Prove that $2.71828182845904523536 < e < 2.71828182845904523537$ [on hold]How to analyze limit of function sequences?Does it make sense to talk about the concatenation of infinite series?Show that $sum_n frac1a_nlt90$Prove that there are infinitely many numbers between two real numbers. (example from Hardy's book)Convergence of $1 +frac15+frac19 +frac113+dots$Prove that the series $sumlimits_k=1^infty[ln(ak+b)- ln(ak)]$ divergesFind the Taylor series for $f(x) = sqrt x-2$ centered at $3$?Let $a_n$ be a sequence of reals going to $0$ but whose series diverges. Prove that we can flip the signs of each term to get a convergent seriesHow to prove that $e^k > k$?Finding integers $N$ and $M$ such that $|Nalpha-M-x|<epsilon$.
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Prove that $2.71828182845904523536
How to analyze limit of function sequences?Does it make sense to talk about the concatenation of infinite series?Show that $sum_n frac1a_nlt90$Prove that there are infinitely many numbers between two real numbers. (example from Hardy's book)Convergence of $1 +frac15+frac19 +frac113+dots$Prove that the series $sumlimits_k=1^infty[ln(ak+b)- ln(ak)]$ divergesFind the Taylor series for $f(x) = sqrt x-2$ centered at $3$?Let $a_n$ be a sequence of reals going to $0$ but whose series diverges. Prove that we can flip the signs of each term to get a convergent seriesHow to prove that $e^k > k$?Finding integers $N$ and $M$ such that $|Nalpha-M-x|<epsilon$.
$begingroup$
For smaller rational numbers or integers, I can just write the first few elements of the series $1/n!$ but am stuck at this.
real-analysis calculus sequences-and-series
edited yesterday
Rócherz
2,8912821
asked yesterday
Tuna BayerTuna Bayer
42
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Tuna Bayer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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put on hold as off-topic by Travis, John B, YiFan, Eevee Trainer, Saad yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Travis, John B, Eevee Trainer, Saad
add a comment |
$begingroup$
For smaller rational numbers or integers, I can just write the first few elements of the series $1/n!$ but am stuck at this.
real-analysis calculus sequences-and-series
edited yesterday
Rócherz
2,8912821
asked yesterday
Tuna BayerTuna Bayer
42
New contributor
Tuna Bayer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
put on hold as off-topic by Travis, John B, YiFan, Eevee Trainer, Saad yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Travis, John B, Eevee Trainer, Saad
5
$begingroup$
The series $1 / n!$ converges very quickly, so you can use the same type of argument as for coarser approximations. You just need a few more terms.
$endgroup$
– Travis
yesterday
add a comment |
$begingroup$
For smaller rational numbers or integers, I can just write the first few elements of the series $1/n!$ but am stuck at this.
real-analysis calculus sequences-and-series
edited yesterday
Rócherz
2,8912821
asked yesterday
Tuna BayerTuna Bayer
42
New contributor
Tuna Bayer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
For smaller rational numbers or integers, I can just write the first few elements of the series $1/n!$ but am stuck at this.
real-analysis calculus sequences-and-series
real-analysis calculus sequences-and-series
edited yesterday
Rócherz
2,8912821
asked yesterday
Tuna BayerTuna Bayer
42
New contributor
Tuna Bayer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Rócherz
2,8912821
asked yesterday
Tuna BayerTuna Bayer
42
New contributor
Tuna Bayer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Rócherz
2,8912821
edited yesterday
Rócherz
2,8912821
edited yesterday
Rócherz
2,8912821
2,8912821
asked yesterday
Tuna BayerTuna Bayer
42
New contributor
Tuna Bayer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Tuna BayerTuna Bayer
42
asked yesterday
Tuna BayerTuna Bayer
42
42
New contributor
Tuna Bayer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Tuna Bayer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Tuna Bayer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Travis, John B, YiFan, Eevee Trainer, Saad yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Travis, John B, Eevee Trainer, Saad
put on hold as off-topic by Travis, John B, YiFan, Eevee Trainer, Saad yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Travis, John B, Eevee Trainer, Saad
5
$begingroup$
The series $1 / n!$ converges very quickly, so you can use the same type of argument as for coarser approximations. You just need a few more terms.
$endgroup$
– Travis
yesterday
add a comment |
5
$begingroup$
The series $1 / n!$ converges very quickly, so you can use the same type of argument as for coarser approximations. You just need a few more terms.
$endgroup$
– Travis
yesterday
5
5
$begingroup$
The series $1 / n!$ converges very quickly, so you can use the same type of argument as for coarser approximations. You just need a few more terms.
$endgroup$
– Travis
yesterday
$begingroup$
The series $1 / n!$ converges very quickly, so you can use the same type of argument as for coarser approximations. You just need a few more terms.
$endgroup$
– Travis
yesterday
add a comment |
2 Answers
2
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oldest
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$begingroup$
Hint: We have $$eapprox left( 1+frac110^42right)^10^42$$ by the standard limit definition of $e$. (Perhaps a better choice of natural number than $10^42$ would be more feasible though.)
answered yesterday
ShaunShaun
9,334113684
$endgroup$
add a comment |
$begingroup$
You could use the continued fraction for $e$. The successive convergents alternate between being less than and greater than $e$. So grind out $25$ or so convergents and you know that $e$ lies between any two successive convergents. As soon as the convergents are as accurate as you want, you'll be done.
answered yesterday
B. GoddardB. Goddard
19.5k21442
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: We have $$eapprox left( 1+frac110^42right)^10^42$$ by the standard limit definition of $e$. (Perhaps a better choice of natural number than $10^42$ would be more feasible though.)
answered yesterday
ShaunShaun
9,334113684
$endgroup$
add a comment |
$begingroup$
Hint: We have $$eapprox left( 1+frac110^42right)^10^42$$ by the standard limit definition of $e$. (Perhaps a better choice of natural number than $10^42$ would be more feasible though.)
answered yesterday
ShaunShaun
9,334113684
$endgroup$
add a comment |
$begingroup$
Hint: We have $$eapprox left( 1+frac110^42right)^10^42$$ by the standard limit definition of $e$. (Perhaps a better choice of natural number than $10^42$ would be more feasible though.)
answered yesterday
ShaunShaun
9,334113684
$endgroup$
Hint: We have $$eapprox left( 1+frac110^42right)^10^42$$ by the standard limit definition of $e$. (Perhaps a better choice of natural number than $10^42$ would be more feasible though.)
answered yesterday
ShaunShaun
9,334113684
answered yesterday
ShaunShaun
9,334113684
answered yesterday
ShaunShaun
9,334113684
answered yesterday
ShaunShaun
9,334113684
9,334113684
add a comment |
add a comment |
$begingroup$
You could use the continued fraction for $e$. The successive convergents alternate between being less than and greater than $e$. So grind out $25$ or so convergents and you know that $e$ lies between any two successive convergents. As soon as the convergents are as accurate as you want, you'll be done.
answered yesterday
B. GoddardB. Goddard
19.5k21442
$endgroup$
add a comment |
$begingroup$
You could use the continued fraction for $e$. The successive convergents alternate between being less than and greater than $e$. So grind out $25$ or so convergents and you know that $e$ lies between any two successive convergents. As soon as the convergents are as accurate as you want, you'll be done.
answered yesterday
B. GoddardB. Goddard
19.5k21442
$endgroup$
add a comment |
$begingroup$
You could use the continued fraction for $e$. The successive convergents alternate between being less than and greater than $e$. So grind out $25$ or so convergents and you know that $e$ lies between any two successive convergents. As soon as the convergents are as accurate as you want, you'll be done.
answered yesterday
B. GoddardB. Goddard
19.5k21442
$endgroup$
You could use the continued fraction for $e$. The successive convergents alternate between being less than and greater than $e$. So grind out $25$ or so convergents and you know that $e$ lies between any two successive convergents. As soon as the convergents are as accurate as you want, you'll be done.
answered yesterday
B. GoddardB. Goddard
19.5k21442
answered yesterday
B. GoddardB. Goddard
19.5k21442
answered yesterday
B. GoddardB. Goddard
19.5k21442
answered yesterday
B. GoddardB. Goddard
19.5k21442
19.5k21442
add a comment |
add a comment |
5
$begingroup$
The series $1 / n!$ converges very quickly, so you can use the same type of argument as for coarser approximations. You just need a few more terms.
$endgroup$
– Travis
yesterday