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Can someone shed some light on this inequality?
The Next CEO of Stack OverflowShow that the sequence $left(frac2^nn!right)$ has a limit.Determine value the following: $L=sum_k=1^inftyfrac1k^k$Could someone help me clarify the steps for this solution?Understanding how to use $epsilon-delta$ definition of a limitHow can an imaginary equation give a real answer?Can someone claify on the work that was done in this question on Maclaurin SeriesConvergence of series $nq^n$.How does this limit converge to zeroUnderstanding part of a proof for Stolz-Cesaro TheoremAbout a statement of partial fraction in an answer
$begingroup$
I have a question relating to image that I've attached. It is a proof that the sequence is increasing. I don't understand the logic behind the third equation $$fraca_n+1a_n>left (1-frac1n+1right ) left (fracn+1nright)$$
where does the equation in the first and second parenthesis come from?
Ok, I have another relating question:
why $$fraca_n+1a_n> (1+frac1n)$$ ( The expression of third line.
![!The proof[1]](https://i.stack.imgur.com/wgz75.png)
sequences-and-series limits eulers-constant
edited yesterday
Rodrigo de Azevedo
13.2k41960
asked yesterday
Ieva BrakmaneIeva Brakmane
316
$endgroup$
add a comment |
$begingroup$
I have a question relating to image that I've attached. It is a proof that the sequence is increasing. I don't understand the logic behind the third equation $$fraca_n+1a_n>left (1-frac1n+1right ) left (fracn+1nright)$$
where does the equation in the first and second parenthesis come from?
Ok, I have another relating question:
why $$fraca_n+1a_n> (1+frac1n)$$ ( The expression of third line.
sequences-and-series limits eulers-constant
edited yesterday
Rodrigo de Azevedo
13.2k41960
asked yesterday
Ieva BrakmaneIeva Brakmane
316
$endgroup$
$begingroup$
Please do not post necessary information only in a picture, not everyone can display and read it properly.
$endgroup$
– Carsten S
yesterday
add a comment |
$begingroup$
I have a question relating to image that I've attached. It is a proof that the sequence is increasing. I don't understand the logic behind the third equation $$fraca_n+1a_n>left (1-frac1n+1right ) left (fracn+1nright)$$
where does the equation in the first and second parenthesis come from?
Ok, I have another relating question:
why $$fraca_n+1a_n> (1+frac1n)$$ ( The expression of third line.
sequences-and-series limits eulers-constant
edited yesterday
Rodrigo de Azevedo
13.2k41960
asked yesterday
Ieva BrakmaneIeva Brakmane
316
$endgroup$
I have a question relating to image that I've attached. It is a proof that the sequence is increasing. I don't understand the logic behind the third equation $$fraca_n+1a_n>left (1-frac1n+1right ) left (fracn+1nright)$$
where does the equation in the first and second parenthesis come from?
Ok, I have another relating question:
why $$fraca_n+1a_n> (1+frac1n)$$ ( The expression of third line.
sequences-and-series limits eulers-constant
sequences-and-series limits eulers-constant
edited yesterday
Rodrigo de Azevedo
13.2k41960
asked yesterday
Ieva BrakmaneIeva Brakmane
316
edited yesterday
Rodrigo de Azevedo
13.2k41960
asked yesterday
Ieva BrakmaneIeva Brakmane
316
edited yesterday
Rodrigo de Azevedo
13.2k41960
edited yesterday
Rodrigo de Azevedo
13.2k41960
edited yesterday
Rodrigo de Azevedo
13.2k41960
13.2k41960
asked yesterday
Ieva BrakmaneIeva Brakmane
316
asked yesterday
Ieva BrakmaneIeva Brakmane
316
asked yesterday
Ieva BrakmaneIeva Brakmane
316
316
$begingroup$
Please do not post necessary information only in a picture, not everyone can display and read it properly.
$endgroup$
– Carsten S
yesterday
add a comment |
$begingroup$
Please do not post necessary information only in a picture, not everyone can display and read it properly.
$endgroup$
– Carsten S
yesterday
$begingroup$
Please do not post necessary information only in a picture, not everyone can display and read it properly.
$endgroup$
– Carsten S
yesterday
$begingroup$
Please do not post necessary information only in a picture, not everyone can display and read it properly.
$endgroup$
– Carsten S
yesterday
add a comment |
3 Answers
3
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oldest
votes
$begingroup$
It is putting together the result from the first red box with the second one:
- $fraca_n+1a_n = colorblueleft(1- frac1(n+1)^2 right)^n+1left( fracn+1nright)$
- $colorblueleft(1- frac1(n+1)^2 right)^n+1 > colorgreen1 + (n+1)left( frac-1(n+1)^2right)$
$$Rightarrow fraca_n+1a_n > left(colorgreen1 + (n+1)left( frac-1(n+1)^2right)right)left( fracn+1nright) = left(underbrace1- frac1n+1_=fracnn+1right)left( fracn+1nright)$$
answered yesterday
trancelocationtrancelocation
13.3k1827
$endgroup$
add a comment |
$begingroup$
From Bernoulli's inequality, we have
$$left( 1- frac1(n+1)^2right) > 1+(n+1) left(frac-1(n+1)^2 right)=1-frac1n+1$$
Hence,
$$fraca_n+1a_n>left( 1- frac1(n+1)^2right)left( fracn+1nright)>left(1-frac1n+1 right)left( fracn+1nright)$$
answered yesterday
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
add a comment |
$begingroup$
So, we have
$$fraca_n+1a_n = left(1 - frac1(n+1)^2right)^n+1left(fracn+1nright).$$
The author then applies Bernoulli's inequality to the first term on the RHS:
$$left(1 - frac1(n+1)^2right)^n+1 > 1 + (n+1)left(frac-1(n+1)^2right) = 1 - frac1n+1.$$
We can now return to the first equation and utilize this estimate; namely, we have
$$fraca_n+1a_n = left(1 - frac1(n+1)^2right)^n+1left(fracn+1nright) > left(1-frac1n+1right)left(fracn+1nright).$$
Finally, we multiply out the RHS of the inequality
$$fraca_n+1a_n > left(1-frac1n+1right)left(fracn+1nright) = fracn+1n - frac1n = 1.$$
So, we have
$$fraca_n+1a_n > 1 implies a_n+1 > a_n,$$
which means that $a_n$ is an increasing sequence.
answered yesterday
Gary MoonGary Moon
89627
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is putting together the result from the first red box with the second one:
- $fraca_n+1a_n = colorblueleft(1- frac1(n+1)^2 right)^n+1left( fracn+1nright)$
- $colorblueleft(1- frac1(n+1)^2 right)^n+1 > colorgreen1 + (n+1)left( frac-1(n+1)^2right)$
$$Rightarrow fraca_n+1a_n > left(colorgreen1 + (n+1)left( frac-1(n+1)^2right)right)left( fracn+1nright) = left(underbrace1- frac1n+1_=fracnn+1right)left( fracn+1nright)$$
answered yesterday
trancelocationtrancelocation
13.3k1827
$endgroup$
add a comment |
$begingroup$
It is putting together the result from the first red box with the second one:
- $fraca_n+1a_n = colorblueleft(1- frac1(n+1)^2 right)^n+1left( fracn+1nright)$
- $colorblueleft(1- frac1(n+1)^2 right)^n+1 > colorgreen1 + (n+1)left( frac-1(n+1)^2right)$
$$Rightarrow fraca_n+1a_n > left(colorgreen1 + (n+1)left( frac-1(n+1)^2right)right)left( fracn+1nright) = left(underbrace1- frac1n+1_=fracnn+1right)left( fracn+1nright)$$
answered yesterday
trancelocationtrancelocation
13.3k1827
$endgroup$
add a comment |
$begingroup$
It is putting together the result from the first red box with the second one:
- $fraca_n+1a_n = colorblueleft(1- frac1(n+1)^2 right)^n+1left( fracn+1nright)$
- $colorblueleft(1- frac1(n+1)^2 right)^n+1 > colorgreen1 + (n+1)left( frac-1(n+1)^2right)$
$$Rightarrow fraca_n+1a_n > left(colorgreen1 + (n+1)left( frac-1(n+1)^2right)right)left( fracn+1nright) = left(underbrace1- frac1n+1_=fracnn+1right)left( fracn+1nright)$$
answered yesterday
trancelocationtrancelocation
13.3k1827
$endgroup$
It is putting together the result from the first red box with the second one:
- $fraca_n+1a_n = colorblueleft(1- frac1(n+1)^2 right)^n+1left( fracn+1nright)$
- $colorblueleft(1- frac1(n+1)^2 right)^n+1 > colorgreen1 + (n+1)left( frac-1(n+1)^2right)$
$$Rightarrow fraca_n+1a_n > left(colorgreen1 + (n+1)left( frac-1(n+1)^2right)right)left( fracn+1nright) = left(underbrace1- frac1n+1_=fracnn+1right)left( fracn+1nright)$$
answered yesterday
trancelocationtrancelocation
13.3k1827
answered yesterday
trancelocationtrancelocation
13.3k1827
answered yesterday
trancelocationtrancelocation
13.3k1827
answered yesterday
trancelocationtrancelocation
13.3k1827
13.3k1827
add a comment |
add a comment |
$begingroup$
From Bernoulli's inequality, we have
$$left( 1- frac1(n+1)^2right) > 1+(n+1) left(frac-1(n+1)^2 right)=1-frac1n+1$$
Hence,
$$fraca_n+1a_n>left( 1- frac1(n+1)^2right)left( fracn+1nright)>left(1-frac1n+1 right)left( fracn+1nright)$$
answered yesterday
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
add a comment |
$begingroup$
From Bernoulli's inequality, we have
$$left( 1- frac1(n+1)^2right) > 1+(n+1) left(frac-1(n+1)^2 right)=1-frac1n+1$$
Hence,
$$fraca_n+1a_n>left( 1- frac1(n+1)^2right)left( fracn+1nright)>left(1-frac1n+1 right)left( fracn+1nright)$$
answered yesterday
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
add a comment |
$begingroup$
From Bernoulli's inequality, we have
$$left( 1- frac1(n+1)^2right) > 1+(n+1) left(frac-1(n+1)^2 right)=1-frac1n+1$$
Hence,
$$fraca_n+1a_n>left( 1- frac1(n+1)^2right)left( fracn+1nright)>left(1-frac1n+1 right)left( fracn+1nright)$$
answered yesterday
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
From Bernoulli's inequality, we have
$$left( 1- frac1(n+1)^2right) > 1+(n+1) left(frac-1(n+1)^2 right)=1-frac1n+1$$
Hence,
$$fraca_n+1a_n>left( 1- frac1(n+1)^2right)left( fracn+1nright)>left(1-frac1n+1 right)left( fracn+1nright)$$
answered yesterday
Siong Thye GohSiong Thye Goh
103k1468119
answered yesterday
Siong Thye GohSiong Thye Goh
103k1468119
answered yesterday
Siong Thye GohSiong Thye Goh
103k1468119
answered yesterday
Siong Thye GohSiong Thye Goh
103k1468119
103k1468119
add a comment |
add a comment |
$begingroup$
So, we have
$$fraca_n+1a_n = left(1 - frac1(n+1)^2right)^n+1left(fracn+1nright).$$
The author then applies Bernoulli's inequality to the first term on the RHS:
$$left(1 - frac1(n+1)^2right)^n+1 > 1 + (n+1)left(frac-1(n+1)^2right) = 1 - frac1n+1.$$
We can now return to the first equation and utilize this estimate; namely, we have
$$fraca_n+1a_n = left(1 - frac1(n+1)^2right)^n+1left(fracn+1nright) > left(1-frac1n+1right)left(fracn+1nright).$$
Finally, we multiply out the RHS of the inequality
$$fraca_n+1a_n > left(1-frac1n+1right)left(fracn+1nright) = fracn+1n - frac1n = 1.$$
So, we have
$$fraca_n+1a_n > 1 implies a_n+1 > a_n,$$
which means that $a_n$ is an increasing sequence.
answered yesterday
Gary MoonGary Moon
89627
$endgroup$
add a comment |
$begingroup$
So, we have
$$fraca_n+1a_n = left(1 - frac1(n+1)^2right)^n+1left(fracn+1nright).$$
The author then applies Bernoulli's inequality to the first term on the RHS:
$$left(1 - frac1(n+1)^2right)^n+1 > 1 + (n+1)left(frac-1(n+1)^2right) = 1 - frac1n+1.$$
We can now return to the first equation and utilize this estimate; namely, we have
$$fraca_n+1a_n = left(1 - frac1(n+1)^2right)^n+1left(fracn+1nright) > left(1-frac1n+1right)left(fracn+1nright).$$
Finally, we multiply out the RHS of the inequality
$$fraca_n+1a_n > left(1-frac1n+1right)left(fracn+1nright) = fracn+1n - frac1n = 1.$$
So, we have
$$fraca_n+1a_n > 1 implies a_n+1 > a_n,$$
which means that $a_n$ is an increasing sequence.
answered yesterday
Gary MoonGary Moon
89627
$endgroup$
add a comment |
$begingroup$
So, we have
$$fraca_n+1a_n = left(1 - frac1(n+1)^2right)^n+1left(fracn+1nright).$$
The author then applies Bernoulli's inequality to the first term on the RHS:
$$left(1 - frac1(n+1)^2right)^n+1 > 1 + (n+1)left(frac-1(n+1)^2right) = 1 - frac1n+1.$$
We can now return to the first equation and utilize this estimate; namely, we have
$$fraca_n+1a_n = left(1 - frac1(n+1)^2right)^n+1left(fracn+1nright) > left(1-frac1n+1right)left(fracn+1nright).$$
Finally, we multiply out the RHS of the inequality
$$fraca_n+1a_n > left(1-frac1n+1right)left(fracn+1nright) = fracn+1n - frac1n = 1.$$
So, we have
$$fraca_n+1a_n > 1 implies a_n+1 > a_n,$$
which means that $a_n$ is an increasing sequence.
answered yesterday
Gary MoonGary Moon
89627
$endgroup$
So, we have
$$fraca_n+1a_n = left(1 - frac1(n+1)^2right)^n+1left(fracn+1nright).$$
The author then applies Bernoulli's inequality to the first term on the RHS:
$$left(1 - frac1(n+1)^2right)^n+1 > 1 + (n+1)left(frac-1(n+1)^2right) = 1 - frac1n+1.$$
We can now return to the first equation and utilize this estimate; namely, we have
$$fraca_n+1a_n = left(1 - frac1(n+1)^2right)^n+1left(fracn+1nright) > left(1-frac1n+1right)left(fracn+1nright).$$
Finally, we multiply out the RHS of the inequality
$$fraca_n+1a_n > left(1-frac1n+1right)left(fracn+1nright) = fracn+1n - frac1n = 1.$$
So, we have
$$fraca_n+1a_n > 1 implies a_n+1 > a_n,$$
which means that $a_n$ is an increasing sequence.
answered yesterday
Gary MoonGary Moon
89627
answered yesterday
Gary MoonGary Moon
89627
answered yesterday
Gary MoonGary Moon
89627
answered yesterday
Gary MoonGary Moon
89627
89627
add a comment |
add a comment |
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$begingroup$
Please do not post necessary information only in a picture, not everyone can display and read it properly.
$endgroup$
– Carsten S
yesterday