Simple series question. $sum_{n=0}^infty frac{1}{((n^5)+1)^frac{1}{3}}$
1
$begingroup$
I think this converges due to direct comparison with $frac{1}{n^{5/3}}$ but I can't double check this anywhere.
sequences-and-series
edited Jan 18 at 2:16
Thomas Shelby
4,7832727
asked Jan 18 at 1:25
FinegoldFinegold
62
$endgroup$
add a comment |
1
$begingroup$
I think this converges due to direct comparison with $frac{1}{n^{5/3}}$ but I can't double check this anywhere.
sequences-and-series
edited Jan 18 at 2:16
Thomas Shelby
4,7832727
asked Jan 18 at 1:25
FinegoldFinegold
62
$endgroup$
$begingroup$
You're on the right track. What exactly does the direct comparison test say, and how can you prove it's true that $a_n leq b_n$
$endgroup$
– Tom Himler
Jan 18 at 1:33
2
$begingroup$
I thought that considering there was a plus 1 in the denominator of the problem question it would always be smaller than 1/n^5/3 so since 1/n^5/3 converges due to the p-test and is bigger than a similar smaller function know that the smaller converges as well do to direct comparison. Do you agree?
$endgroup$
– Finegold
Jan 18 at 1:47
$begingroup$
You're exactly right.
$endgroup$
– Tom Himler
Jan 18 at 2:21
add a comment |
1
1
1
$begingroup$
I think this converges due to direct comparison with $frac{1}{n^{5/3}}$ but I can't double check this anywhere.
sequences-and-series
edited Jan 18 at 2:16
Thomas Shelby
4,7832727
asked Jan 18 at 1:25
FinegoldFinegold
62
$endgroup$
I think this converges due to direct comparison with $frac{1}{n^{5/3}}$ but I can't double check this anywhere.
sequences-and-series
sequences-and-series
edited Jan 18 at 2:16
Thomas Shelby
4,7832727
asked Jan 18 at 1:25
FinegoldFinegold
62
edited Jan 18 at 2:16
Thomas Shelby
4,7832727
asked Jan 18 at 1:25
FinegoldFinegold
62
edited Jan 18 at 2:16
Thomas Shelby
4,7832727
edited Jan 18 at 2:16
Thomas Shelby
4,7832727
edited Jan 18 at 2:16
Thomas Shelby
4,7832727
4,7832727
asked Jan 18 at 1:25
FinegoldFinegold
62
asked Jan 18 at 1:25
FinegoldFinegold
62
asked Jan 18 at 1:25
FinegoldFinegold
62
62
$begingroup$
You're on the right track. What exactly does the direct comparison test say, and how can you prove it's true that $a_n leq b_n$
$endgroup$
– Tom Himler
Jan 18 at 1:33
2
$begingroup$
I thought that considering there was a plus 1 in the denominator of the problem question it would always be smaller than 1/n^5/3 so since 1/n^5/3 converges due to the p-test and is bigger than a similar smaller function know that the smaller converges as well do to direct comparison. Do you agree?
$endgroup$
– Finegold
Jan 18 at 1:47
$begingroup$
You're exactly right.
$endgroup$
– Tom Himler
Jan 18 at 2:21
add a comment |
$begingroup$
You're on the right track. What exactly does the direct comparison test say, and how can you prove it's true that $a_n leq b_n$
$endgroup$
– Tom Himler
Jan 18 at 1:33
2
$begingroup$
I thought that considering there was a plus 1 in the denominator of the problem question it would always be smaller than 1/n^5/3 so since 1/n^5/3 converges due to the p-test and is bigger than a similar smaller function know that the smaller converges as well do to direct comparison. Do you agree?
$endgroup$
– Finegold
Jan 18 at 1:47
$begingroup$
You're exactly right.
$endgroup$
– Tom Himler
Jan 18 at 2:21
$begingroup$
You're on the right track. What exactly does the direct comparison test say, and how can you prove it's true that $a_n leq b_n$
$endgroup$
– Tom Himler
Jan 18 at 1:33
$begingroup$
You're on the right track. What exactly does the direct comparison test say, and how can you prove it's true that $a_n leq b_n$
$endgroup$
– Tom Himler
Jan 18 at 1:33
2
2
$begingroup$
I thought that considering there was a plus 1 in the denominator of the problem question it would always be smaller than 1/n^5/3 so since 1/n^5/3 converges due to the p-test and is bigger than a similar smaller function know that the smaller converges as well do to direct comparison. Do you agree?
$endgroup$
– Finegold
Jan 18 at 1:47
$begingroup$
I thought that considering there was a plus 1 in the denominator of the problem question it would always be smaller than 1/n^5/3 so since 1/n^5/3 converges due to the p-test and is bigger than a similar smaller function know that the smaller converges as well do to direct comparison. Do you agree?
$endgroup$
– Finegold
Jan 18 at 1:47
$begingroup$
You're exactly right.
$endgroup$
– Tom Himler
Jan 18 at 2:21
$begingroup$
You're exactly right.
$endgroup$
– Tom Himler
Jan 18 at 2:21
add a comment |
1 Answer
1
active
oldest
votes
2
$begingroup$
Right.
$(n^5+1)^{1/3}
gt (n^5)^{1/3}
=n^{5/3}
$
so the sum converges by the $p$-test:
$sum dfrac1{n^p}$
converges for
$p > 1$
(easily proved by the
integral test)
and diverges for
$p le 1$.
answered Jan 18 at 2:08
marty cohenmarty cohen
75.7k549130
$endgroup$
add a comment |
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2
$begingroup$
Right.
$(n^5+1)^{1/3}
gt (n^5)^{1/3}
=n^{5/3}
$
so the sum converges by the $p$-test:
$sum dfrac1{n^p}$
converges for
$p > 1$
(easily proved by the
integral test)
and diverges for
$p le 1$.
answered Jan 18 at 2:08
marty cohenmarty cohen
75.7k549130
$endgroup$
add a comment |
2
$begingroup$
Right.
$(n^5+1)^{1/3}
gt (n^5)^{1/3}
=n^{5/3}
$
so the sum converges by the $p$-test:
$sum dfrac1{n^p}$
converges for
$p > 1$
(easily proved by the
integral test)
and diverges for
$p le 1$.
answered Jan 18 at 2:08
marty cohenmarty cohen
75.7k549130
$endgroup$
add a comment |
2
2
2
$begingroup$
Right.
$(n^5+1)^{1/3}
gt (n^5)^{1/3}
=n^{5/3}
$
so the sum converges by the $p$-test:
$sum dfrac1{n^p}$
converges for
$p > 1$
(easily proved by the
integral test)
and diverges for
$p le 1$.
answered Jan 18 at 2:08
marty cohenmarty cohen
75.7k549130
$endgroup$
Right.
$(n^5+1)^{1/3}
gt (n^5)^{1/3}
=n^{5/3}
$
so the sum converges by the $p$-test:
$sum dfrac1{n^p}$
converges for
$p > 1$
(easily proved by the
integral test)
and diverges for
$p le 1$.
answered Jan 18 at 2:08
marty cohenmarty cohen
75.7k549130
answered Jan 18 at 2:08
marty cohenmarty cohen
75.7k549130
answered Jan 18 at 2:08
marty cohenmarty cohen
75.7k549130
answered Jan 18 at 2:08
marty cohenmarty cohen
75.7k549130
75.7k549130
add a comment |
add a comment |
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$begingroup$
You're on the right track. What exactly does the direct comparison test say, and how can you prove it's true that $a_n leq b_n$
$endgroup$
– Tom Himler
Jan 18 at 1:33
2
$begingroup$
I thought that considering there was a plus 1 in the denominator of the problem question it would always be smaller than 1/n^5/3 so since 1/n^5/3 converges due to the p-test and is bigger than a similar smaller function know that the smaller converges as well do to direct comparison. Do you agree?
$endgroup$
– Finegold
Jan 18 at 1:47
$begingroup$
You're exactly right.
$endgroup$
– Tom Himler
Jan 18 at 2:21