Does series with factorials converge/diverge: $sumlimits_{n=1}^infty frac{4^n n!n!}{(2n)!}$?
$$sum_{n=1}^infty {{4^n n!n!}over{(2n)!}}$$
I tried the ratio test but got that the limit is equal to 1, this tells me nothing of whether the series diverges or converges. if I didn't make any errors when doing the ratio test, it may diverge, but I need help proving that. Is there any other test I could try.
calculus sequences-and-series convergence factorial divergent-series
edited Nov 25 '18 at 12:16
Martin Sleziak
44.7k7115270
asked Nov 7 '15 at 21:09
user287927
142
add a comment |
$$sum_{n=1}^infty {{4^n n!n!}over{(2n)!}}$$
I tried the ratio test but got that the limit is equal to 1, this tells me nothing of whether the series diverges or converges. if I didn't make any errors when doing the ratio test, it may diverge, but I need help proving that. Is there any other test I could try.
calculus sequences-and-series convergence factorial divergent-series
edited Nov 25 '18 at 12:16
Martin Sleziak
44.7k7115270
asked Nov 7 '15 at 21:09
user287927
142
math.stackexchange.com/questions/78533/…
– R.N
Nov 7 '15 at 21:18
the searched limit is infinity
– Dr. Sonnhard Graubner
Nov 7 '15 at 21:22
See also central binomial coefficient.
– Lucian
Nov 8 '15 at 8:56
add a comment |
$$sum_{n=1}^infty {{4^n n!n!}over{(2n)!}}$$
I tried the ratio test but got that the limit is equal to 1, this tells me nothing of whether the series diverges or converges. if I didn't make any errors when doing the ratio test, it may diverge, but I need help proving that. Is there any other test I could try.
calculus sequences-and-series convergence factorial divergent-series
edited Nov 25 '18 at 12:16
Martin Sleziak
44.7k7115270
asked Nov 7 '15 at 21:09
user287927
142
$$sum_{n=1}^infty {{4^n n!n!}over{(2n)!}}$$
I tried the ratio test but got that the limit is equal to 1, this tells me nothing of whether the series diverges or converges. if I didn't make any errors when doing the ratio test, it may diverge, but I need help proving that. Is there any other test I could try.
calculus sequences-and-series convergence factorial divergent-series
calculus sequences-and-series convergence factorial divergent-series
edited Nov 25 '18 at 12:16
Martin Sleziak
44.7k7115270
asked Nov 7 '15 at 21:09
user287927
142
edited Nov 25 '18 at 12:16
Martin Sleziak
44.7k7115270
asked Nov 7 '15 at 21:09
user287927
142
edited Nov 25 '18 at 12:16
Martin Sleziak
44.7k7115270
edited Nov 25 '18 at 12:16
Martin Sleziak
44.7k7115270
edited Nov 25 '18 at 12:16
Martin Sleziak
44.7k7115270
44.7k7115270
asked Nov 7 '15 at 21:09
user287927
142
asked Nov 7 '15 at 21:09
user287927
142
asked Nov 7 '15 at 21:09
user287927
142
142
math.stackexchange.com/questions/78533/…
– R.N
Nov 7 '15 at 21:18
the searched limit is infinity
– Dr. Sonnhard Graubner
Nov 7 '15 at 21:22
See also central binomial coefficient.
– Lucian
Nov 8 '15 at 8:56
add a comment |
math.stackexchange.com/questions/78533/…
– R.N
Nov 7 '15 at 21:18
the searched limit is infinity
– Dr. Sonnhard Graubner
Nov 7 '15 at 21:22
See also central binomial coefficient.
– Lucian
Nov 8 '15 at 8:56
math.stackexchange.com/questions/78533/…
– R.N
Nov 7 '15 at 21:18
math.stackexchange.com/questions/78533/…
– R.N
Nov 7 '15 at 21:18
the searched limit is infinity
– Dr. Sonnhard Graubner
Nov 7 '15 at 21:22
the searched limit is infinity
– Dr. Sonnhard Graubner
Nov 7 '15 at 21:22
See also central binomial coefficient.
– Lucian
Nov 8 '15 at 8:56
See also central binomial coefficient.
– Lucian
Nov 8 '15 at 8:56
add a comment |
2 Answers
2
active
oldest
votes
Note that $$frac{(2n)!}{n!n!}leqsum_0^{2n}binom{2n}{k}=(1+1)^{2n}=4^n$$
So $$frac{4^n n!n!}{(2n)!}geq 1$$
answered Nov 7 '15 at 21:24
Wojowu
17.1k22565
Do you think the series comparison test would help ?
– user287927
Nov 7 '15 at 21:31
It would, but it's not required. The terms of your series don't tend to $0$, so the series cannot be convergent.
– Wojowu
Nov 7 '15 at 21:32
how would you show that it diverges
– user287927
Nov 7 '15 at 21:35
add a comment |
One approach is a comparison test, approximating the function using Stirling's approximation. Since $n!= sqrt{2pi}n^{n+1/2}text{e}^{-n}(1+mathcal{O}(tfrac{1}{n}))$ we have $$frac{4^n n!^2}{(2n)!}approxfrac{4^{n}2pi cdot n^{2n+1}text{e}^{-2n}}{sqrt{2pi} 2^{2n+1/2} n^{2n+1/2}text{e}^{-2n}}=sqrt{pi n}.$$ The series therefore diverges.
answered Nov 7 '15 at 21:57
J.G.
22.8k22136
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Note that $$frac{(2n)!}{n!n!}leqsum_0^{2n}binom{2n}{k}=(1+1)^{2n}=4^n$$
So $$frac{4^n n!n!}{(2n)!}geq 1$$
answered Nov 7 '15 at 21:24
Wojowu
17.1k22565
Do you think the series comparison test would help ?
– user287927
Nov 7 '15 at 21:31
It would, but it's not required. The terms of your series don't tend to $0$, so the series cannot be convergent.
– Wojowu
Nov 7 '15 at 21:32
how would you show that it diverges
– user287927
Nov 7 '15 at 21:35
add a comment |
Note that $$frac{(2n)!}{n!n!}leqsum_0^{2n}binom{2n}{k}=(1+1)^{2n}=4^n$$
So $$frac{4^n n!n!}{(2n)!}geq 1$$
answered Nov 7 '15 at 21:24
Wojowu
17.1k22565
Do you think the series comparison test would help ?
– user287927
Nov 7 '15 at 21:31
It would, but it's not required. The terms of your series don't tend to $0$, so the series cannot be convergent.
– Wojowu
Nov 7 '15 at 21:32
how would you show that it diverges
– user287927
Nov 7 '15 at 21:35
add a comment |
Note that $$frac{(2n)!}{n!n!}leqsum_0^{2n}binom{2n}{k}=(1+1)^{2n}=4^n$$
So $$frac{4^n n!n!}{(2n)!}geq 1$$
answered Nov 7 '15 at 21:24
Wojowu
17.1k22565
Note that $$frac{(2n)!}{n!n!}leqsum_0^{2n}binom{2n}{k}=(1+1)^{2n}=4^n$$
So $$frac{4^n n!n!}{(2n)!}geq 1$$
answered Nov 7 '15 at 21:24
Wojowu
17.1k22565
answered Nov 7 '15 at 21:24
Wojowu
17.1k22565
answered Nov 7 '15 at 21:24
Wojowu
17.1k22565
answered Nov 7 '15 at 21:24
Wojowu
17.1k22565
17.1k22565
Do you think the series comparison test would help ?
– user287927
Nov 7 '15 at 21:31
It would, but it's not required. The terms of your series don't tend to $0$, so the series cannot be convergent.
– Wojowu
Nov 7 '15 at 21:32
how would you show that it diverges
– user287927
Nov 7 '15 at 21:35
add a comment |
Do you think the series comparison test would help ?
– user287927
Nov 7 '15 at 21:31
It would, but it's not required. The terms of your series don't tend to $0$, so the series cannot be convergent.
– Wojowu
Nov 7 '15 at 21:32
how would you show that it diverges
– user287927
Nov 7 '15 at 21:35
Do you think the series comparison test would help ?
– user287927
Nov 7 '15 at 21:31
Do you think the series comparison test would help ?
– user287927
Nov 7 '15 at 21:31
It would, but it's not required. The terms of your series don't tend to $0$, so the series cannot be convergent.
– Wojowu
Nov 7 '15 at 21:32
It would, but it's not required. The terms of your series don't tend to $0$, so the series cannot be convergent.
– Wojowu
Nov 7 '15 at 21:32
how would you show that it diverges
– user287927
Nov 7 '15 at 21:35
how would you show that it diverges
– user287927
Nov 7 '15 at 21:35
add a comment |
One approach is a comparison test, approximating the function using Stirling's approximation. Since $n!= sqrt{2pi}n^{n+1/2}text{e}^{-n}(1+mathcal{O}(tfrac{1}{n}))$ we have $$frac{4^n n!^2}{(2n)!}approxfrac{4^{n}2pi cdot n^{2n+1}text{e}^{-2n}}{sqrt{2pi} 2^{2n+1/2} n^{2n+1/2}text{e}^{-2n}}=sqrt{pi n}.$$ The series therefore diverges.
answered Nov 7 '15 at 21:57
J.G.
22.8k22136
add a comment |
One approach is a comparison test, approximating the function using Stirling's approximation. Since $n!= sqrt{2pi}n^{n+1/2}text{e}^{-n}(1+mathcal{O}(tfrac{1}{n}))$ we have $$frac{4^n n!^2}{(2n)!}approxfrac{4^{n}2pi cdot n^{2n+1}text{e}^{-2n}}{sqrt{2pi} 2^{2n+1/2} n^{2n+1/2}text{e}^{-2n}}=sqrt{pi n}.$$ The series therefore diverges.
answered Nov 7 '15 at 21:57
J.G.
22.8k22136
add a comment |
One approach is a comparison test, approximating the function using Stirling's approximation. Since $n!= sqrt{2pi}n^{n+1/2}text{e}^{-n}(1+mathcal{O}(tfrac{1}{n}))$ we have $$frac{4^n n!^2}{(2n)!}approxfrac{4^{n}2pi cdot n^{2n+1}text{e}^{-2n}}{sqrt{2pi} 2^{2n+1/2} n^{2n+1/2}text{e}^{-2n}}=sqrt{pi n}.$$ The series therefore diverges.
answered Nov 7 '15 at 21:57
J.G.
22.8k22136
One approach is a comparison test, approximating the function using Stirling's approximation. Since $n!= sqrt{2pi}n^{n+1/2}text{e}^{-n}(1+mathcal{O}(tfrac{1}{n}))$ we have $$frac{4^n n!^2}{(2n)!}approxfrac{4^{n}2pi cdot n^{2n+1}text{e}^{-2n}}{sqrt{2pi} 2^{2n+1/2} n^{2n+1/2}text{e}^{-2n}}=sqrt{pi n}.$$ The series therefore diverges.
answered Nov 7 '15 at 21:57
J.G.
22.8k22136
edited Nov 25 '18 at 13:55
answered Nov 7 '15 at 21:57
J.G.
22.8k22136
answered Nov 7 '15 at 21:57
J.G.
22.8k22136
answered Nov 7 '15 at 21:57
J.G.
22.8k22136
22.8k22136
add a comment |
add a comment |
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math.stackexchange.com/questions/78533/…
– R.N
Nov 7 '15 at 21:18
the searched limit is infinity
– Dr. Sonnhard Graubner
Nov 7 '15 at 21:22
See also central binomial coefficient.
– Lucian
Nov 8 '15 at 8:56