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Can someone shed some light on this inequality?
Show 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.
sequences-and-series limits eulers-constant
$endgroup$
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
$endgroup$
$begingroup$
Please do not post necessary information only in a picture, not everyone can display and read it properly.
$endgroup$
– Carsten S
Mar 28 at 16:04
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
$endgroup$
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 Mar 28 at 16:34
Rodrigo de Azevedo
13.1k41962
13.1k41962
asked Mar 28 at 11:44
Ieva BrakmaneIeva Brakmane
556
556
$begingroup$
Please do not post necessary information only in a picture, not everyone can display and read it properly.
$endgroup$
– Carsten S
Mar 28 at 16:04
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
Mar 28 at 16:04
$begingroup$
Please do not post necessary information only in a picture, not everyone can display and read it properly.
$endgroup$
– Carsten S
Mar 28 at 16:04
$begingroup$
Please do not post necessary information only in a picture, not everyone can display and read it properly.
$endgroup$
– Carsten S
Mar 28 at 16:04
add a comment |
3 Answers
3
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oldest
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$begingroup$
It is putting together the result from the first red box with the second one:
- $fraca_n+1a_n = colorblueleft(1- frac1(n+1)^2 right)^n+1left( fracn+1nright)$
- $colorblueleft(1- frac1(n+1)^2 right)^n+1 > colorgreen1 + (n+1)left( frac-1(n+1)^2right)$
$$Rightarrow fraca_n+1a_n > left(colorgreen1 + (n+1)left( frac-1(n+1)^2right)right)left( fracn+1nright) = left(underbrace1- frac1n+1_=fracnn+1right)left( fracn+1nright)$$
$endgroup$
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,
$$fraca_n+1a_n>left( 1- frac1(n+1)^2right)left( fracn+1nright)>left(1-frac1n+1 right)left( fracn+1nright)$$
$endgroup$
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.
$endgroup$
add a comment |
Your Answer
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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:
- $fraca_n+1a_n = colorblueleft(1- frac1(n+1)^2 right)^n+1left( fracn+1nright)$
- $colorblueleft(1- frac1(n+1)^2 right)^n+1 > colorgreen1 + (n+1)left( frac-1(n+1)^2right)$
$$Rightarrow fraca_n+1a_n > left(colorgreen1 + (n+1)left( frac-1(n+1)^2right)right)left( fracn+1nright) = left(underbrace1- frac1n+1_=fracnn+1right)left( fracn+1nright)$$
$endgroup$
add a comment |
$begingroup$
It is putting together the result from the first red box with the second one:
- $fraca_n+1a_n = colorblueleft(1- frac1(n+1)^2 right)^n+1left( fracn+1nright)$
- $colorblueleft(1- frac1(n+1)^2 right)^n+1 > colorgreen1 + (n+1)left( frac-1(n+1)^2right)$
$$Rightarrow fraca_n+1a_n > left(colorgreen1 + (n+1)left( frac-1(n+1)^2right)right)left( fracn+1nright) = left(underbrace1- frac1n+1_=fracnn+1right)left( fracn+1nright)$$
$endgroup$
add a comment |
$begingroup$
It is putting together the result from the first red box with the second one:
- $fraca_n+1a_n = colorblueleft(1- frac1(n+1)^2 right)^n+1left( fracn+1nright)$
- $colorblueleft(1- frac1(n+1)^2 right)^n+1 > colorgreen1 + (n+1)left( frac-1(n+1)^2right)$
$$Rightarrow fraca_n+1a_n > left(colorgreen1 + (n+1)left( frac-1(n+1)^2right)right)left( fracn+1nright) = left(underbrace1- frac1n+1_=fracnn+1right)left( fracn+1nright)$$
$endgroup$
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 Mar 28 at 11:57
trancelocationtrancelocation
13.7k1829
13.7k1829
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,
$$fraca_n+1a_n>left( 1- frac1(n+1)^2right)left( fracn+1nright)>left(1-frac1n+1 right)left( fracn+1nright)$$
$endgroup$
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,
$$fraca_n+1a_n>left( 1- frac1(n+1)^2right)left( fracn+1nright)>left(1-frac1n+1 right)left( fracn+1nright)$$
$endgroup$
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,
$$fraca_n+1a_n>left( 1- frac1(n+1)^2right)left( fracn+1nright)>left(1-frac1n+1 right)left( fracn+1nright)$$
$endgroup$
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 Mar 28 at 11:52
Siong Thye GohSiong Thye Goh
104k1468120
104k1468120
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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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 Mar 28 at 12:10
Gary MoonGary Moon
921127
921127
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
Mar 28 at 16:04