Prove that $(1/n^n)leq(1/n!)$ for every $ngeq1$












6












$begingroup$


I want to know if my answer is correct:



1) For n=1: $ 1=1$ Correct!

2) Let n=k is an inductive assumption which is correct:
$$frac{1}{k^k}leq frac{1}{k!}$$
3) For n=k+1, we should prove that:
$$frac{1}{(k+1)^{k+1}}leq frac{1}{(k+1)!}$$
So,
$$frac{1}{(k+1)!}=
frac{1}{(k+1)k!}geq
frac{1}{(k+1)k^k}geq
frac{1}{(k+1)(k+1)^k}=
frac{1}{(k+1)^{k+1}}$$



It's correct also for $n=k+1$, so the inequality $1/(k+1)^{k+1}leq1/(k+1)!$ is correct for every number $ngeq1$






discrete-mathematics proof-verification induction







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$endgroup$












  • $begingroup$
    Hello, and welcome to MSE. I don't see anything in particular wrong with your solution. Did you have any particular reason to think your answer may not be correct?
    $endgroup$
    – John Omielan
    Jan 14 at 3:01












  • $begingroup$
    It's okay isn't it?
    $endgroup$
    – Viktor
    Jan 14 at 3:02










  • $begingroup$
    If by "okay" you mean is there anything wrong with it, I don't see any mistakes.
    $endgroup$
    – John Omielan
    Jan 14 at 3:03






  • 1




    $begingroup$
    You are welcome for the confirmation. However, this is a fairly minor point, but as you're asking for confirmation of accuracy, well I just noticed that your last part should have $k + 1$ replaced by $n$, plus you may wish to make it clear you're dealing with natural numbers. Thus, you may wish to have it end with "so the inequality $frac{1}{n^n} leq frac{1}{n!}$ is correct for every natural number $n ge 1$".
    $endgroup$
    – John Omielan
    Jan 14 at 3:10








  • 2




    $begingroup$
    The inequality says $n! leq n^{n}$. Isn't this obvious since $n! =(1)(2)...(n) leq (n)(n)...(n)=n^{n}$?
    $endgroup$
    – Kavi Rama Murthy
    Jan 14 at 5:40
















6












$begingroup$


I want to know if my answer is correct:



1) For n=1: $ 1=1$ Correct!

2) Let n=k is an inductive assumption which is correct:
$$frac{1}{k^k}leq frac{1}{k!}$$
3) For n=k+1, we should prove that:
$$frac{1}{(k+1)^{k+1}}leq frac{1}{(k+1)!}$$
So,
$$frac{1}{(k+1)!}=
frac{1}{(k+1)k!}geq
frac{1}{(k+1)k^k}geq
frac{1}{(k+1)(k+1)^k}=
frac{1}{(k+1)^{k+1}}$$



It's correct also for $n=k+1$, so the inequality $1/(k+1)^{k+1}leq1/(k+1)!$ is correct for every number $ngeq1$






discrete-mathematics proof-verification induction







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hello, and welcome to MSE. I don't see anything in particular wrong with your solution. Did you have any particular reason to think your answer may not be correct?
    $endgroup$
    – John Omielan
    Jan 14 at 3:01












  • $begingroup$
    It's okay isn't it?
    $endgroup$
    – Viktor
    Jan 14 at 3:02










  • $begingroup$
    If by "okay" you mean is there anything wrong with it, I don't see any mistakes.
    $endgroup$
    – John Omielan
    Jan 14 at 3:03






  • 1




    $begingroup$
    You are welcome for the confirmation. However, this is a fairly minor point, but as you're asking for confirmation of accuracy, well I just noticed that your last part should have $k + 1$ replaced by $n$, plus you may wish to make it clear you're dealing with natural numbers. Thus, you may wish to have it end with "so the inequality $frac{1}{n^n} leq frac{1}{n!}$ is correct for every natural number $n ge 1$".
    $endgroup$
    – John Omielan
    Jan 14 at 3:10








  • 2




    $begingroup$
    The inequality says $n! leq n^{n}$. Isn't this obvious since $n! =(1)(2)...(n) leq (n)(n)...(n)=n^{n}$?
    $endgroup$
    – Kavi Rama Murthy
    Jan 14 at 5:40














6












6








6





$begingroup$


I want to know if my answer is correct:



1) For n=1: $ 1=1$ Correct!

2) Let n=k is an inductive assumption which is correct:
$$frac{1}{k^k}leq frac{1}{k!}$$
3) For n=k+1, we should prove that:
$$frac{1}{(k+1)^{k+1}}leq frac{1}{(k+1)!}$$
So,
$$frac{1}{(k+1)!}=
frac{1}{(k+1)k!}geq
frac{1}{(k+1)k^k}geq
frac{1}{(k+1)(k+1)^k}=
frac{1}{(k+1)^{k+1}}$$



It's correct also for $n=k+1$, so the inequality $1/(k+1)^{k+1}leq1/(k+1)!$ is correct for every number $ngeq1$






discrete-mathematics proof-verification induction







share|cite|improve this question











$endgroup$




I want to know if my answer is correct:



1) For n=1: $ 1=1$ Correct!

2) Let n=k is an inductive assumption which is correct:
$$frac{1}{k^k}leq frac{1}{k!}$$
3) For n=k+1, we should prove that:
$$frac{1}{(k+1)^{k+1}}leq frac{1}{(k+1)!}$$
So,
$$frac{1}{(k+1)!}=
frac{1}{(k+1)k!}geq
frac{1}{(k+1)k^k}geq
frac{1}{(k+1)(k+1)^k}=
frac{1}{(k+1)^{k+1}}$$



It's correct also for $n=k+1$, so the inequality $1/(k+1)^{k+1}leq1/(k+1)!$ is correct for every number $ngeq1$






discrete-mathematics proof-verification induction




discrete-mathematics proof-verification induction






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 14 at 10:01









rtybase

11.6k31534




11.6k31534










asked Jan 14 at 2:58









ViktorViktor

427




427












  • $begingroup$
    Hello, and welcome to MSE. I don't see anything in particular wrong with your solution. Did you have any particular reason to think your answer may not be correct?
    $endgroup$
    – John Omielan
    Jan 14 at 3:01












  • $begingroup$
    It's okay isn't it?
    $endgroup$
    – Viktor
    Jan 14 at 3:02










  • $begingroup$
    If by "okay" you mean is there anything wrong with it, I don't see any mistakes.
    $endgroup$
    – John Omielan
    Jan 14 at 3:03






  • 1




    $begingroup$
    You are welcome for the confirmation. However, this is a fairly minor point, but as you're asking for confirmation of accuracy, well I just noticed that your last part should have $k + 1$ replaced by $n$, plus you may wish to make it clear you're dealing with natural numbers. Thus, you may wish to have it end with "so the inequality $frac{1}{n^n} leq frac{1}{n!}$ is correct for every natural number $n ge 1$".
    $endgroup$
    – John Omielan
    Jan 14 at 3:10








  • 2




    $begingroup$
    The inequality says $n! leq n^{n}$. Isn't this obvious since $n! =(1)(2)...(n) leq (n)(n)...(n)=n^{n}$?
    $endgroup$
    – Kavi Rama Murthy
    Jan 14 at 5:40


















  • $begingroup$
    Hello, and welcome to MSE. I don't see anything in particular wrong with your solution. Did you have any particular reason to think your answer may not be correct?
    $endgroup$
    – John Omielan
    Jan 14 at 3:01












  • $begingroup$
    It's okay isn't it?
    $endgroup$
    – Viktor
    Jan 14 at 3:02










  • $begingroup$
    If by "okay" you mean is there anything wrong with it, I don't see any mistakes.
    $endgroup$
    – John Omielan
    Jan 14 at 3:03






  • 1




    $begingroup$
    You are welcome for the confirmation. However, this is a fairly minor point, but as you're asking for confirmation of accuracy, well I just noticed that your last part should have $k + 1$ replaced by $n$, plus you may wish to make it clear you're dealing with natural numbers. Thus, you may wish to have it end with "so the inequality $frac{1}{n^n} leq frac{1}{n!}$ is correct for every natural number $n ge 1$".
    $endgroup$
    – John Omielan
    Jan 14 at 3:10








  • 2




    $begingroup$
    The inequality says $n! leq n^{n}$. Isn't this obvious since $n! =(1)(2)...(n) leq (n)(n)...(n)=n^{n}$?
    $endgroup$
    – Kavi Rama Murthy
    Jan 14 at 5:40
















$begingroup$
Hello, and welcome to MSE. I don't see anything in particular wrong with your solution. Did you have any particular reason to think your answer may not be correct?
$endgroup$
– John Omielan
Jan 14 at 3:01






$begingroup$
Hello, and welcome to MSE. I don't see anything in particular wrong with your solution. Did you have any particular reason to think your answer may not be correct?
$endgroup$
– John Omielan
Jan 14 at 3:01














$begingroup$
It's okay isn't it?
$endgroup$
– Viktor
Jan 14 at 3:02




$begingroup$
It's okay isn't it?
$endgroup$
– Viktor
Jan 14 at 3:02












$begingroup$
If by "okay" you mean is there anything wrong with it, I don't see any mistakes.
$endgroup$
– John Omielan
Jan 14 at 3:03




$begingroup$
If by "okay" you mean is there anything wrong with it, I don't see any mistakes.
$endgroup$
– John Omielan
Jan 14 at 3:03




1




1




$begingroup$
You are welcome for the confirmation. However, this is a fairly minor point, but as you're asking for confirmation of accuracy, well I just noticed that your last part should have $k + 1$ replaced by $n$, plus you may wish to make it clear you're dealing with natural numbers. Thus, you may wish to have it end with "so the inequality $frac{1}{n^n} leq frac{1}{n!}$ is correct for every natural number $n ge 1$".
$endgroup$
– John Omielan
Jan 14 at 3:10






$begingroup$
You are welcome for the confirmation. However, this is a fairly minor point, but as you're asking for confirmation of accuracy, well I just noticed that your last part should have $k + 1$ replaced by $n$, plus you may wish to make it clear you're dealing with natural numbers. Thus, you may wish to have it end with "so the inequality $frac{1}{n^n} leq frac{1}{n!}$ is correct for every natural number $n ge 1$".
$endgroup$
– John Omielan
Jan 14 at 3:10






2




2




$begingroup$
The inequality says $n! leq n^{n}$. Isn't this obvious since $n! =(1)(2)...(n) leq (n)(n)...(n)=n^{n}$?
$endgroup$
– Kavi Rama Murthy
Jan 14 at 5:40




$begingroup$
The inequality says $n! leq n^{n}$. Isn't this obvious since $n! =(1)(2)...(n) leq (n)(n)...(n)=n^{n}$?
$endgroup$
– Kavi Rama Murthy
Jan 14 at 5:40










1 Answer
1






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oldest

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1












$begingroup$

Your answer is correct! I see nothing wrong with it as far as I know about mathematical induction. Keep up the good work!






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Please avoid posting just "You are correct"-like answers. They should be in the comments instead. To make this post into an answer, you could say that the OP is correct, then suggest an alternative solution to prove the inequality. For example, $n^n=ncdot ncdots n$ and $n!=ncdot(n-1)cdots1$, which means that...
    $endgroup$
    – TheSimpliFire
    Feb 13 at 9:29














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1 Answer
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1 Answer
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active

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active

oldest

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active

oldest

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1












$begingroup$

Your answer is correct! I see nothing wrong with it as far as I know about mathematical induction. Keep up the good work!






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Please avoid posting just "You are correct"-like answers. They should be in the comments instead. To make this post into an answer, you could say that the OP is correct, then suggest an alternative solution to prove the inequality. For example, $n^n=ncdot ncdots n$ and $n!=ncdot(n-1)cdots1$, which means that...
    $endgroup$
    – TheSimpliFire
    Feb 13 at 9:29


















1












$begingroup$

Your answer is correct! I see nothing wrong with it as far as I know about mathematical induction. Keep up the good work!






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Please avoid posting just "You are correct"-like answers. They should be in the comments instead. To make this post into an answer, you could say that the OP is correct, then suggest an alternative solution to prove the inequality. For example, $n^n=ncdot ncdots n$ and $n!=ncdot(n-1)cdots1$, which means that...
    $endgroup$
    – TheSimpliFire
    Feb 13 at 9:29
















1












1








1





$begingroup$

Your answer is correct! I see nothing wrong with it as far as I know about mathematical induction. Keep up the good work!






share|cite|improve this answer









$endgroup$



Your answer is correct! I see nothing wrong with it as far as I know about mathematical induction. Keep up the good work!







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 14 at 5:02









BadAtGeometryBadAtGeometry

254217




254217












  • $begingroup$
    Please avoid posting just "You are correct"-like answers. They should be in the comments instead. To make this post into an answer, you could say that the OP is correct, then suggest an alternative solution to prove the inequality. For example, $n^n=ncdot ncdots n$ and $n!=ncdot(n-1)cdots1$, which means that...
    $endgroup$
    – TheSimpliFire
    Feb 13 at 9:29




















  • $begingroup$
    Please avoid posting just "You are correct"-like answers. They should be in the comments instead. To make this post into an answer, you could say that the OP is correct, then suggest an alternative solution to prove the inequality. For example, $n^n=ncdot ncdots n$ and $n!=ncdot(n-1)cdots1$, which means that...
    $endgroup$
    – TheSimpliFire
    Feb 13 at 9:29


















$begingroup$
Please avoid posting just "You are correct"-like answers. They should be in the comments instead. To make this post into an answer, you could say that the OP is correct, then suggest an alternative solution to prove the inequality. For example, $n^n=ncdot ncdots n$ and $n!=ncdot(n-1)cdots1$, which means that...
$endgroup$
– TheSimpliFire
Feb 13 at 9:29






$begingroup$
Please avoid posting just "You are correct"-like answers. They should be in the comments instead. To make this post into an answer, you could say that the OP is correct, then suggest an alternative solution to prove the inequality. For example, $n^n=ncdot ncdots n$ and $n!=ncdot(n-1)cdots1$, which means that...
$endgroup$
– TheSimpliFire
Feb 13 at 9:29




















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