Mixing
Calculate the limit (Squeeze Theorem?)
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I have to calculate the limit of this formula as $nto infty$.
$$a_n = frac1sqrtnbigl(frac1sqrtn+1+cdots+frac1sqrt2nbigl)$$
I tried the Squeeze Theorem, but I get something like this:
$$frac1sqrt2leftarrowfracnsqrt2n^2lefrac1sqrtnbigl(frac1sqrtn+1+cdots+frac1sqrt2nbigl) le fracnsqrtn^2+nto1$$
As you can see, the limits of two other sequences aren't the same. Can you give me some hints? Thank you in advance.
sequences-and-series
asked 2 hours ago
iforgotmypass
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up vote
1
down vote
favorite
I have to calculate the limit of this formula as $nto infty$.
$$a_n = frac1sqrtnbigl(frac1sqrtn+1+cdots+frac1sqrt2nbigl)$$
I tried the Squeeze Theorem, but I get something like this:
$$frac1sqrt2leftarrowfracnsqrt2n^2lefrac1sqrtnbigl(frac1sqrtn+1+cdots+frac1sqrt2nbigl) le fracnsqrtn^2+nto1$$
As you can see, the limits of two other sequences aren't the same. Can you give me some hints? Thank you in advance.
sequences-and-series
asked 2 hours ago
iforgotmypass
283
New contributor
iforgotmypass is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have to calculate the limit of this formula as $nto infty$.
$$a_n = frac1sqrtnbigl(frac1sqrtn+1+cdots+frac1sqrt2nbigl)$$
I tried the Squeeze Theorem, but I get something like this:
$$frac1sqrt2leftarrowfracnsqrt2n^2lefrac1sqrtnbigl(frac1sqrtn+1+cdots+frac1sqrt2nbigl) le fracnsqrtn^2+nto1$$
As you can see, the limits of two other sequences aren't the same. Can you give me some hints? Thank you in advance.
sequences-and-series
asked 2 hours ago
iforgotmypass
283
New contributor
iforgotmypass is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have to calculate the limit of this formula as $nto infty$.
$$a_n = frac1sqrtnbigl(frac1sqrtn+1+cdots+frac1sqrt2nbigl)$$
I tried the Squeeze Theorem, but I get something like this:
$$frac1sqrt2leftarrowfracnsqrt2n^2lefrac1sqrtnbigl(frac1sqrtn+1+cdots+frac1sqrt2nbigl) le fracnsqrtn^2+nto1$$
As you can see, the limits of two other sequences aren't the same. Can you give me some hints? Thank you in advance.
sequences-and-series
sequences-and-series
asked 2 hours ago
iforgotmypass
283
New contributor
iforgotmypass is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
iforgotmypass
283
New contributor
iforgotmypass is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
iforgotmypass
283
New contributor
iforgotmypass is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
iforgotmypass
283
asked 2 hours ago
iforgotmypass
283
283
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iforgotmypass is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
iforgotmypass is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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3 Answers
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Rearrange it as
$$ frac1nleft(sqrtfracnn+1+sqrtfracnn+2+ldots+sqrtfracnn+1right) = frac1nsum_k=1^nfrac1sqrt1+frackn$$
which is a Riemann sum for
$$ int_0^1fracdxsqrt1+x=2sqrt2-2.$$
Since $frac1sqrt1+x$ is a convex function on $[0,1]$, the Hermite-Hadamard and Karamata's inequalities give us that $a_n_ngeq 1$ is an increasing sequence convergent to $2sqrt2-2$. Additionally it is not difficult to check that $a_n= 2sqrt2-2-Thetaleft(frac1nright)$ as $nto +infty$.
answered 2 hours ago
Jack D'Aurizio
280k33272651
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up vote
2
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As an alternative by Stolz-Cesaro
$$fracb_nc_n = fracfrac1sqrtn+1+cdots+frac1sqrt2nsqrt n$$
$$fracb_n+1-b_nc_n+1-c_n = fracfrac1sqrt2n+2+frac1sqrt2n+1-frac1sqrtn+1sqrtn+1-sqrt n$$
and
$$fracfrac1sqrt2n+2+frac1sqrt2n+1-frac1sqrtn+1sqrtn+1-sqrt nfracsqrtn+1+sqrt nsqrtn+1+sqrt n=$$
$$fracsqrtn+1+sqrt nsqrt2n+2+fracsqrtn+1+sqrt nsqrt2n+1-fracsqrtn+1+sqrt nsqrtn+1tofrac4sqrt 2-2=2sqrt 2-2$$
answered 1 hour ago
gimusi
81.3k74090
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up vote
0
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for a decreasing function such as $1/sqrt x$ with $x$ positive, a simple picture shows
$$ int_a^b+1 ; f(x) ; dx < sum_k=a^b f(k) < int_a-1^b ; f(x) ; dx $$
$$ int_n+1^2n+1 ; frac1sqrt x ; dx < sum_k=n+1^2n frac1sqrt k < int_n^2n ; frac1sqrt x ; dx $$
getting there
$$ 2 sqrt 2n+1 - 2 sqrt n+1 < sum_k=n+1^2n frac1sqrt k < 2 sqrt 2n - 2 sqrt n $$
$$ 2 sqrt 2+frac1n - 2 sqrt 1+frac1n < frac1sqrt n sum_k=n+1^2n frac1sqrt k < 2 sqrt 2 - 2 sqrt 1 $$
answered 50 mins ago
Will Jagy
99.5k597198
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Rearrange it as
$$ frac1nleft(sqrtfracnn+1+sqrtfracnn+2+ldots+sqrtfracnn+1right) = frac1nsum_k=1^nfrac1sqrt1+frackn$$
which is a Riemann sum for
$$ int_0^1fracdxsqrt1+x=2sqrt2-2.$$
Since $frac1sqrt1+x$ is a convex function on $[0,1]$, the Hermite-Hadamard and Karamata's inequalities give us that $a_n_ngeq 1$ is an increasing sequence convergent to $2sqrt2-2$. Additionally it is not difficult to check that $a_n= 2sqrt2-2-Thetaleft(frac1nright)$ as $nto +infty$.
answered 2 hours ago
Jack D'Aurizio
280k33272651
add a comment |Â
up vote
3
down vote
Rearrange it as
$$ frac1nleft(sqrtfracnn+1+sqrtfracnn+2+ldots+sqrtfracnn+1right) = frac1nsum_k=1^nfrac1sqrt1+frackn$$
which is a Riemann sum for
$$ int_0^1fracdxsqrt1+x=2sqrt2-2.$$
Since $frac1sqrt1+x$ is a convex function on $[0,1]$, the Hermite-Hadamard and Karamata's inequalities give us that $a_n_ngeq 1$ is an increasing sequence convergent to $2sqrt2-2$. Additionally it is not difficult to check that $a_n= 2sqrt2-2-Thetaleft(frac1nright)$ as $nto +infty$.
answered 2 hours ago
Jack D'Aurizio
280k33272651
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Rearrange it as
$$ frac1nleft(sqrtfracnn+1+sqrtfracnn+2+ldots+sqrtfracnn+1right) = frac1nsum_k=1^nfrac1sqrt1+frackn$$
which is a Riemann sum for
$$ int_0^1fracdxsqrt1+x=2sqrt2-2.$$
Since $frac1sqrt1+x$ is a convex function on $[0,1]$, the Hermite-Hadamard and Karamata's inequalities give us that $a_n_ngeq 1$ is an increasing sequence convergent to $2sqrt2-2$. Additionally it is not difficult to check that $a_n= 2sqrt2-2-Thetaleft(frac1nright)$ as $nto +infty$.
answered 2 hours ago
Jack D'Aurizio
280k33272651
Rearrange it as
$$ frac1nleft(sqrtfracnn+1+sqrtfracnn+2+ldots+sqrtfracnn+1right) = frac1nsum_k=1^nfrac1sqrt1+frackn$$
which is a Riemann sum for
$$ int_0^1fracdxsqrt1+x=2sqrt2-2.$$
Since $frac1sqrt1+x$ is a convex function on $[0,1]$, the Hermite-Hadamard and Karamata's inequalities give us that $a_n_ngeq 1$ is an increasing sequence convergent to $2sqrt2-2$. Additionally it is not difficult to check that $a_n= 2sqrt2-2-Thetaleft(frac1nright)$ as $nto +infty$.
answered 2 hours ago
Jack D'Aurizio
280k33272651
edited 1 hour ago
answered 2 hours ago
Jack D'Aurizio
280k33272651
answered 2 hours ago
Jack D'Aurizio
280k33272651
answered 2 hours ago
Jack D'Aurizio
280k33272651
280k33272651
add a comment |Â
add a comment |Â
up vote
2
down vote
As an alternative by Stolz-Cesaro
$$fracb_nc_n = fracfrac1sqrtn+1+cdots+frac1sqrt2nsqrt n$$
$$fracb_n+1-b_nc_n+1-c_n = fracfrac1sqrt2n+2+frac1sqrt2n+1-frac1sqrtn+1sqrtn+1-sqrt n$$
and
$$fracfrac1sqrt2n+2+frac1sqrt2n+1-frac1sqrtn+1sqrtn+1-sqrt nfracsqrtn+1+sqrt nsqrtn+1+sqrt n=$$
$$fracsqrtn+1+sqrt nsqrt2n+2+fracsqrtn+1+sqrt nsqrt2n+1-fracsqrtn+1+sqrt nsqrtn+1tofrac4sqrt 2-2=2sqrt 2-2$$
answered 1 hour ago
gimusi
81.3k74090
add a comment |Â
up vote
2
down vote
As an alternative by Stolz-Cesaro
$$fracb_nc_n = fracfrac1sqrtn+1+cdots+frac1sqrt2nsqrt n$$
$$fracb_n+1-b_nc_n+1-c_n = fracfrac1sqrt2n+2+frac1sqrt2n+1-frac1sqrtn+1sqrtn+1-sqrt n$$
and
$$fracfrac1sqrt2n+2+frac1sqrt2n+1-frac1sqrtn+1sqrtn+1-sqrt nfracsqrtn+1+sqrt nsqrtn+1+sqrt n=$$
$$fracsqrtn+1+sqrt nsqrt2n+2+fracsqrtn+1+sqrt nsqrt2n+1-fracsqrtn+1+sqrt nsqrtn+1tofrac4sqrt 2-2=2sqrt 2-2$$
answered 1 hour ago
gimusi
81.3k74090
add a comment |Â
up vote
2
down vote
up vote
2
down vote
As an alternative by Stolz-Cesaro
$$fracb_nc_n = fracfrac1sqrtn+1+cdots+frac1sqrt2nsqrt n$$
$$fracb_n+1-b_nc_n+1-c_n = fracfrac1sqrt2n+2+frac1sqrt2n+1-frac1sqrtn+1sqrtn+1-sqrt n$$
and
$$fracfrac1sqrt2n+2+frac1sqrt2n+1-frac1sqrtn+1sqrtn+1-sqrt nfracsqrtn+1+sqrt nsqrtn+1+sqrt n=$$
$$fracsqrtn+1+sqrt nsqrt2n+2+fracsqrtn+1+sqrt nsqrt2n+1-fracsqrtn+1+sqrt nsqrtn+1tofrac4sqrt 2-2=2sqrt 2-2$$
answered 1 hour ago
gimusi
81.3k74090
As an alternative by Stolz-Cesaro
$$fracb_nc_n = fracfrac1sqrtn+1+cdots+frac1sqrt2nsqrt n$$
$$fracb_n+1-b_nc_n+1-c_n = fracfrac1sqrt2n+2+frac1sqrt2n+1-frac1sqrtn+1sqrtn+1-sqrt n$$
and
$$fracfrac1sqrt2n+2+frac1sqrt2n+1-frac1sqrtn+1sqrtn+1-sqrt nfracsqrtn+1+sqrt nsqrtn+1+sqrt n=$$
$$fracsqrtn+1+sqrt nsqrt2n+2+fracsqrtn+1+sqrt nsqrt2n+1-fracsqrtn+1+sqrt nsqrtn+1tofrac4sqrt 2-2=2sqrt 2-2$$
answered 1 hour ago
gimusi
81.3k74090
answered 1 hour ago
gimusi
81.3k74090
answered 1 hour ago
gimusi
81.3k74090
answered 1 hour ago
gimusi
81.3k74090
81.3k74090
add a comment |Â
add a comment |Â
up vote
0
down vote
for a decreasing function such as $1/sqrt x$ with $x$ positive, a simple picture shows
$$ int_a^b+1 ; f(x) ; dx < sum_k=a^b f(k) < int_a-1^b ; f(x) ; dx $$
$$ int_n+1^2n+1 ; frac1sqrt x ; dx < sum_k=n+1^2n frac1sqrt k < int_n^2n ; frac1sqrt x ; dx $$
getting there
$$ 2 sqrt 2n+1 - 2 sqrt n+1 < sum_k=n+1^2n frac1sqrt k < 2 sqrt 2n - 2 sqrt n $$
$$ 2 sqrt 2+frac1n - 2 sqrt 1+frac1n < frac1sqrt n sum_k=n+1^2n frac1sqrt k < 2 sqrt 2 - 2 sqrt 1 $$
answered 50 mins ago
Will Jagy
99.5k597198
add a comment |Â
up vote
0
down vote
for a decreasing function such as $1/sqrt x$ with $x$ positive, a simple picture shows
$$ int_a^b+1 ; f(x) ; dx < sum_k=a^b f(k) < int_a-1^b ; f(x) ; dx $$
$$ int_n+1^2n+1 ; frac1sqrt x ; dx < sum_k=n+1^2n frac1sqrt k < int_n^2n ; frac1sqrt x ; dx $$
getting there
$$ 2 sqrt 2n+1 - 2 sqrt n+1 < sum_k=n+1^2n frac1sqrt k < 2 sqrt 2n - 2 sqrt n $$
$$ 2 sqrt 2+frac1n - 2 sqrt 1+frac1n < frac1sqrt n sum_k=n+1^2n frac1sqrt k < 2 sqrt 2 - 2 sqrt 1 $$
answered 50 mins ago
Will Jagy
99.5k597198
add a comment |Â
up vote
0
down vote
up vote
0
down vote
for a decreasing function such as $1/sqrt x$ with $x$ positive, a simple picture shows
$$ int_a^b+1 ; f(x) ; dx < sum_k=a^b f(k) < int_a-1^b ; f(x) ; dx $$
$$ int_n+1^2n+1 ; frac1sqrt x ; dx < sum_k=n+1^2n frac1sqrt k < int_n^2n ; frac1sqrt x ; dx $$
getting there
$$ 2 sqrt 2n+1 - 2 sqrt n+1 < sum_k=n+1^2n frac1sqrt k < 2 sqrt 2n - 2 sqrt n $$
$$ 2 sqrt 2+frac1n - 2 sqrt 1+frac1n < frac1sqrt n sum_k=n+1^2n frac1sqrt k < 2 sqrt 2 - 2 sqrt 1 $$
answered 50 mins ago
Will Jagy
99.5k597198
for a decreasing function such as $1/sqrt x$ with $x$ positive, a simple picture shows
$$ int_a^b+1 ; f(x) ; dx < sum_k=a^b f(k) < int_a-1^b ; f(x) ; dx $$
$$ int_n+1^2n+1 ; frac1sqrt x ; dx < sum_k=n+1^2n frac1sqrt k < int_n^2n ; frac1sqrt x ; dx $$
getting there
$$ 2 sqrt 2n+1 - 2 sqrt n+1 < sum_k=n+1^2n frac1sqrt k < 2 sqrt 2n - 2 sqrt n $$
$$ 2 sqrt 2+frac1n - 2 sqrt 1+frac1n < frac1sqrt n sum_k=n+1^2n frac1sqrt k < 2 sqrt 2 - 2 sqrt 1 $$
answered 50 mins ago
Will Jagy
99.5k597198
edited 42 mins ago
answered 50 mins ago
Will Jagy
99.5k597198
answered 50 mins ago
Will Jagy
99.5k597198
answered 50 mins ago
Will Jagy
99.5k597198
99.5k597198
add a comment |Â
add a comment |Â
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