Convergence of $sum_n=1^infty(-n)^lflooralpharfloorfracsin^3left(frac1nright)left(log(n)right)^alpha-1$
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Convergence of the following series as $alphainmathbbR$
$$sum_n=1^infty(-n)^lflooralpharfloorfracsin^3left(frac1nright)left(log(n)right)^alpha-1$$
As $n to +infty$ we have that $fracsin^3left(frac1nright)left(log(n)right)^alpha-1 = mathcalOleft(frac1n^3log^alpha-1(n)right)$
How to handle $(-n)^lflooralpharfloor$ with that result?
real-analysis sequences-and-series convergence
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up vote
2
down vote
favorite
Convergence of the following series as $alphainmathbbR$
$$sum_n=1^infty(-n)^lflooralpharfloorfracsin^3left(frac1nright)left(log(n)right)^alpha-1$$
As $n to +infty$ we have that $fracsin^3left(frac1nright)left(log(n)right)^alpha-1 = mathcalOleft(frac1n^3log^alpha-1(n)right)$
How to handle $(-n)^lflooralpharfloor$ with that result?
real-analysis sequences-and-series convergence
i think you should take two cases, for $lflooralpharfloor$ to be odd or even
â Subhajit Halder
Sep 6 at 12:49
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Convergence of the following series as $alphainmathbbR$
$$sum_n=1^infty(-n)^lflooralpharfloorfracsin^3left(frac1nright)left(log(n)right)^alpha-1$$
As $n to +infty$ we have that $fracsin^3left(frac1nright)left(log(n)right)^alpha-1 = mathcalOleft(frac1n^3log^alpha-1(n)right)$
How to handle $(-n)^lflooralpharfloor$ with that result?
real-analysis sequences-and-series convergence
Convergence of the following series as $alphainmathbbR$
$$sum_n=1^infty(-n)^lflooralpharfloorfracsin^3left(frac1nright)left(log(n)right)^alpha-1$$
As $n to +infty$ we have that $fracsin^3left(frac1nright)left(log(n)right)^alpha-1 = mathcalOleft(frac1n^3log^alpha-1(n)right)$
How to handle $(-n)^lflooralpharfloor$ with that result?
real-analysis sequences-and-series convergence
real-analysis sequences-and-series convergence
edited Sep 6 at 13:27
gimusi
73.6k73889
73.6k73889
asked Sep 6 at 11:52
F.inc
3908
3908
i think you should take two cases, for $lflooralpharfloor$ to be odd or even
â Subhajit Halder
Sep 6 at 12:49
add a comment |Â
i think you should take two cases, for $lflooralpharfloor$ to be odd or even
â Subhajit Halder
Sep 6 at 12:49
i think you should take two cases, for $lflooralpharfloor$ to be odd or even
â Subhajit Halder
Sep 6 at 12:49
i think you should take two cases, for $lflooralpharfloor$ to be odd or even
â Subhajit Halder
Sep 6 at 12:49
add a comment |Â
1 Answer
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Yes the key point is that
$$fracsin^3left(frac1nright)left(log(n)right)^alpha-1 sim frac1n^3log^alpha-1n$$
then we need to distinguish the cases, for example for $0le alpha <1$ we have $lflooralpha rfloor=0$ and then
$$(-n)^lflooralpharfloorfracsin^3left(frac1nright)left(log(n)right)^alpha-1 sim fraclog nn^3$$
which converges by limit comparison test with $sum frac1n^2$.
And you can proceed similarly for the others values.
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Yes the key point is that
$$fracsin^3left(frac1nright)left(log(n)right)^alpha-1 sim frac1n^3log^alpha-1n$$
then we need to distinguish the cases, for example for $0le alpha <1$ we have $lflooralpha rfloor=0$ and then
$$(-n)^lflooralpharfloorfracsin^3left(frac1nright)left(log(n)right)^alpha-1 sim fraclog nn^3$$
which converges by limit comparison test with $sum frac1n^2$.
And you can proceed similarly for the others values.
add a comment |Â
up vote
2
down vote
Yes the key point is that
$$fracsin^3left(frac1nright)left(log(n)right)^alpha-1 sim frac1n^3log^alpha-1n$$
then we need to distinguish the cases, for example for $0le alpha <1$ we have $lflooralpha rfloor=0$ and then
$$(-n)^lflooralpharfloorfracsin^3left(frac1nright)left(log(n)right)^alpha-1 sim fraclog nn^3$$
which converges by limit comparison test with $sum frac1n^2$.
And you can proceed similarly for the others values.
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Yes the key point is that
$$fracsin^3left(frac1nright)left(log(n)right)^alpha-1 sim frac1n^3log^alpha-1n$$
then we need to distinguish the cases, for example for $0le alpha <1$ we have $lflooralpha rfloor=0$ and then
$$(-n)^lflooralpharfloorfracsin^3left(frac1nright)left(log(n)right)^alpha-1 sim fraclog nn^3$$
which converges by limit comparison test with $sum frac1n^2$.
And you can proceed similarly for the others values.
Yes the key point is that
$$fracsin^3left(frac1nright)left(log(n)right)^alpha-1 sim frac1n^3log^alpha-1n$$
then we need to distinguish the cases, for example for $0le alpha <1$ we have $lflooralpha rfloor=0$ and then
$$(-n)^lflooralpharfloorfracsin^3left(frac1nright)left(log(n)right)^alpha-1 sim fraclog nn^3$$
which converges by limit comparison test with $sum frac1n^2$.
And you can proceed similarly for the others values.
edited Sep 6 at 14:08
answered Sep 6 at 13:23
gimusi
73.6k73889
73.6k73889
add a comment |Â
add a comment |Â
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i think you should take two cases, for $lflooralpharfloor$ to be odd or even
â Subhajit Halder
Sep 6 at 12:49