infty frac(2n)!4^n(n!)^2$ I'm trying to prove its divergent but with no luck. It doesn't tend to $infty$ as $n$ grows large and ratio test is inconclusive.
I have been trying to do a comparison test but all series less than $S$ that I was able to come up with were convergent. So I'm starting to suspect that the series might actually be convergent, any hints?
Thanks!
sequences-and-series
asked Jul 20 at 18:05
Shadowfirex
456
Given the series $S=sum_n=1^infty frac(2n)!4^n(n!)^2$ I'm trying to prove its divergent but with no luck. It doesn't tend to $infty$ as $n$ grows large and ratio test is inconclusive.
I have been trying to do a comparison test but all series less than $S$ that I was able to come up with were convergent. So I'm starting to suspect that the series might actually be convergent, any hints?
Thanks!
sequences-and-series
asked Jul 20 at 18:05
Shadowfirex
456
asked Jul 20 at 18:05
Shadowfirex
456
asked Jul 20 at 18:05
Shadowfirex
456
asked Jul 20 at 18:05
Shadowfirex
456
456
Hint: The series is divergent.
– Cornman
Jul 20 at 18:07
2
@Comman - the OP declares he wants to prove that the series is divergent. How does "the series is divergent" provide a hint?
– uniquesolution
Jul 20 at 18:08
1
Did you try Stirling's formula?
– uniquesolution
Jul 20 at 18:08
@Cornman: how is that a hint in this context? Care to elaborate?
– Clayton
Jul 20 at 18:10
2
@uniquesolution I do not know what you are reading, but I read "So I'm starting to suspect that the series might actually be convergent"
– Cornman
Jul 20 at 18:10
add a comment |Â
Hint: The series is divergent.
– Cornman
Jul 20 at 18:07
2
@Comman - the OP declares he wants to prove that the series is divergent. How does "the series is divergent" provide a hint?
– uniquesolution
Jul 20 at 18:08
1
Did you try Stirling's formula?
– uniquesolution
Jul 20 at 18:08
@Cornman: how is that a hint in this context? Care to elaborate?
– Clayton
Jul 20 at 18:10
2
@uniquesolution I do not know what you are reading, but I read "So I'm starting to suspect that the series might actually be convergent"
– Cornman
Jul 20 at 18:10
Hint: The series is divergent.
– Cornman
Jul 20 at 18:07
Hint: The series is divergent.
– Cornman
Jul 20 at 18:07
2
2
@Comman - the OP declares he wants to prove that the series is divergent. How does "the series is divergent" provide a hint?
– uniquesolution
Jul 20 at 18:08
@Comman - the OP declares he wants to prove that the series is divergent. How does "the series is divergent" provide a hint?
– uniquesolution
Jul 20 at 18:08
1
1
Did you try Stirling's formula?
– uniquesolution
Jul 20 at 18:08
Did you try Stirling's formula?
– uniquesolution
Jul 20 at 18:08
@Cornman: how is that a hint in this context? Care to elaborate?
– Clayton
Jul 20 at 18:10
@Cornman: how is that a hint in this context? Care to elaborate?
– Clayton
Jul 20 at 18:10
2
2
@uniquesolution I do not know what you are reading, but I read "So I'm starting to suspect that the series might actually be convergent"
– Cornman
Jul 20 at 18:10
@uniquesolution I do not know what you are reading, but I read "So I'm starting to suspect that the series might actually be convergent"
– Cornman
Jul 20 at 18:10
add a comment |Â
5 Answers
5
active
oldest
votes
up vote
5
down vote
accepted
$$frac(2n)!4^n(n!)^2=fracC_2n^n4^n=fracsum (C_n^i)^24^ngeqfracfrac(sum C_n^i)^2n+14^n=frac1n+1$$ by vandermonde identity and QM-AM inequality.
and harmonic series is divergent so the series in question is also divergent.
answered Jul 20 at 18:13
Roronoa Zoro
234211
Can you expand a little on how we got the last three inequalities/equalities?
– Shadowfirex
Jul 20 at 19:12
Sorry I don't know vandermonde identity and QM-AM, I'm reading about them right now. But any more details would be helpful too!
– Shadowfirex
Jul 20 at 19:13
Makes sense thanks!
– Shadowfirex
Jul 22 at 22:57
add a comment |Â
up vote
3
down vote
Consider $$S(x)= sum_n=0^infty binom2nn left(fracx 2 right) ^2 n $$ Since $$binomfrac-12n=(-1)^nfracbinom2nn4^n$$ By the binomial series we have that $$sum_n=0^inftybinomfrac-12n(-x^2)^n=frac1sqrt1-x^2$$ thus your series is just $$S(1)=frac1sqrt1-1rightarrowinfty$$
answered Jul 20 at 18:18
Zacky
2,2251326
add a comment |Â
up vote
2
down vote
Since in a comment you said you're not familiar with QM-AM and Vandermonde's identity, you may find the following useful.
Hint: Note that$$s_n=frac(2n)!4^n(n!)^2=frac14^nfrac(n+1)(n+2)cdots(2n)n!$$so one has $$s_n+1=s_nfrac(2n+1)(2n+2)4(n+1)^2=fracs_n2frac2n+1n+1.$$Use this to prove $s_ngefrac1n+1$ by induction.
answered Jul 20 at 19:50
Vincenzo Oliva
5,16711133
$fracs_n2frac2n+1n+1 geq frac1n+1frac2n+12n+2 < frac1n+1$ so $s_n+1 geq frac1n+1$ is not true.
– Shadowfirex
Jul 20 at 21:08
1
@Shadowfirex: Yeah, but you want $s_n+1gefrac1n+2$.
– Vincenzo Oliva
Jul 20 at 21:20
@Shadowfirex Did you see what I pointed out to you?
– Vincenzo Oliva
Jul 23 at 20:29
add a comment |Â
up vote
0
down vote
Use Raabe's Test with
$$dfraca_n+1a_n=1-dfracAn+dfracA_nn=1-dfracfrac12n+dfracA_nn$$
where $A_n=dfrac12(n+1)to0$ as $ntoinfty$, then $A=dfrac12<1$ shows the series diverges!
answered Jul 20 at 18:17
Nosrati
19.5k41544
add a comment |Â
up vote
0
down vote
Whatare you allowed to use? With Stirling's approximation
$$
binom2nn sim frac4^nsqrt pi n
$$
so the summand is $O(frac1sqrtn)$ and
$$
sum_k=1^n frac1sqrtpi k
$$
and this diverges by comparing it to the Harmonic series.
Once again, this works if you are allowed to use the divergence of Harmonic series and Stirling's approximation.
answered Jul 20 at 21:13
Alex
13.9k41932
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
$$frac(2n)!4^n(n!)^2=fracC_2n^n4^n=fracsum (C_n^i)^24^ngeqfracfrac(sum C_n^i)^2n+14^n=frac1n+1$$ by vandermonde identity and QM-AM inequality.
and harmonic series is divergent so the series in question is also divergent.
answered Jul 20 at 18:13
Roronoa Zoro
234211
Can you expand a little on how we got the last three inequalities/equalities?
– Shadowfirex
Jul 20 at 19:12
Sorry I don't know vandermonde identity and QM-AM, I'm reading about them right now. But any more details would be helpful too!
– Shadowfirex
Jul 20 at 19:13
Makes sense thanks!
– Shadowfirex
Jul 22 at 22:57
add a comment |Â
up vote
5
down vote
accepted
$$frac(2n)!4^n(n!)^2=fracC_2n^n4^n=fracsum (C_n^i)^24^ngeqfracfrac(sum C_n^i)^2n+14^n=frac1n+1$$ by vandermonde identity and QM-AM inequality.
and harmonic series is divergent so the series in question is also divergent.
answered Jul 20 at 18:13
Roronoa Zoro
234211
Can you expand a little on how we got the last three inequalities/equalities?
– Shadowfirex
Jul 20 at 19:12
Sorry I don't know vandermonde identity and QM-AM, I'm reading about them right now. But any more details would be helpful too!
– Shadowfirex
Jul 20 at 19:13
Makes sense thanks!
– Shadowfirex
Jul 22 at 22:57
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
$$frac(2n)!4^n(n!)^2=fracC_2n^n4^n=fracsum (C_n^i)^24^ngeqfracfrac(sum C_n^i)^2n+14^n=frac1n+1$$ by vandermonde identity and QM-AM inequality.
and harmonic series is divergent so the series in question is also divergent.
answered Jul 20 at 18:13
Roronoa Zoro
234211
$$frac(2n)!4^n(n!)^2=fracC_2n^n4^n=fracsum (C_n^i)^24^ngeqfracfrac(sum C_n^i)^2n+14^n=frac1n+1$$ by vandermonde identity and QM-AM inequality.
and harmonic series is divergent so the series in question is also divergent.
answered Jul 20 at 18:13
Roronoa Zoro
234211
answered Jul 20 at 18:13
Roronoa Zoro
234211
answered Jul 20 at 18:13
Roronoa Zoro
234211
answered Jul 20 at 18:13
Roronoa Zoro
234211
234211
Can you expand a little on how we got the last three inequalities/equalities?
– Shadowfirex
Jul 20 at 19:12
Sorry I don't know vandermonde identity and QM-AM, I'm reading about them right now. But any more details would be helpful too!
– Shadowfirex
Jul 20 at 19:13
Makes sense thanks!
– Shadowfirex
Jul 22 at 22:57
add a comment |Â
Can you expand a little on how we got the last three inequalities/equalities?
– Shadowfirex
Jul 20 at 19:12
Sorry I don't know vandermonde identity and QM-AM, I'm reading about them right now. But any more details would be helpful too!
– Shadowfirex
Jul 20 at 19:13
Makes sense thanks!
– Shadowfirex
Jul 22 at 22:57
Can you expand a little on how we got the last three inequalities/equalities?
– Shadowfirex
Jul 20 at 19:12
Can you expand a little on how we got the last three inequalities/equalities?
– Shadowfirex
Jul 20 at 19:12
Sorry I don't know vandermonde identity and QM-AM, I'm reading about them right now. But any more details would be helpful too!
– Shadowfirex
Jul 20 at 19:13
Sorry I don't know vandermonde identity and QM-AM, I'm reading about them right now. But any more details would be helpful too!
– Shadowfirex
Jul 20 at 19:13
Makes sense thanks!
– Shadowfirex
Jul 22 at 22:57
Makes sense thanks!
– Shadowfirex
Jul 22 at 22:57
add a comment |Â
up vote
3
down vote
Consider $$S(x)= sum_n=0^infty binom2nn left(fracx 2 right) ^2 n $$ Since $$binomfrac-12n=(-1)^nfracbinom2nn4^n$$ By the binomial series we have that $$sum_n=0^inftybinomfrac-12n(-x^2)^n=frac1sqrt1-x^2$$ thus your series is just $$S(1)=frac1sqrt1-1rightarrowinfty$$
answered Jul 20 at 18:18
Zacky
2,2251326
add a comment |Â
up vote
3
down vote
Consider $$S(x)= sum_n=0^infty binom2nn left(fracx 2 right) ^2 n $$ Since $$binomfrac-12n=(-1)^nfracbinom2nn4^n$$ By the binomial series we have that $$sum_n=0^inftybinomfrac-12n(-x^2)^n=frac1sqrt1-x^2$$ thus your series is just $$S(1)=frac1sqrt1-1rightarrowinfty$$
answered Jul 20 at 18:18
Zacky
2,2251326
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Consider $$S(x)= sum_n=0^infty binom2nn left(fracx 2 right) ^2 n $$ Since $$binomfrac-12n=(-1)^nfracbinom2nn4^n$$ By the binomial series we have that $$sum_n=0^inftybinomfrac-12n(-x^2)^n=frac1sqrt1-x^2$$ thus your series is just $$S(1)=frac1sqrt1-1rightarrowinfty$$
answered Jul 20 at 18:18
Zacky
2,2251326
Consider $$S(x)= sum_n=0^infty binom2nn left(fracx 2 right) ^2 n $$ Since $$binomfrac-12n=(-1)^nfracbinom2nn4^n$$ By the binomial series we have that $$sum_n=0^inftybinomfrac-12n(-x^2)^n=frac1sqrt1-x^2$$ thus your series is just $$S(1)=frac1sqrt1-1rightarrowinfty$$
answered Jul 20 at 18:18
Zacky
2,2251326
edited Jul 20 at 18:29
answered Jul 20 at 18:18
Zacky
2,2251326
answered Jul 20 at 18:18
Zacky
2,2251326
answered Jul 20 at 18:18
Zacky
2,2251326
2,2251326
add a comment |Â
add a comment |Â
up vote
2
down vote
Since in a comment you said you're not familiar with QM-AM and Vandermonde's identity, you may find the following useful.
Hint: Note that$$s_n=frac(2n)!4^n(n!)^2=frac14^nfrac(n+1)(n+2)cdots(2n)n!$$so one has $$s_n+1=s_nfrac(2n+1)(2n+2)4(n+1)^2=fracs_n2frac2n+1n+1.$$Use this to prove $s_ngefrac1n+1$ by induction.
answered Jul 20 at 19:50
Vincenzo Oliva
5,16711133
$fracs_n2frac2n+1n+1 geq frac1n+1frac2n+12n+2 < frac1n+1$ so $s_n+1 geq frac1n+1$ is not true.
– Shadowfirex
Jul 20 at 21:08
1
@Shadowfirex: Yeah, but you want $s_n+1gefrac1n+2$.
– Vincenzo Oliva
Jul 20 at 21:20
@Shadowfirex Did you see what I pointed out to you?
– Vincenzo Oliva
Jul 23 at 20:29
add a comment |Â
up vote
2
down vote
Since in a comment you said you're not familiar with QM-AM and Vandermonde's identity, you may find the following useful.
Hint: Note that$$s_n=frac(2n)!4^n(n!)^2=frac14^nfrac(n+1)(n+2)cdots(2n)n!$$so one has $$s_n+1=s_nfrac(2n+1)(2n+2)4(n+1)^2=fracs_n2frac2n+1n+1.$$Use this to prove $s_ngefrac1n+1$ by induction.
answered Jul 20 at 19:50
Vincenzo Oliva
5,16711133
$fracs_n2frac2n+1n+1 geq frac1n+1frac2n+12n+2 < frac1n+1$ so $s_n+1 geq frac1n+1$ is not true.
– Shadowfirex
Jul 20 at 21:08
1
@Shadowfirex: Yeah, but you want $s_n+1gefrac1n+2$.
– Vincenzo Oliva
Jul 20 at 21:20
@Shadowfirex Did you see what I pointed out to you?
– Vincenzo Oliva
Jul 23 at 20:29
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Since in a comment you said you're not familiar with QM-AM and Vandermonde's identity, you may find the following useful.
Hint: Note that$$s_n=frac(2n)!4^n(n!)^2=frac14^nfrac(n+1)(n+2)cdots(2n)n!$$so one has $$s_n+1=s_nfrac(2n+1)(2n+2)4(n+1)^2=fracs_n2frac2n+1n+1.$$Use this to prove $s_ngefrac1n+1$ by induction.
answered Jul 20 at 19:50
Vincenzo Oliva
5,16711133
Since in a comment you said you're not familiar with QM-AM and Vandermonde's identity, you may find the following useful.
Hint: Note that$$s_n=frac(2n)!4^n(n!)^2=frac14^nfrac(n+1)(n+2)cdots(2n)n!$$so one has $$s_n+1=s_nfrac(2n+1)(2n+2)4(n+1)^2=fracs_n2frac2n+1n+1.$$Use this to prove $s_ngefrac1n+1$ by induction.
answered Jul 20 at 19:50
Vincenzo Oliva
5,16711133
answered Jul 20 at 19:50
Vincenzo Oliva
5,16711133
answered Jul 20 at 19:50
Vincenzo Oliva
5,16711133
answered Jul 20 at 19:50
Vincenzo Oliva
5,16711133
5,16711133
$fracs_n2frac2n+1n+1 geq frac1n+1frac2n+12n+2 < frac1n+1$ so $s_n+1 geq frac1n+1$ is not true.
– Shadowfirex
Jul 20 at 21:08
1
@Shadowfirex: Yeah, but you want $s_n+1gefrac1n+2$.
– Vincenzo Oliva
Jul 20 at 21:20
@Shadowfirex Did you see what I pointed out to you?
– Vincenzo Oliva
Jul 23 at 20:29
add a comment |Â
$fracs_n2frac2n+1n+1 geq frac1n+1frac2n+12n+2 < frac1n+1$ so $s_n+1 geq frac1n+1$ is not true.
– Shadowfirex
Jul 20 at 21:08
1
@Shadowfirex: Yeah, but you want $s_n+1gefrac1n+2$.
– Vincenzo Oliva
Jul 20 at 21:20
@Shadowfirex Did you see what I pointed out to you?
– Vincenzo Oliva
Jul 23 at 20:29
$fracs_n2frac2n+1n+1 geq frac1n+1frac2n+12n+2 < frac1n+1$ so $s_n+1 geq frac1n+1$ is not true.
– Shadowfirex
Jul 20 at 21:08
$fracs_n2frac2n+1n+1 geq frac1n+1frac2n+12n+2 < frac1n+1$ so $s_n+1 geq frac1n+1$ is not true.
– Shadowfirex
Jul 20 at 21:08
1
1
@Shadowfirex: Yeah, but you want $s_n+1gefrac1n+2$.
– Vincenzo Oliva
Jul 20 at 21:20
@Shadowfirex: Yeah, but you want $s_n+1gefrac1n+2$.
– Vincenzo Oliva
Jul 20 at 21:20
@Shadowfirex Did you see what I pointed out to you?
– Vincenzo Oliva
Jul 23 at 20:29
@Shadowfirex Did you see what I pointed out to you?
– Vincenzo Oliva
Jul 23 at 20:29
add a comment |Â
up vote
0
down vote
Use Raabe's Test with
$$dfraca_n+1a_n=1-dfracAn+dfracA_nn=1-dfracfrac12n+dfracA_nn$$
where $A_n=dfrac12(n+1)to0$ as $ntoinfty$, then $A=dfrac12<1$ shows the series diverges!
answered Jul 20 at 18:17
Nosrati
19.5k41544
add a comment |Â
up vote
0
down vote
Use Raabe's Test with
$$dfraca_n+1a_n=1-dfracAn+dfracA_nn=1-dfracfrac12n+dfracA_nn$$
where $A_n=dfrac12(n+1)to0$ as $ntoinfty$, then $A=dfrac12<1$ shows the series diverges!
answered Jul 20 at 18:17
Nosrati
19.5k41544
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Use Raabe's Test with
$$dfraca_n+1a_n=1-dfracAn+dfracA_nn=1-dfracfrac12n+dfracA_nn$$
where $A_n=dfrac12(n+1)to0$ as $ntoinfty$, then $A=dfrac12<1$ shows the series diverges!
answered Jul 20 at 18:17
Nosrati
19.5k41544
Use Raabe's Test with
$$dfraca_n+1a_n=1-dfracAn+dfracA_nn=1-dfracfrac12n+dfracA_nn$$
where $A_n=dfrac12(n+1)to0$ as $ntoinfty$, then $A=dfrac12<1$ shows the series diverges!
answered Jul 20 at 18:17
Nosrati
19.5k41544
edited Jul 20 at 19:00
answered Jul 20 at 18:17
Nosrati
19.5k41544
answered Jul 20 at 18:17
Nosrati
19.5k41544
answered Jul 20 at 18:17
Nosrati
19.5k41544
19.5k41544
add a comment |Â
add a comment |Â
up vote
0
down vote
Whatare you allowed to use? With Stirling's approximation
$$
binom2nn sim frac4^nsqrt pi n
$$
so the summand is $O(frac1sqrtn)$ and
$$
sum_k=1^n frac1sqrtpi k
$$
and this diverges by comparing it to the Harmonic series.
Once again, this works if you are allowed to use the divergence of Harmonic series and Stirling's approximation.
answered Jul 20 at 21:13
Alex
13.9k41932
add a comment |Â
up vote
0
down vote
Whatare you allowed to use? With Stirling's approximation
$$
binom2nn sim frac4^nsqrt pi n
$$
so the summand is $O(frac1sqrtn)$ and
$$
sum_k=1^n frac1sqrtpi k
$$
and this diverges by comparing it to the Harmonic series.
Once again, this works if you are allowed to use the divergence of Harmonic series and Stirling's approximation.
answered Jul 20 at 21:13
Alex
13.9k41932
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Whatare you allowed to use? With Stirling's approximation
$$
binom2nn sim frac4^nsqrt pi n
$$
so the summand is $O(frac1sqrtn)$ and
$$
sum_k=1^n frac1sqrtpi k
$$
and this diverges by comparing it to the Harmonic series.
Once again, this works if you are allowed to use the divergence of Harmonic series and Stirling's approximation.
answered Jul 20 at 21:13
Alex
13.9k41932
Whatare you allowed to use? With Stirling's approximation
$$
binom2nn sim frac4^nsqrt pi n
$$
so the summand is $O(frac1sqrtn)$ and
$$
sum_k=1^n frac1sqrtpi k
$$
and this diverges by comparing it to the Harmonic series.
Once again, this works if you are allowed to use the divergence of Harmonic series and Stirling's approximation.
answered Jul 20 at 21:13
Alex
13.9k41932
answered Jul 20 at 21:13
Alex
13.9k41932
answered Jul 20 at 21:13
Alex
13.9k41932
answered Jul 20 at 21:13
Alex
13.9k41932
13.9k41932
add a comment |Â
add a comment |Â
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Hint: The series is divergent.
– Cornman
Jul 20 at 18:07
2
@Comman - the OP declares he wants to prove that the series is divergent. How does "the series is divergent" provide a hint?
– uniquesolution
Jul 20 at 18:08
1
Did you try Stirling's formula?
– uniquesolution
Jul 20 at 18:08
@Cornman: how is that a hint in this context? Care to elaborate?
– Clayton
Jul 20 at 18:10
2
@uniquesolution I do not know what you are reading, but I read "So I'm starting to suspect that the series might actually be convergent"
– Cornman
Jul 20 at 18:10