Mixing
How to prove that $a(1-a^N)/(1-a)<N$?
$begingroup$
I derive the expectation for a problem and get
$$N-frac{a(1-a^N)}{1-a}$$
where $0<a<1$, and $N>0$.
The physical meaning of this value determines it should be positive. But I intuitively think $frac{a(1-a^N)}{1-a}$ might exceed $N$ when $a$ is close to $1$.
However, I have enumerated some numerical values of $a$, i.e., $0.99999$, $0.999999$ and find
$$frac{a(1-a^N)}{1-a}<N$$ (where $0<a<1$, and $N>0$) holds true, so I conjecture it is always true, but am not sure how to prove it.
limits proof-verification inequality
asked Jan 24 at 5:49
Guoyang QinGuoyang Qin
1327
$endgroup$
add a comment |
$begingroup$
I derive the expectation for a problem and get
$$N-frac{a(1-a^N)}{1-a}$$
where $0<a<1$, and $N>0$.
The physical meaning of this value determines it should be positive. But I intuitively think $frac{a(1-a^N)}{1-a}$ might exceed $N$ when $a$ is close to $1$.
However, I have enumerated some numerical values of $a$, i.e., $0.99999$, $0.999999$ and find
$$frac{a(1-a^N)}{1-a}<N$$ (where $0<a<1$, and $N>0$) holds true, so I conjecture it is always true, but am not sure how to prove it.
limits proof-verification inequality
asked Jan 24 at 5:49
Guoyang QinGuoyang Qin
1327
$endgroup$
add a comment |
$begingroup$
I derive the expectation for a problem and get
$$N-frac{a(1-a^N)}{1-a}$$
where $0<a<1$, and $N>0$.
The physical meaning of this value determines it should be positive. But I intuitively think $frac{a(1-a^N)}{1-a}$ might exceed $N$ when $a$ is close to $1$.
However, I have enumerated some numerical values of $a$, i.e., $0.99999$, $0.999999$ and find
$$frac{a(1-a^N)}{1-a}<N$$ (where $0<a<1$, and $N>0$) holds true, so I conjecture it is always true, but am not sure how to prove it.
limits proof-verification inequality
asked Jan 24 at 5:49
Guoyang QinGuoyang Qin
1327
$endgroup$
I derive the expectation for a problem and get
$$N-frac{a(1-a^N)}{1-a}$$
where $0<a<1$, and $N>0$.
The physical meaning of this value determines it should be positive. But I intuitively think $frac{a(1-a^N)}{1-a}$ might exceed $N$ when $a$ is close to $1$.
However, I have enumerated some numerical values of $a$, i.e., $0.99999$, $0.999999$ and find
$$frac{a(1-a^N)}{1-a}<N$$ (where $0<a<1$, and $N>0$) holds true, so I conjecture it is always true, but am not sure how to prove it.
limits proof-verification inequality
limits proof-verification inequality
asked Jan 24 at 5:49
Guoyang QinGuoyang Qin
1327
asked Jan 24 at 5:49
Guoyang QinGuoyang Qin
1327
asked Jan 24 at 5:49
Guoyang QinGuoyang Qin
1327
asked Jan 24 at 5:49
Guoyang QinGuoyang Qin
1327
asked Jan 24 at 5:49
Guoyang QinGuoyang Qin
1327
1327
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
When $N$ is a positive integer, and $0<a<1$, then
$$frac{a(1-a^N)}{1-a}=a+a^2+a^3+cdots+a^N<1+1+1+cdots+1=N.$$
answered Jan 24 at 5:51
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
$begingroup$
Thanks. The thing is I actually derived $frac{a(1-a^N}{1-a}$ just from the geometric sequence, however I didn't realize it.
$endgroup$
– Guoyang Qin
Jan 24 at 5:59
add a comment |
$begingroup$
Well, $1-a^N=(1-a)(1+a+a^2+a^3+...+a^{N-1})$, and each of the terms in the second bracket is less than or equal to 1, so $1-a^Nleq(1-a)N$
Therefore, $frac{a(1-a^N)}{1-a}leqfrac{a(1-a)N}{1-a}leq Na<N$ as long as $0<a<1$
Does that help?
answered Jan 24 at 5:54
Michael HartleyMichael Hartley
1,10555
$endgroup$
1
$begingroup$
Thanks, it does help. Actually I just derived it from the sequence but I didn't realize it could tell me why.
$endgroup$
– Guoyang Qin
Jan 24 at 6:00
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
$begingroup$
When $N$ is a positive integer, and $0<a<1$, then
$$frac{a(1-a^N)}{1-a}=a+a^2+a^3+cdots+a^N<1+1+1+cdots+1=N.$$
answered Jan 24 at 5:51
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
$begingroup$
Thanks. The thing is I actually derived $frac{a(1-a^N}{1-a}$ just from the geometric sequence, however I didn't realize it.
$endgroup$
– Guoyang Qin
Jan 24 at 5:59
add a comment |
$begingroup$
When $N$ is a positive integer, and $0<a<1$, then
$$frac{a(1-a^N)}{1-a}=a+a^2+a^3+cdots+a^N<1+1+1+cdots+1=N.$$
answered Jan 24 at 5:51
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
$begingroup$
Thanks. The thing is I actually derived $frac{a(1-a^N}{1-a}$ just from the geometric sequence, however I didn't realize it.
$endgroup$
– Guoyang Qin
Jan 24 at 5:59
add a comment |
$begingroup$
When $N$ is a positive integer, and $0<a<1$, then
$$frac{a(1-a^N)}{1-a}=a+a^2+a^3+cdots+a^N<1+1+1+cdots+1=N.$$
answered Jan 24 at 5:51
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
When $N$ is a positive integer, and $0<a<1$, then
$$frac{a(1-a^N)}{1-a}=a+a^2+a^3+cdots+a^N<1+1+1+cdots+1=N.$$
answered Jan 24 at 5:51
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered Jan 24 at 5:51
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered Jan 24 at 5:51
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered Jan 24 at 5:51
Lord Shark the UnknownLord Shark the Unknown
106k1161133
106k1161133
$begingroup$
Thanks. The thing is I actually derived $frac{a(1-a^N}{1-a}$ just from the geometric sequence, however I didn't realize it.
$endgroup$
– Guoyang Qin
Jan 24 at 5:59
add a comment |
$begingroup$
Thanks. The thing is I actually derived $frac{a(1-a^N}{1-a}$ just from the geometric sequence, however I didn't realize it.
$endgroup$
– Guoyang Qin
Jan 24 at 5:59
$begingroup$
Thanks. The thing is I actually derived $frac{a(1-a^N}{1-a}$ just from the geometric sequence, however I didn't realize it.
$endgroup$
– Guoyang Qin
Jan 24 at 5:59
$begingroup$
Thanks. The thing is I actually derived $frac{a(1-a^N}{1-a}$ just from the geometric sequence, however I didn't realize it.
$endgroup$
– Guoyang Qin
Jan 24 at 5:59
add a comment |
$begingroup$
Well, $1-a^N=(1-a)(1+a+a^2+a^3+...+a^{N-1})$, and each of the terms in the second bracket is less than or equal to 1, so $1-a^Nleq(1-a)N$
Therefore, $frac{a(1-a^N)}{1-a}leqfrac{a(1-a)N}{1-a}leq Na<N$ as long as $0<a<1$
Does that help?
answered Jan 24 at 5:54
Michael HartleyMichael Hartley
1,10555
$endgroup$
1
$begingroup$
Thanks, it does help. Actually I just derived it from the sequence but I didn't realize it could tell me why.
$endgroup$
– Guoyang Qin
Jan 24 at 6:00
add a comment |
$begingroup$
Well, $1-a^N=(1-a)(1+a+a^2+a^3+...+a^{N-1})$, and each of the terms in the second bracket is less than or equal to 1, so $1-a^Nleq(1-a)N$
Therefore, $frac{a(1-a^N)}{1-a}leqfrac{a(1-a)N}{1-a}leq Na<N$ as long as $0<a<1$
Does that help?
answered Jan 24 at 5:54
Michael HartleyMichael Hartley
1,10555
$endgroup$
1
$begingroup$
Thanks, it does help. Actually I just derived it from the sequence but I didn't realize it could tell me why.
$endgroup$
– Guoyang Qin
Jan 24 at 6:00
add a comment |
$begingroup$
Well, $1-a^N=(1-a)(1+a+a^2+a^3+...+a^{N-1})$, and each of the terms in the second bracket is less than or equal to 1, so $1-a^Nleq(1-a)N$
Therefore, $frac{a(1-a^N)}{1-a}leqfrac{a(1-a)N}{1-a}leq Na<N$ as long as $0<a<1$
Does that help?
answered Jan 24 at 5:54
Michael HartleyMichael Hartley
1,10555
$endgroup$
Well, $1-a^N=(1-a)(1+a+a^2+a^3+...+a^{N-1})$, and each of the terms in the second bracket is less than or equal to 1, so $1-a^Nleq(1-a)N$
Therefore, $frac{a(1-a^N)}{1-a}leqfrac{a(1-a)N}{1-a}leq Na<N$ as long as $0<a<1$
Does that help?
answered Jan 24 at 5:54
Michael HartleyMichael Hartley
1,10555
answered Jan 24 at 5:54
Michael HartleyMichael Hartley
1,10555
answered Jan 24 at 5:54
Michael HartleyMichael Hartley
1,10555
answered Jan 24 at 5:54
Michael HartleyMichael Hartley
1,10555
1,10555
1
$begingroup$
Thanks, it does help. Actually I just derived it from the sequence but I didn't realize it could tell me why.
$endgroup$
– Guoyang Qin
Jan 24 at 6:00
add a comment |
1
$begingroup$
Thanks, it does help. Actually I just derived it from the sequence but I didn't realize it could tell me why.
$endgroup$
– Guoyang Qin
Jan 24 at 6:00
1
1
$begingroup$
Thanks, it does help. Actually I just derived it from the sequence but I didn't realize it could tell me why.
$endgroup$
– Guoyang Qin
Jan 24 at 6:00
$begingroup$
Thanks, it does help. Actually I just derived it from the sequence but I didn't realize it could tell me why.
$endgroup$
– Guoyang Qin
Jan 24 at 6:00
add a comment |
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