Mixing
Limit of a sequence with a difference of cosines
Define the following sequence:
$$x_n = sqrt[3]{n+1}cos{sqrt{n+1}} - sqrt[3]{n}cos{sqrt{n}},forall nin mathbb{N} $$
Does the limit
$$lim_{n to infty} x_n$$
exist?
I believe that it does not, mainly because (i think that) $cossqrt{n+1}$ and $cos{sqrt{n}}$ have opposite signs an infinite number of times.
More exactly, I tried showing that there exists an infinite sequence $a_n$ of natural numbers such that
$$ sqrt{a_n} text{ mod } 2pi in left(frac{3pi}{2} + frac{pi}{4} ,2piright) text{ and } sqrt{a_n + 1} text{ mod } 2pi in left( 0, frac{pi}{4} right)$$
and a sequence $b_n$ such that
$$sqrt{b_n} text{ mod } 2pi in left(frac{pi}{2} + frac{pi}{4}, pi right) text{ and } sqrt{b_n + 1} text{ mod }in left(pi, pi + frac{pi}{4} right),$$
because then the subsequences $x_{a_n}$ and $x_{b_n}$ would both be bounded and of different signs, showing that $x_n$ is divergent.
Would this approach work? Is the sequence actually somehow convergent?
real-analysis sequences-and-series limits
asked Nov 20 '18 at 17:49
Tanny Sieben
30118
add a comment |
Define the following sequence:
$$x_n = sqrt[3]{n+1}cos{sqrt{n+1}} - sqrt[3]{n}cos{sqrt{n}},forall nin mathbb{N} $$
Does the limit
$$lim_{n to infty} x_n$$
exist?
I believe that it does not, mainly because (i think that) $cossqrt{n+1}$ and $cos{sqrt{n}}$ have opposite signs an infinite number of times.
More exactly, I tried showing that there exists an infinite sequence $a_n$ of natural numbers such that
$$ sqrt{a_n} text{ mod } 2pi in left(frac{3pi}{2} + frac{pi}{4} ,2piright) text{ and } sqrt{a_n + 1} text{ mod } 2pi in left( 0, frac{pi}{4} right)$$
and a sequence $b_n$ such that
$$sqrt{b_n} text{ mod } 2pi in left(frac{pi}{2} + frac{pi}{4}, pi right) text{ and } sqrt{b_n + 1} text{ mod }in left(pi, pi + frac{pi}{4} right),$$
because then the subsequences $x_{a_n}$ and $x_{b_n}$ would both be bounded and of different signs, showing that $x_n$ is divergent.
Would this approach work? Is the sequence actually somehow convergent?
real-analysis sequences-and-series limits
asked Nov 20 '18 at 17:49
Tanny Sieben
30118
Use mean value theorem on $f(x) =x^{1/3}cossqrt{x}$ on interval $[n, n+1]$
– Paramanand Singh
Nov 21 '18 at 11:03
@TannySieben Please let me know if something is not clear in my derivation.
– gimusi
Nov 21 '18 at 20:47
add a comment |
Define the following sequence:
$$x_n = sqrt[3]{n+1}cos{sqrt{n+1}} - sqrt[3]{n}cos{sqrt{n}},forall nin mathbb{N} $$
Does the limit
$$lim_{n to infty} x_n$$
exist?
I believe that it does not, mainly because (i think that) $cossqrt{n+1}$ and $cos{sqrt{n}}$ have opposite signs an infinite number of times.
More exactly, I tried showing that there exists an infinite sequence $a_n$ of natural numbers such that
$$ sqrt{a_n} text{ mod } 2pi in left(frac{3pi}{2} + frac{pi}{4} ,2piright) text{ and } sqrt{a_n + 1} text{ mod } 2pi in left( 0, frac{pi}{4} right)$$
and a sequence $b_n$ such that
$$sqrt{b_n} text{ mod } 2pi in left(frac{pi}{2} + frac{pi}{4}, pi right) text{ and } sqrt{b_n + 1} text{ mod }in left(pi, pi + frac{pi}{4} right),$$
because then the subsequences $x_{a_n}$ and $x_{b_n}$ would both be bounded and of different signs, showing that $x_n$ is divergent.
Would this approach work? Is the sequence actually somehow convergent?
real-analysis sequences-and-series limits
asked Nov 20 '18 at 17:49
Tanny Sieben
30118
Define the following sequence:
$$x_n = sqrt[3]{n+1}cos{sqrt{n+1}} - sqrt[3]{n}cos{sqrt{n}},forall nin mathbb{N} $$
Does the limit
$$lim_{n to infty} x_n$$
exist?
I believe that it does not, mainly because (i think that) $cossqrt{n+1}$ and $cos{sqrt{n}}$ have opposite signs an infinite number of times.
More exactly, I tried showing that there exists an infinite sequence $a_n$ of natural numbers such that
$$ sqrt{a_n} text{ mod } 2pi in left(frac{3pi}{2} + frac{pi}{4} ,2piright) text{ and } sqrt{a_n + 1} text{ mod } 2pi in left( 0, frac{pi}{4} right)$$
and a sequence $b_n$ such that
$$sqrt{b_n} text{ mod } 2pi in left(frac{pi}{2} + frac{pi}{4}, pi right) text{ and } sqrt{b_n + 1} text{ mod }in left(pi, pi + frac{pi}{4} right),$$
because then the subsequences $x_{a_n}$ and $x_{b_n}$ would both be bounded and of different signs, showing that $x_n$ is divergent.
Would this approach work? Is the sequence actually somehow convergent?
real-analysis sequences-and-series limits
real-analysis sequences-and-series limits
asked Nov 20 '18 at 17:49
Tanny Sieben
30118
asked Nov 20 '18 at 17:49
Tanny Sieben
30118
asked Nov 20 '18 at 17:49
Tanny Sieben
30118
asked Nov 20 '18 at 17:49
Tanny Sieben
30118
asked Nov 20 '18 at 17:49
Tanny Sieben
30118
30118
Use mean value theorem on $f(x) =x^{1/3}cossqrt{x}$ on interval $[n, n+1]$
– Paramanand Singh
Nov 21 '18 at 11:03
@TannySieben Please let me know if something is not clear in my derivation.
– gimusi
Nov 21 '18 at 20:47
add a comment |
Use mean value theorem on $f(x) =x^{1/3}cossqrt{x}$ on interval $[n, n+1]$
– Paramanand Singh
Nov 21 '18 at 11:03
@TannySieben Please let me know if something is not clear in my derivation.
– gimusi
Nov 21 '18 at 20:47
Use mean value theorem on $f(x) =x^{1/3}cossqrt{x}$ on interval $[n, n+1]$
– Paramanand Singh
Nov 21 '18 at 11:03
Use mean value theorem on $f(x) =x^{1/3}cossqrt{x}$ on interval $[n, n+1]$
– Paramanand Singh
Nov 21 '18 at 11:03
@TannySieben Please let me know if something is not clear in my derivation.
– gimusi
Nov 21 '18 at 20:47
@TannySieben Please let me know if something is not clear in my derivation.
– gimusi
Nov 21 '18 at 20:47
add a comment |
1 Answer
1
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oldest
votes
By binomial series
$$sqrt[3]{n+1}=sqrt[3]{n}left(1+frac1nright)^frac13=sqrt[3]{n}+Oleft(frac1{n^frac23}right)$$
then
$$sqrt[3]{n+1}cos{sqrt{n+1}} - sqrt[3]{n}cos{sqrt{n}}=$$$$=sqrt[3]{n}(cos{sqrt{n+1}}-cos{sqrt{n}})+Oleft(frac{cos{sqrt{n+1}}}{n^frac23}right)$$
and by sum to product formula
$$cos{sqrt{n+1}}-cos{sqrt{n}}=-2sinleft(frac{sqrt{n+1}+sqrt{n}}{2}right)sinleft(frac{sqrt{n+1}-sqrt{n}}{2}right)=$$
$$=-2sinleft(frac{sqrt{n+1}+sqrt{n}}{2}right)sinleft(frac{1}{2(sqrt{n+1}+sqrt{n})}right)sim -frac{sin(sqrt n)}{sqrt n}$$
therefore the given sequence converges to zero.
answered Nov 20 '18 at 18:10
gimusi
1
add a comment |
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1 Answer
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By binomial series
$$sqrt[3]{n+1}=sqrt[3]{n}left(1+frac1nright)^frac13=sqrt[3]{n}+Oleft(frac1{n^frac23}right)$$
then
$$sqrt[3]{n+1}cos{sqrt{n+1}} - sqrt[3]{n}cos{sqrt{n}}=$$$$=sqrt[3]{n}(cos{sqrt{n+1}}-cos{sqrt{n}})+Oleft(frac{cos{sqrt{n+1}}}{n^frac23}right)$$
and by sum to product formula
$$cos{sqrt{n+1}}-cos{sqrt{n}}=-2sinleft(frac{sqrt{n+1}+sqrt{n}}{2}right)sinleft(frac{sqrt{n+1}-sqrt{n}}{2}right)=$$
$$=-2sinleft(frac{sqrt{n+1}+sqrt{n}}{2}right)sinleft(frac{1}{2(sqrt{n+1}+sqrt{n})}right)sim -frac{sin(sqrt n)}{sqrt n}$$
therefore the given sequence converges to zero.
answered Nov 20 '18 at 18:10
gimusi
1
add a comment |
By binomial series
$$sqrt[3]{n+1}=sqrt[3]{n}left(1+frac1nright)^frac13=sqrt[3]{n}+Oleft(frac1{n^frac23}right)$$
then
$$sqrt[3]{n+1}cos{sqrt{n+1}} - sqrt[3]{n}cos{sqrt{n}}=$$$$=sqrt[3]{n}(cos{sqrt{n+1}}-cos{sqrt{n}})+Oleft(frac{cos{sqrt{n+1}}}{n^frac23}right)$$
and by sum to product formula
$$cos{sqrt{n+1}}-cos{sqrt{n}}=-2sinleft(frac{sqrt{n+1}+sqrt{n}}{2}right)sinleft(frac{sqrt{n+1}-sqrt{n}}{2}right)=$$
$$=-2sinleft(frac{sqrt{n+1}+sqrt{n}}{2}right)sinleft(frac{1}{2(sqrt{n+1}+sqrt{n})}right)sim -frac{sin(sqrt n)}{sqrt n}$$
therefore the given sequence converges to zero.
answered Nov 20 '18 at 18:10
gimusi
1
add a comment |
By binomial series
$$sqrt[3]{n+1}=sqrt[3]{n}left(1+frac1nright)^frac13=sqrt[3]{n}+Oleft(frac1{n^frac23}right)$$
then
$$sqrt[3]{n+1}cos{sqrt{n+1}} - sqrt[3]{n}cos{sqrt{n}}=$$$$=sqrt[3]{n}(cos{sqrt{n+1}}-cos{sqrt{n}})+Oleft(frac{cos{sqrt{n+1}}}{n^frac23}right)$$
and by sum to product formula
$$cos{sqrt{n+1}}-cos{sqrt{n}}=-2sinleft(frac{sqrt{n+1}+sqrt{n}}{2}right)sinleft(frac{sqrt{n+1}-sqrt{n}}{2}right)=$$
$$=-2sinleft(frac{sqrt{n+1}+sqrt{n}}{2}right)sinleft(frac{1}{2(sqrt{n+1}+sqrt{n})}right)sim -frac{sin(sqrt n)}{sqrt n}$$
therefore the given sequence converges to zero.
answered Nov 20 '18 at 18:10
gimusi
1
By binomial series
$$sqrt[3]{n+1}=sqrt[3]{n}left(1+frac1nright)^frac13=sqrt[3]{n}+Oleft(frac1{n^frac23}right)$$
then
$$sqrt[3]{n+1}cos{sqrt{n+1}} - sqrt[3]{n}cos{sqrt{n}}=$$$$=sqrt[3]{n}(cos{sqrt{n+1}}-cos{sqrt{n}})+Oleft(frac{cos{sqrt{n+1}}}{n^frac23}right)$$
and by sum to product formula
$$cos{sqrt{n+1}}-cos{sqrt{n}}=-2sinleft(frac{sqrt{n+1}+sqrt{n}}{2}right)sinleft(frac{sqrt{n+1}-sqrt{n}}{2}right)=$$
$$=-2sinleft(frac{sqrt{n+1}+sqrt{n}}{2}right)sinleft(frac{1}{2(sqrt{n+1}+sqrt{n})}right)sim -frac{sin(sqrt n)}{sqrt n}$$
therefore the given sequence converges to zero.
answered Nov 20 '18 at 18:10
gimusi
1
edited Nov 21 '18 at 8:55
answered Nov 20 '18 at 18:10
gimusi
1
answered Nov 20 '18 at 18:10
gimusi
1
answered Nov 20 '18 at 18:10
gimusi
1
1
add a comment |
add a comment |
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Use mean value theorem on $f(x) =x^{1/3}cossqrt{x}$ on interval $[n, n+1]$
– Paramanand Singh
Nov 21 '18 at 11:03
@TannySieben Please let me know if something is not clear in my derivation.
– gimusi
Nov 21 '18 at 20:47