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Prove that $left(1+frac1 nright)^n > 2$
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I'm trying to demonstrate that $left( 1+frac1 n right)^n$ is bigger than $2$. I have tried to prove that $left( 1+frac1 n right)^n$ is smaller than $left( 1+frac1n+1 right)^n+1$ by expanding
$left( 1+frac1n right)^n = sumlimits_i=0^n left( fracnk right) frac1n^k$ and $left( 1+frac1n+1 right)^n+1 = sumlimits_i=0^n+1 left( frac(n+1)k right) frac1(n+1)^k$ but it doesn't seem to work.
What am I missing? Also, is there a method to demonstrate that without induction?
calculus
edited Jul 17 at 0:56
David G. Stork
7,6912929
asked Jul 17 at 0:01
Sergi boquera fuentes
141
add a comment |Â
up vote
1
down vote
favorite
I'm trying to demonstrate that $left( 1+frac1 n right)^n$ is bigger than $2$. I have tried to prove that $left( 1+frac1 n right)^n$ is smaller than $left( 1+frac1n+1 right)^n+1$ by expanding
$left( 1+frac1n right)^n = sumlimits_i=0^n left( fracnk right) frac1n^k$ and $left( 1+frac1n+1 right)^n+1 = sumlimits_i=0^n+1 left( frac(n+1)k right) frac1(n+1)^k$ but it doesn't seem to work.
What am I missing? Also, is there a method to demonstrate that without induction?
calculus
edited Jul 17 at 0:56
David G. Stork
7,6912929
asked Jul 17 at 0:01
Sergi boquera fuentes
141
Induction is going to be there, although it could be hiding in the proof of some other result. If you know the "expansion" (Binomial formula), then look at the first two terms $1+nfrac1n+...$. They already add up to $2$. Finally notice that the other terms are all non-negative. In this case induction is hiding in proving the binomial theorem.
– user574889
Jul 17 at 0:08
In many books you can also find your inequality proven as a direct consequence of Bernoulli's inequality
– user574889
Jul 17 at 0:10
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm trying to demonstrate that $left( 1+frac1 n right)^n$ is bigger than $2$. I have tried to prove that $left( 1+frac1 n right)^n$ is smaller than $left( 1+frac1n+1 right)^n+1$ by expanding
$left( 1+frac1n right)^n = sumlimits_i=0^n left( fracnk right) frac1n^k$ and $left( 1+frac1n+1 right)^n+1 = sumlimits_i=0^n+1 left( frac(n+1)k right) frac1(n+1)^k$ but it doesn't seem to work.
What am I missing? Also, is there a method to demonstrate that without induction?
calculus
edited Jul 17 at 0:56
David G. Stork
7,6912929
asked Jul 17 at 0:01
Sergi boquera fuentes
141
I'm trying to demonstrate that $left( 1+frac1 n right)^n$ is bigger than $2$. I have tried to prove that $left( 1+frac1 n right)^n$ is smaller than $left( 1+frac1n+1 right)^n+1$ by expanding
$left( 1+frac1n right)^n = sumlimits_i=0^n left( fracnk right) frac1n^k$ and $left( 1+frac1n+1 right)^n+1 = sumlimits_i=0^n+1 left( frac(n+1)k right) frac1(n+1)^k$ but it doesn't seem to work.
What am I missing? Also, is there a method to demonstrate that without induction?
calculus
edited Jul 17 at 0:56
David G. Stork
7,6912929
asked Jul 17 at 0:01
Sergi boquera fuentes
141
edited Jul 17 at 0:56
David G. Stork
7,6912929
edited Jul 17 at 0:56
David G. Stork
7,6912929
edited Jul 17 at 0:56
David G. Stork
7,6912929
7,6912929
asked Jul 17 at 0:01
Sergi boquera fuentes
141
asked Jul 17 at 0:01
Sergi boquera fuentes
141
asked Jul 17 at 0:01
Sergi boquera fuentes
141
141
Induction is going to be there, although it could be hiding in the proof of some other result. If you know the "expansion" (Binomial formula), then look at the first two terms $1+nfrac1n+...$. They already add up to $2$. Finally notice that the other terms are all non-negative. In this case induction is hiding in proving the binomial theorem.
– user574889
Jul 17 at 0:08
In many books you can also find your inequality proven as a direct consequence of Bernoulli's inequality
– user574889
Jul 17 at 0:10
add a comment |Â
Induction is going to be there, although it could be hiding in the proof of some other result. If you know the "expansion" (Binomial formula), then look at the first two terms $1+nfrac1n+...$. They already add up to $2$. Finally notice that the other terms are all non-negative. In this case induction is hiding in proving the binomial theorem.
– user574889
Jul 17 at 0:08
In many books you can also find your inequality proven as a direct consequence of Bernoulli's inequality
– user574889
Jul 17 at 0:10
Induction is going to be there, although it could be hiding in the proof of some other result. If you know the "expansion" (Binomial formula), then look at the first two terms $1+nfrac1n+...$. They already add up to $2$. Finally notice that the other terms are all non-negative. In this case induction is hiding in proving the binomial theorem.
– user574889
Jul 17 at 0:08
Induction is going to be there, although it could be hiding in the proof of some other result. If you know the "expansion" (Binomial formula), then look at the first two terms $1+nfrac1n+...$. They already add up to $2$. Finally notice that the other terms are all non-negative. In this case induction is hiding in proving the binomial theorem.
– user574889
Jul 17 at 0:08
In many books you can also find your inequality proven as a direct consequence of Bernoulli's inequality
– user574889
Jul 17 at 0:10
In many books you can also find your inequality proven as a direct consequence of Bernoulli's inequality
– user574889
Jul 17 at 0:10
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
8
down vote
You just need to consider the first $2$ terms in the binomial expansion
begineqnarray*
left( 1+ frac1n right)^n = sum_i=0^nbinomnifrac1n^i=1+n frac1n +cdots geq 2.
endeqnarray*
answered Jul 17 at 0:07
Donald Splutterwit
21.3k21243
Aww, darn, you beat me to it! :-P
– Brian Tung
Jul 17 at 0:07
1
@BrianTung ... Sorry about that ... I owe you one!
– Donald Splutterwit
Jul 17 at 0:09
Haha, not at all! GMTA.
– Brian Tung
Jul 17 at 0:09
add a comment |Â
up vote
3
down vote
Let $f(x) = (1+x)^n$. Note that $f$ is convex for $x ge 0$ and so
$f(x) ge f(0)+f'(0) x = 1+xn$. Hence $f(1 over n) ge 2$.
answered Jul 17 at 0:48
copper.hat
122k557156
add a comment |Â
up vote
1
down vote
beginalign
left(1+frac1nright)^n &= sum_k=0^n n choose k frac1n^k \
&= sum_k=0^n fracn(n-1)cdots(n-k+1)k!n^k \
&= sum_k=0^n frac1k!left(1-frac1nright)left(1-frac2nright)cdots left(1-frack-1nright)
endalign
so
beginalign
left(1+frac1n+1right)^n+1 &= sum_k=0^n+1 frac1k!left(1-frac1n+1right)left(1-frac2n+1right)cdots left(1-frack-1n+1right)\
&> sum_k=0^n frac1k!left(1-frac1nright)left(1-frac2nright)cdots left(1-frack-1nright)\
&= left(1+frac1nright)^n
endalign
On the other hand, for $n = 1$ we have
$$left(1+frac11right)^1 = 2$$
so $left(1+frac1nright)^n > 2, forall n ge 2$.
answered Jul 17 at 0:29
mechanodroid
22.3k52041
add a comment |Â
up vote
1
down vote
Another way is to prove first that your sequence is monotonically increasing like has been done here:
I have to show $(1+frac1n)^n$ is monotonically increasing sequence
... and since your first term is $2$, it follows that the subsequent ones are larger than $2$.
answered Jul 17 at 0:41
Momo
11.9k21330
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
You just need to consider the first $2$ terms in the binomial expansion
begineqnarray*
left( 1+ frac1n right)^n = sum_i=0^nbinomnifrac1n^i=1+n frac1n +cdots geq 2.
endeqnarray*
answered Jul 17 at 0:07
Donald Splutterwit
21.3k21243
Aww, darn, you beat me to it! :-P
– Brian Tung
Jul 17 at 0:07
1
@BrianTung ... Sorry about that ... I owe you one!
– Donald Splutterwit
Jul 17 at 0:09
Haha, not at all! GMTA.
– Brian Tung
Jul 17 at 0:09
add a comment |Â
up vote
8
down vote
You just need to consider the first $2$ terms in the binomial expansion
begineqnarray*
left( 1+ frac1n right)^n = sum_i=0^nbinomnifrac1n^i=1+n frac1n +cdots geq 2.
endeqnarray*
answered Jul 17 at 0:07
Donald Splutterwit
21.3k21243
Aww, darn, you beat me to it! :-P
– Brian Tung
Jul 17 at 0:07
1
@BrianTung ... Sorry about that ... I owe you one!
– Donald Splutterwit
Jul 17 at 0:09
Haha, not at all! GMTA.
– Brian Tung
Jul 17 at 0:09
add a comment |Â
up vote
8
down vote
up vote
8
down vote
You just need to consider the first $2$ terms in the binomial expansion
begineqnarray*
left( 1+ frac1n right)^n = sum_i=0^nbinomnifrac1n^i=1+n frac1n +cdots geq 2.
endeqnarray*
answered Jul 17 at 0:07
Donald Splutterwit
21.3k21243
You just need to consider the first $2$ terms in the binomial expansion
begineqnarray*
left( 1+ frac1n right)^n = sum_i=0^nbinomnifrac1n^i=1+n frac1n +cdots geq 2.
endeqnarray*
answered Jul 17 at 0:07
Donald Splutterwit
21.3k21243
edited Jul 17 at 0:08
answered Jul 17 at 0:07
Donald Splutterwit
21.3k21243
answered Jul 17 at 0:07
Donald Splutterwit
21.3k21243
answered Jul 17 at 0:07
Donald Splutterwit
21.3k21243
21.3k21243
Aww, darn, you beat me to it! :-P
– Brian Tung
Jul 17 at 0:07
1
@BrianTung ... Sorry about that ... I owe you one!
– Donald Splutterwit
Jul 17 at 0:09
Haha, not at all! GMTA.
– Brian Tung
Jul 17 at 0:09
add a comment |Â
Aww, darn, you beat me to it! :-P
– Brian Tung
Jul 17 at 0:07
1
@BrianTung ... Sorry about that ... I owe you one!
– Donald Splutterwit
Jul 17 at 0:09
Haha, not at all! GMTA.
– Brian Tung
Jul 17 at 0:09
Aww, darn, you beat me to it! :-P
– Brian Tung
Jul 17 at 0:07
Aww, darn, you beat me to it! :-P
– Brian Tung
Jul 17 at 0:07
1
1
@BrianTung ... Sorry about that ... I owe you one!
– Donald Splutterwit
Jul 17 at 0:09
@BrianTung ... Sorry about that ... I owe you one!
– Donald Splutterwit
Jul 17 at 0:09
Haha, not at all! GMTA.
– Brian Tung
Jul 17 at 0:09
Haha, not at all! GMTA.
– Brian Tung
Jul 17 at 0:09
add a comment |Â
up vote
3
down vote
Let $f(x) = (1+x)^n$. Note that $f$ is convex for $x ge 0$ and so
$f(x) ge f(0)+f'(0) x = 1+xn$. Hence $f(1 over n) ge 2$.
answered Jul 17 at 0:48
copper.hat
122k557156
add a comment |Â
up vote
3
down vote
Let $f(x) = (1+x)^n$. Note that $f$ is convex for $x ge 0$ and so
$f(x) ge f(0)+f'(0) x = 1+xn$. Hence $f(1 over n) ge 2$.
answered Jul 17 at 0:48
copper.hat
122k557156
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Let $f(x) = (1+x)^n$. Note that $f$ is convex for $x ge 0$ and so
$f(x) ge f(0)+f'(0) x = 1+xn$. Hence $f(1 over n) ge 2$.
answered Jul 17 at 0:48
copper.hat
122k557156
Let $f(x) = (1+x)^n$. Note that $f$ is convex for $x ge 0$ and so
$f(x) ge f(0)+f'(0) x = 1+xn$. Hence $f(1 over n) ge 2$.
answered Jul 17 at 0:48
copper.hat
122k557156
answered Jul 17 at 0:48
copper.hat
122k557156
answered Jul 17 at 0:48
copper.hat
122k557156
answered Jul 17 at 0:48
copper.hat
122k557156
122k557156
add a comment |Â
add a comment |Â
up vote
1
down vote
beginalign
left(1+frac1nright)^n &= sum_k=0^n n choose k frac1n^k \
&= sum_k=0^n fracn(n-1)cdots(n-k+1)k!n^k \
&= sum_k=0^n frac1k!left(1-frac1nright)left(1-frac2nright)cdots left(1-frack-1nright)
endalign
so
beginalign
left(1+frac1n+1right)^n+1 &= sum_k=0^n+1 frac1k!left(1-frac1n+1right)left(1-frac2n+1right)cdots left(1-frack-1n+1right)\
&> sum_k=0^n frac1k!left(1-frac1nright)left(1-frac2nright)cdots left(1-frack-1nright)\
&= left(1+frac1nright)^n
endalign
On the other hand, for $n = 1$ we have
$$left(1+frac11right)^1 = 2$$
so $left(1+frac1nright)^n > 2, forall n ge 2$.
answered Jul 17 at 0:29
mechanodroid
22.3k52041
add a comment |Â
up vote
1
down vote
beginalign
left(1+frac1nright)^n &= sum_k=0^n n choose k frac1n^k \
&= sum_k=0^n fracn(n-1)cdots(n-k+1)k!n^k \
&= sum_k=0^n frac1k!left(1-frac1nright)left(1-frac2nright)cdots left(1-frack-1nright)
endalign
so
beginalign
left(1+frac1n+1right)^n+1 &= sum_k=0^n+1 frac1k!left(1-frac1n+1right)left(1-frac2n+1right)cdots left(1-frack-1n+1right)\
&> sum_k=0^n frac1k!left(1-frac1nright)left(1-frac2nright)cdots left(1-frack-1nright)\
&= left(1+frac1nright)^n
endalign
On the other hand, for $n = 1$ we have
$$left(1+frac11right)^1 = 2$$
so $left(1+frac1nright)^n > 2, forall n ge 2$.
answered Jul 17 at 0:29
mechanodroid
22.3k52041
add a comment |Â
up vote
1
down vote
up vote
1
down vote
beginalign
left(1+frac1nright)^n &= sum_k=0^n n choose k frac1n^k \
&= sum_k=0^n fracn(n-1)cdots(n-k+1)k!n^k \
&= sum_k=0^n frac1k!left(1-frac1nright)left(1-frac2nright)cdots left(1-frack-1nright)
endalign
so
beginalign
left(1+frac1n+1right)^n+1 &= sum_k=0^n+1 frac1k!left(1-frac1n+1right)left(1-frac2n+1right)cdots left(1-frack-1n+1right)\
&> sum_k=0^n frac1k!left(1-frac1nright)left(1-frac2nright)cdots left(1-frack-1nright)\
&= left(1+frac1nright)^n
endalign
On the other hand, for $n = 1$ we have
$$left(1+frac11right)^1 = 2$$
so $left(1+frac1nright)^n > 2, forall n ge 2$.
answered Jul 17 at 0:29
mechanodroid
22.3k52041
beginalign
left(1+frac1nright)^n &= sum_k=0^n n choose k frac1n^k \
&= sum_k=0^n fracn(n-1)cdots(n-k+1)k!n^k \
&= sum_k=0^n frac1k!left(1-frac1nright)left(1-frac2nright)cdots left(1-frack-1nright)
endalign
so
beginalign
left(1+frac1n+1right)^n+1 &= sum_k=0^n+1 frac1k!left(1-frac1n+1right)left(1-frac2n+1right)cdots left(1-frack-1n+1right)\
&> sum_k=0^n frac1k!left(1-frac1nright)left(1-frac2nright)cdots left(1-frack-1nright)\
&= left(1+frac1nright)^n
endalign
On the other hand, for $n = 1$ we have
$$left(1+frac11right)^1 = 2$$
so $left(1+frac1nright)^n > 2, forall n ge 2$.
answered Jul 17 at 0:29
mechanodroid
22.3k52041
answered Jul 17 at 0:29
mechanodroid
22.3k52041
answered Jul 17 at 0:29
mechanodroid
22.3k52041
answered Jul 17 at 0:29
mechanodroid
22.3k52041
22.3k52041
add a comment |Â
add a comment |Â
up vote
1
down vote
Another way is to prove first that your sequence is monotonically increasing like has been done here:
I have to show $(1+frac1n)^n$ is monotonically increasing sequence
... and since your first term is $2$, it follows that the subsequent ones are larger than $2$.
answered Jul 17 at 0:41
Momo
11.9k21330
add a comment |Â
up vote
1
down vote
Another way is to prove first that your sequence is monotonically increasing like has been done here:
I have to show $(1+frac1n)^n$ is monotonically increasing sequence
... and since your first term is $2$, it follows that the subsequent ones are larger than $2$.
answered Jul 17 at 0:41
Momo
11.9k21330
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Another way is to prove first that your sequence is monotonically increasing like has been done here:
I have to show $(1+frac1n)^n$ is monotonically increasing sequence
... and since your first term is $2$, it follows that the subsequent ones are larger than $2$.
answered Jul 17 at 0:41
Momo
11.9k21330
Another way is to prove first that your sequence is monotonically increasing like has been done here:
I have to show $(1+frac1n)^n$ is monotonically increasing sequence
... and since your first term is $2$, it follows that the subsequent ones are larger than $2$.
answered Jul 17 at 0:41
Momo
11.9k21330
answered Jul 17 at 0:41
Momo
11.9k21330
answered Jul 17 at 0:41
Momo
11.9k21330
answered Jul 17 at 0:41
Momo
11.9k21330
11.9k21330
add a comment |Â
add a comment |Â
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Induction is going to be there, although it could be hiding in the proof of some other result. If you know the "expansion" (Binomial formula), then look at the first two terms $1+nfrac1n+...$. They already add up to $2$. Finally notice that the other terms are all non-negative. In this case induction is hiding in proving the binomial theorem.
– user574889
Jul 17 at 0:08
In many books you can also find your inequality proven as a direct consequence of Bernoulli's inequality
– user574889
Jul 17 at 0:10