Mixing
Limit of $frac{n!}{(n+1)!}$ as n approaches infinity.
$begingroup$
I know that factorials grow faster than any exponential function, but what if you put two factorials up against each other? My problem is finding the limit of:
$$frac{n!}{(n+1)!}$$
as $n$ approaches infinity.
Any ways on how to proceed would be appreciated.
sequences-and-series limits
edited Feb 1 at 19:49
José Carlos Santos
174k23133242
asked Nov 28 '14 at 16:20
martinmartin
235414
$endgroup$
add a comment |
$begingroup$
I know that factorials grow faster than any exponential function, but what if you put two factorials up against each other? My problem is finding the limit of:
$$frac{n!}{(n+1)!}$$
as $n$ approaches infinity.
Any ways on how to proceed would be appreciated.
sequences-and-series limits
edited Feb 1 at 19:49
José Carlos Santos
174k23133242
asked Nov 28 '14 at 16:20
martinmartin
235414
$endgroup$
10
$begingroup$
$$(n+1)!=(n+1)cdot n!$$
$endgroup$
– lab bhattacharjee
Nov 28 '14 at 16:21
add a comment |
$begingroup$
I know that factorials grow faster than any exponential function, but what if you put two factorials up against each other? My problem is finding the limit of:
$$frac{n!}{(n+1)!}$$
as $n$ approaches infinity.
Any ways on how to proceed would be appreciated.
sequences-and-series limits
edited Feb 1 at 19:49
José Carlos Santos
174k23133242
asked Nov 28 '14 at 16:20
martinmartin
235414
$endgroup$
I know that factorials grow faster than any exponential function, but what if you put two factorials up against each other? My problem is finding the limit of:
$$frac{n!}{(n+1)!}$$
as $n$ approaches infinity.
Any ways on how to proceed would be appreciated.
sequences-and-series limits
sequences-and-series limits
edited Feb 1 at 19:49
José Carlos Santos
174k23133242
asked Nov 28 '14 at 16:20
martinmartin
235414
edited Feb 1 at 19:49
José Carlos Santos
174k23133242
asked Nov 28 '14 at 16:20
martinmartin
235414
edited Feb 1 at 19:49
José Carlos Santos
174k23133242
edited Feb 1 at 19:49
José Carlos Santos
174k23133242
edited Feb 1 at 19:49
José Carlos Santos
174k23133242
174k23133242
asked Nov 28 '14 at 16:20
martinmartin
235414
asked Nov 28 '14 at 16:20
martinmartin
235414
asked Nov 28 '14 at 16:20
martinmartin
235414
235414
10
$begingroup$
$$(n+1)!=(n+1)cdot n!$$
$endgroup$
– lab bhattacharjee
Nov 28 '14 at 16:21
add a comment |
10
$begingroup$
$$(n+1)!=(n+1)cdot n!$$
$endgroup$
– lab bhattacharjee
Nov 28 '14 at 16:21
10
10
$begingroup$
$$(n+1)!=(n+1)cdot n!$$
$endgroup$
– lab bhattacharjee
Nov 28 '14 at 16:21
$begingroup$
$$(n+1)!=(n+1)cdot n!$$
$endgroup$
– lab bhattacharjee
Nov 28 '14 at 16:21
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$frac{n!}{(n+1)!} = frac{require{cancel}cancel{n!}}{(n+1)cancel{n!}}=frac 1{n+1}$$
Can you take it from here?
answered Nov 28 '14 at 16:25
NamasteNamaste
1
$endgroup$
$begingroup$
Indeed, thank you! :)
$endgroup$
– martin
Nov 28 '14 at 16:27
$begingroup$
You're welcome, fadaes!
$endgroup$
– Namaste
Nov 28 '14 at 16:27
add a comment |
$begingroup$
Hint:
$$frac{n!}{(n+1)!}=frac{n!}{(n+1)(n!)}=frac{1}{n+1}$$
Game Over!
answered Nov 28 '14 at 16:24
IuʇǝƃɹɐʇoɹIuʇǝƃɹɐʇoɹ
8,36723550
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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$begingroup$
$$frac{n!}{(n+1)!} = frac{require{cancel}cancel{n!}}{(n+1)cancel{n!}}=frac 1{n+1}$$
Can you take it from here?
answered Nov 28 '14 at 16:25
NamasteNamaste
1
$endgroup$
$begingroup$
Indeed, thank you! :)
$endgroup$
– martin
Nov 28 '14 at 16:27
$begingroup$
You're welcome, fadaes!
$endgroup$
– Namaste
Nov 28 '14 at 16:27
add a comment |
$begingroup$
$$frac{n!}{(n+1)!} = frac{require{cancel}cancel{n!}}{(n+1)cancel{n!}}=frac 1{n+1}$$
Can you take it from here?
answered Nov 28 '14 at 16:25
NamasteNamaste
1
$endgroup$
$begingroup$
Indeed, thank you! :)
$endgroup$
– martin
Nov 28 '14 at 16:27
$begingroup$
You're welcome, fadaes!
$endgroup$
– Namaste
Nov 28 '14 at 16:27
add a comment |
$begingroup$
$$frac{n!}{(n+1)!} = frac{require{cancel}cancel{n!}}{(n+1)cancel{n!}}=frac 1{n+1}$$
Can you take it from here?
answered Nov 28 '14 at 16:25
NamasteNamaste
1
$endgroup$
$$frac{n!}{(n+1)!} = frac{require{cancel}cancel{n!}}{(n+1)cancel{n!}}=frac 1{n+1}$$
Can you take it from here?
answered Nov 28 '14 at 16:25
NamasteNamaste
1
answered Nov 28 '14 at 16:25
NamasteNamaste
1
answered Nov 28 '14 at 16:25
NamasteNamaste
1
answered Nov 28 '14 at 16:25
NamasteNamaste
1
1
$begingroup$
Indeed, thank you! :)
$endgroup$
– martin
Nov 28 '14 at 16:27
$begingroup$
You're welcome, fadaes!
$endgroup$
– Namaste
Nov 28 '14 at 16:27
add a comment |
$begingroup$
Indeed, thank you! :)
$endgroup$
– martin
Nov 28 '14 at 16:27
$begingroup$
You're welcome, fadaes!
$endgroup$
– Namaste
Nov 28 '14 at 16:27
$begingroup$
Indeed, thank you! :)
$endgroup$
– martin
Nov 28 '14 at 16:27
$begingroup$
Indeed, thank you! :)
$endgroup$
– martin
Nov 28 '14 at 16:27
$begingroup$
You're welcome, fadaes!
$endgroup$
– Namaste
Nov 28 '14 at 16:27
$begingroup$
You're welcome, fadaes!
$endgroup$
– Namaste
Nov 28 '14 at 16:27
add a comment |
$begingroup$
Hint:
$$frac{n!}{(n+1)!}=frac{n!}{(n+1)(n!)}=frac{1}{n+1}$$
Game Over!
answered Nov 28 '14 at 16:24
IuʇǝƃɹɐʇoɹIuʇǝƃɹɐʇoɹ
8,36723550
$endgroup$
add a comment |
$begingroup$
Hint:
$$frac{n!}{(n+1)!}=frac{n!}{(n+1)(n!)}=frac{1}{n+1}$$
Game Over!
answered Nov 28 '14 at 16:24
IuʇǝƃɹɐʇoɹIuʇǝƃɹɐʇoɹ
8,36723550
$endgroup$
add a comment |
$begingroup$
Hint:
$$frac{n!}{(n+1)!}=frac{n!}{(n+1)(n!)}=frac{1}{n+1}$$
Game Over!
answered Nov 28 '14 at 16:24
IuʇǝƃɹɐʇoɹIuʇǝƃɹɐʇoɹ
8,36723550
$endgroup$
Hint:
$$frac{n!}{(n+1)!}=frac{n!}{(n+1)(n!)}=frac{1}{n+1}$$
Game Over!
answered Nov 28 '14 at 16:24
IuʇǝƃɹɐʇoɹIuʇǝƃɹɐʇoɹ
8,36723550
answered Nov 28 '14 at 16:24
IuʇǝƃɹɐʇoɹIuʇǝƃɹɐʇoɹ
8,36723550
answered Nov 28 '14 at 16:24
IuʇǝƃɹɐʇoɹIuʇǝƃɹɐʇoɹ
8,36723550
answered Nov 28 '14 at 16:24
IuʇǝƃɹɐʇoɹIuʇǝƃɹɐʇoɹ
8,36723550
8,36723550
add a comment |
add a comment |
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10
$begingroup$
$$(n+1)!=(n+1)cdot n!$$
$endgroup$
– lab bhattacharjee
Nov 28 '14 at 16:21