Mixing
Use formal definition of a limit of a sequence to prove the following:
$begingroup$
a) lim $(n+1)/(n+2)=1$ as n approaches infinity
b) lim $(n^{3}-1)/(n^{4}-1)=0$ as n approaches infinity
c) lim $n!/n^{n}=0$ as n approaches inifinity
I tried proving those but I'm not sure if i did it right so would appreciate if anyone could point out if i missed any steps or if I can improve them. Thank you
a) |[(n+1)/(n+2)]-1| = |[(n+1)-(n+2)]/(n+2)| = 3/(n+2)
n+2>n
1/(n+2)<1/n
3/(n+2)<3/n
So for any epsilon>0, 3/epsilon>0
By Archimedean property, there exists N in natural numbers such that N>3/epsilon
When n≥N, n≥N>3/epsilon
which implies that 1/n≤1/N<1/(3/epsilon) implying 3/n≤3/N<3/(3/epsilon)
So, $|x_{n}-L|=3/(n+2)$<3/n≤3/N<3/(3/epsilon)=epsilon
Therefore, $|x_{n}-L|$=3/(n+2)
implying that $|x_{n}-L|$
Then by definition of limit, $x_{n}$=(n+1)/(n+2) converges to real number and its limit is 0.
b)for any n≤N in natural numbers, $n^{3}-1<n^{3}$ for any n>2, we have 0<$n^{4}-n^{3}$<$n^{4}-1$
so 1/($n^{4}-1$)<1/($n^{4}-n^{3}$). Hence, for n>2, 0<($n^{3}-1$)/($n^{4}-1$)<$n^{3}$/($n^{4}-1)$<$n^{3}$/($n^{4}-n^{3}$)=1/(n-1)
So given any epsilon>0, it'll be sufficient to choose N such that 1/(N-1)((1/epsilon)+1)
Given epsilon>0, choose N=(1/epsilon)+1. Then N>(1/epsilon)+1, hence 1/(N-1)2, given any n≥N, we have 0<$(n^{3}-1)/(n^{4}-1)$<$n^{3}/(n^{4}-1)$<$n^{3}/(n^{4}-n^{3})$=1/(n-1)<1/(N-1)
Therefore, the stated limit is indeed 0.
c) |n!/$n^{n}$|<1/n for all n in Natural numbers excluding 0
for any epsilon>0, 1/epsilon>0, there exists N in Natural numbers such that N>1/epsilon, when n≥N, n≥N>1/epsilon
implying that 1/n≤1/N<1/(1/epsilon), therefore, 1/n≤1/N
So, |$x_{n}-L$|=n!/$n^{n}$<1/n≤1/N
Therefore, |$x_{n}-L$|
By definition of a limit, $x_{n}=n!/n^{n}$ converges to real number and its limit is 0.
sequences-and-series limits proof-verification proof-writing proof-explanation
asked Jan 26 at 19:46
JennyJenny
625
$endgroup$
add a comment |
$begingroup$
a) lim $(n+1)/(n+2)=1$ as n approaches infinity
b) lim $(n^{3}-1)/(n^{4}-1)=0$ as n approaches infinity
c) lim $n!/n^{n}=0$ as n approaches inifinity
I tried proving those but I'm not sure if i did it right so would appreciate if anyone could point out if i missed any steps or if I can improve them. Thank you
a) |[(n+1)/(n+2)]-1| = |[(n+1)-(n+2)]/(n+2)| = 3/(n+2)
n+2>n
1/(n+2)<1/n
3/(n+2)<3/n
So for any epsilon>0, 3/epsilon>0
By Archimedean property, there exists N in natural numbers such that N>3/epsilon
When n≥N, n≥N>3/epsilon
which implies that 1/n≤1/N<1/(3/epsilon) implying 3/n≤3/N<3/(3/epsilon)
So, $|x_{n}-L|=3/(n+2)$<3/n≤3/N<3/(3/epsilon)=epsilon
Therefore, $|x_{n}-L|$=3/(n+2)
implying that $|x_{n}-L|$
Then by definition of limit, $x_{n}$=(n+1)/(n+2) converges to real number and its limit is 0.
b)for any n≤N in natural numbers, $n^{3}-1<n^{3}$ for any n>2, we have 0<$n^{4}-n^{3}$<$n^{4}-1$
so 1/($n^{4}-1$)<1/($n^{4}-n^{3}$). Hence, for n>2, 0<($n^{3}-1$)/($n^{4}-1$)<$n^{3}$/($n^{4}-1)$<$n^{3}$/($n^{4}-n^{3}$)=1/(n-1)
So given any epsilon>0, it'll be sufficient to choose N such that 1/(N-1)((1/epsilon)+1)
Given epsilon>0, choose N=(1/epsilon)+1. Then N>(1/epsilon)+1, hence 1/(N-1)2, given any n≥N, we have 0<$(n^{3}-1)/(n^{4}-1)$<$n^{3}/(n^{4}-1)$<$n^{3}/(n^{4}-n^{3})$=1/(n-1)<1/(N-1)
Therefore, the stated limit is indeed 0.
c) |n!/$n^{n}$|<1/n for all n in Natural numbers excluding 0
for any epsilon>0, 1/epsilon>0, there exists N in Natural numbers such that N>1/epsilon, when n≥N, n≥N>1/epsilon
implying that 1/n≤1/N<1/(1/epsilon), therefore, 1/n≤1/N
So, |$x_{n}-L$|=n!/$n^{n}$<1/n≤1/N
Therefore, |$x_{n}-L$|
By definition of a limit, $x_{n}=n!/n^{n}$ converges to real number and its limit is 0.
sequences-and-series limits proof-verification proof-writing proof-explanation
asked Jan 26 at 19:46
JennyJenny
625
$endgroup$
$begingroup$
Please use mathjax to typeset any math content.
$endgroup$
– Viktor Glombik
Jan 26 at 19:47
add a comment |
$begingroup$
a) lim $(n+1)/(n+2)=1$ as n approaches infinity
b) lim $(n^{3}-1)/(n^{4}-1)=0$ as n approaches infinity
c) lim $n!/n^{n}=0$ as n approaches inifinity
I tried proving those but I'm not sure if i did it right so would appreciate if anyone could point out if i missed any steps or if I can improve them. Thank you
a) |[(n+1)/(n+2)]-1| = |[(n+1)-(n+2)]/(n+2)| = 3/(n+2)
n+2>n
1/(n+2)<1/n
3/(n+2)<3/n
So for any epsilon>0, 3/epsilon>0
By Archimedean property, there exists N in natural numbers such that N>3/epsilon
When n≥N, n≥N>3/epsilon
which implies that 1/n≤1/N<1/(3/epsilon) implying 3/n≤3/N<3/(3/epsilon)
So, $|x_{n}-L|=3/(n+2)$<3/n≤3/N<3/(3/epsilon)=epsilon
Therefore, $|x_{n}-L|$=3/(n+2)
implying that $|x_{n}-L|$
Then by definition of limit, $x_{n}$=(n+1)/(n+2) converges to real number and its limit is 0.
b)for any n≤N in natural numbers, $n^{3}-1<n^{3}$ for any n>2, we have 0<$n^{4}-n^{3}$<$n^{4}-1$
so 1/($n^{4}-1$)<1/($n^{4}-n^{3}$). Hence, for n>2, 0<($n^{3}-1$)/($n^{4}-1$)<$n^{3}$/($n^{4}-1)$<$n^{3}$/($n^{4}-n^{3}$)=1/(n-1)
So given any epsilon>0, it'll be sufficient to choose N such that 1/(N-1)((1/epsilon)+1)
Given epsilon>0, choose N=(1/epsilon)+1. Then N>(1/epsilon)+1, hence 1/(N-1)2, given any n≥N, we have 0<$(n^{3}-1)/(n^{4}-1)$<$n^{3}/(n^{4}-1)$<$n^{3}/(n^{4}-n^{3})$=1/(n-1)<1/(N-1)
Therefore, the stated limit is indeed 0.
c) |n!/$n^{n}$|<1/n for all n in Natural numbers excluding 0
for any epsilon>0, 1/epsilon>0, there exists N in Natural numbers such that N>1/epsilon, when n≥N, n≥N>1/epsilon
implying that 1/n≤1/N<1/(1/epsilon), therefore, 1/n≤1/N
So, |$x_{n}-L$|=n!/$n^{n}$<1/n≤1/N
Therefore, |$x_{n}-L$|
By definition of a limit, $x_{n}=n!/n^{n}$ converges to real number and its limit is 0.
sequences-and-series limits proof-verification proof-writing proof-explanation
asked Jan 26 at 19:46
JennyJenny
625
$endgroup$
a) lim $(n+1)/(n+2)=1$ as n approaches infinity
b) lim $(n^{3}-1)/(n^{4}-1)=0$ as n approaches infinity
c) lim $n!/n^{n}=0$ as n approaches inifinity
I tried proving those but I'm not sure if i did it right so would appreciate if anyone could point out if i missed any steps or if I can improve them. Thank you
a) |[(n+1)/(n+2)]-1| = |[(n+1)-(n+2)]/(n+2)| = 3/(n+2)
n+2>n
1/(n+2)<1/n
3/(n+2)<3/n
So for any epsilon>0, 3/epsilon>0
By Archimedean property, there exists N in natural numbers such that N>3/epsilon
When n≥N, n≥N>3/epsilon
which implies that 1/n≤1/N<1/(3/epsilon) implying 3/n≤3/N<3/(3/epsilon)
So, $|x_{n}-L|=3/(n+2)$<3/n≤3/N<3/(3/epsilon)=epsilon
Therefore, $|x_{n}-L|$=3/(n+2)
implying that $|x_{n}-L|$
Then by definition of limit, $x_{n}$=(n+1)/(n+2) converges to real number and its limit is 0.
b)for any n≤N in natural numbers, $n^{3}-1<n^{3}$ for any n>2, we have 0<$n^{4}-n^{3}$<$n^{4}-1$
so 1/($n^{4}-1$)<1/($n^{4}-n^{3}$). Hence, for n>2, 0<($n^{3}-1$)/($n^{4}-1$)<$n^{3}$/($n^{4}-1)$<$n^{3}$/($n^{4}-n^{3}$)=1/(n-1)
So given any epsilon>0, it'll be sufficient to choose N such that 1/(N-1)((1/epsilon)+1)
Given epsilon>0, choose N=(1/epsilon)+1. Then N>(1/epsilon)+1, hence 1/(N-1)2, given any n≥N, we have 0<$(n^{3}-1)/(n^{4}-1)$<$n^{3}/(n^{4}-1)$<$n^{3}/(n^{4}-n^{3})$=1/(n-1)<1/(N-1)
Therefore, the stated limit is indeed 0.
c) |n!/$n^{n}$|<1/n for all n in Natural numbers excluding 0
for any epsilon>0, 1/epsilon>0, there exists N in Natural numbers such that N>1/epsilon, when n≥N, n≥N>1/epsilon
implying that 1/n≤1/N<1/(1/epsilon), therefore, 1/n≤1/N
So, |$x_{n}-L$|=n!/$n^{n}$<1/n≤1/N
Therefore, |$x_{n}-L$|
By definition of a limit, $x_{n}=n!/n^{n}$ converges to real number and its limit is 0.
sequences-and-series limits proof-verification proof-writing proof-explanation
sequences-and-series limits proof-verification proof-writing proof-explanation
asked Jan 26 at 19:46
JennyJenny
625
asked Jan 26 at 19:46
JennyJenny
625
asked Jan 26 at 19:46
JennyJenny
625
asked Jan 26 at 19:46
JennyJenny
625
asked Jan 26 at 19:46
JennyJenny
625
625
$begingroup$
Please use mathjax to typeset any math content.
$endgroup$
– Viktor Glombik
Jan 26 at 19:47
add a comment |
$begingroup$
Please use mathjax to typeset any math content.
$endgroup$
– Viktor Glombik
Jan 26 at 19:47
$begingroup$
Please use mathjax to typeset any math content.
$endgroup$
– Viktor Glombik
Jan 26 at 19:47
$begingroup$
Please use mathjax to typeset any math content.
$endgroup$
– Viktor Glombik
Jan 26 at 19:47
add a comment |
1 Answer
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Yes, they seem decent to me.
good job, (Even through it was a bit of a pain to read this as you didn't typeset any maths)
answered Jan 26 at 20:22
AlexandrosAlexandros
9991412
$endgroup$
$begingroup$
Lol I know, sorry! I haven't quite figured out how to use mathjax yet...
$endgroup$
– Jenny
Jan 26 at 21:50
add a comment |
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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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$begingroup$
Yes, they seem decent to me.
good job, (Even through it was a bit of a pain to read this as you didn't typeset any maths)
answered Jan 26 at 20:22
AlexandrosAlexandros
9991412
$endgroup$
$begingroup$
Lol I know, sorry! I haven't quite figured out how to use mathjax yet...
$endgroup$
– Jenny
Jan 26 at 21:50
add a comment |
$begingroup$
Yes, they seem decent to me.
good job, (Even through it was a bit of a pain to read this as you didn't typeset any maths)
answered Jan 26 at 20:22
AlexandrosAlexandros
9991412
$endgroup$
$begingroup$
Lol I know, sorry! I haven't quite figured out how to use mathjax yet...
$endgroup$
– Jenny
Jan 26 at 21:50
add a comment |
$begingroup$
Yes, they seem decent to me.
good job, (Even through it was a bit of a pain to read this as you didn't typeset any maths)
answered Jan 26 at 20:22
AlexandrosAlexandros
9991412
$endgroup$
Yes, they seem decent to me.
good job, (Even through it was a bit of a pain to read this as you didn't typeset any maths)
answered Jan 26 at 20:22
AlexandrosAlexandros
9991412
answered Jan 26 at 20:22
AlexandrosAlexandros
9991412
answered Jan 26 at 20:22
AlexandrosAlexandros
9991412
answered Jan 26 at 20:22
AlexandrosAlexandros
9991412
9991412
$begingroup$
Lol I know, sorry! I haven't quite figured out how to use mathjax yet...
$endgroup$
– Jenny
Jan 26 at 21:50
add a comment |
$begingroup$
Lol I know, sorry! I haven't quite figured out how to use mathjax yet...
$endgroup$
– Jenny
Jan 26 at 21:50
$begingroup$
Lol I know, sorry! I haven't quite figured out how to use mathjax yet...
$endgroup$
– Jenny
Jan 26 at 21:50
$begingroup$
Lol I know, sorry! I haven't quite figured out how to use mathjax yet...
$endgroup$
– Jenny
Jan 26 at 21:50
add a comment |
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$begingroup$
Please use mathjax to typeset any math content.
$endgroup$
– Viktor Glombik
Jan 26 at 19:47