Does the series $sum_{n=1}^infty frac{n}{sqrt[3]{8n^5-1}}$ Converge?
$begingroup$
$$sum_{n=1}^infty frac{n}{sqrt[3]{8n^5-1}}$$
From the tests that I know of:
Divergence Test: The limit is ≠ to a constant, so inconclusive.
Geometric series: I don't think this could be written in that manner.
Comparison Test/Lim Comparison: Compare to $$frac{n}{8n^{frac{5}{3}}}$$
Integral Test: I can't think of a integration method that would work here.
Alternating Series/Root Test don't apply.
Ratio Test: The limit is 1 so inconclusive.
Perhaps I'm making a mistake throughout the methods I've tried, but I'm lost. Using these tests, is it possible to find whether or not it converges or diverges?
sequences-and-series
asked Dec 12 '18 at 5:34
user623028user623028
235
$endgroup$
add a comment |
$begingroup$
$$sum_{n=1}^infty frac{n}{sqrt[3]{8n^5-1}}$$
From the tests that I know of:
Divergence Test: The limit is ≠ to a constant, so inconclusive.
Geometric series: I don't think this could be written in that manner.
Comparison Test/Lim Comparison: Compare to $$frac{n}{8n^{frac{5}{3}}}$$
Integral Test: I can't think of a integration method that would work here.
Alternating Series/Root Test don't apply.
Ratio Test: The limit is 1 so inconclusive.
Perhaps I'm making a mistake throughout the methods I've tried, but I'm lost. Using these tests, is it possible to find whether or not it converges or diverges?
sequences-and-series
asked Dec 12 '18 at 5:34
user623028user623028
235
$endgroup$
$begingroup$
it's same as $frac{1}{n^{2/3}}$, so no, it doesn't converge
$endgroup$
– mathworker21
Dec 12 '18 at 5:36
add a comment |
$begingroup$
$$sum_{n=1}^infty frac{n}{sqrt[3]{8n^5-1}}$$
From the tests that I know of:
Divergence Test: The limit is ≠ to a constant, so inconclusive.
Geometric series: I don't think this could be written in that manner.
Comparison Test/Lim Comparison: Compare to $$frac{n}{8n^{frac{5}{3}}}$$
Integral Test: I can't think of a integration method that would work here.
Alternating Series/Root Test don't apply.
Ratio Test: The limit is 1 so inconclusive.
Perhaps I'm making a mistake throughout the methods I've tried, but I'm lost. Using these tests, is it possible to find whether or not it converges or diverges?
sequences-and-series
asked Dec 12 '18 at 5:34
user623028user623028
235
$endgroup$
$$sum_{n=1}^infty frac{n}{sqrt[3]{8n^5-1}}$$
From the tests that I know of:
Divergence Test: The limit is ≠ to a constant, so inconclusive.
Geometric series: I don't think this could be written in that manner.
Comparison Test/Lim Comparison: Compare to $$frac{n}{8n^{frac{5}{3}}}$$
Integral Test: I can't think of a integration method that would work here.
Alternating Series/Root Test don't apply.
Ratio Test: The limit is 1 so inconclusive.
Perhaps I'm making a mistake throughout the methods I've tried, but I'm lost. Using these tests, is it possible to find whether or not it converges or diverges?
sequences-and-series
sequences-and-series
asked Dec 12 '18 at 5:34
user623028user623028
235
asked Dec 12 '18 at 5:34
user623028user623028
235
asked Dec 12 '18 at 5:34
user623028user623028
235
asked Dec 12 '18 at 5:34
user623028user623028
235
asked Dec 12 '18 at 5:34
user623028user623028
235
235
$begingroup$
it's same as $frac{1}{n^{2/3}}$, so no, it doesn't converge
$endgroup$
– mathworker21
Dec 12 '18 at 5:36
add a comment |
$begingroup$
it's same as $frac{1}{n^{2/3}}$, so no, it doesn't converge
$endgroup$
– mathworker21
Dec 12 '18 at 5:36
$begingroup$
it's same as $frac{1}{n^{2/3}}$, so no, it doesn't converge
$endgroup$
– mathworker21
Dec 12 '18 at 5:36
$begingroup$
it's same as $frac{1}{n^{2/3}}$, so no, it doesn't converge
$endgroup$
– mathworker21
Dec 12 '18 at 5:36
add a comment |
2 Answers
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$begingroup$
Use comparison test,$$frac{n}{sqrt[3]{8n^5-1}} ge frac{n}{sqrt[3]{8n^5}} = frac{1}{2n^{frac53-1}}=frac{1}{2n^{frac23}}$$
Now, use $p$-series to make conclusion that it diverges.
answered Dec 12 '18 at 5:40
Siong Thye GohSiong Thye Goh
102k1466118
$endgroup$
add a comment |
$begingroup$
We have that
$$frac{n}{sqrt[3]{8n^5-1}}sim frac1{2n^{2/3}}$$
therefore we can conclude that the given series diverges by limit comparison test with $sum frac1{n^{2/3}}$.
answered Dec 12 '18 at 7:16
gimusigimusi
92.9k84494
$endgroup$
add a comment |
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2 Answers
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2 Answers
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active
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$begingroup$
Use comparison test,$$frac{n}{sqrt[3]{8n^5-1}} ge frac{n}{sqrt[3]{8n^5}} = frac{1}{2n^{frac53-1}}=frac{1}{2n^{frac23}}$$
Now, use $p$-series to make conclusion that it diverges.
answered Dec 12 '18 at 5:40
Siong Thye GohSiong Thye Goh
102k1466118
$endgroup$
add a comment |
$begingroup$
Use comparison test,$$frac{n}{sqrt[3]{8n^5-1}} ge frac{n}{sqrt[3]{8n^5}} = frac{1}{2n^{frac53-1}}=frac{1}{2n^{frac23}}$$
Now, use $p$-series to make conclusion that it diverges.
answered Dec 12 '18 at 5:40
Siong Thye GohSiong Thye Goh
102k1466118
$endgroup$
add a comment |
$begingroup$
Use comparison test,$$frac{n}{sqrt[3]{8n^5-1}} ge frac{n}{sqrt[3]{8n^5}} = frac{1}{2n^{frac53-1}}=frac{1}{2n^{frac23}}$$
Now, use $p$-series to make conclusion that it diverges.
answered Dec 12 '18 at 5:40
Siong Thye GohSiong Thye Goh
102k1466118
$endgroup$
Use comparison test,$$frac{n}{sqrt[3]{8n^5-1}} ge frac{n}{sqrt[3]{8n^5}} = frac{1}{2n^{frac53-1}}=frac{1}{2n^{frac23}}$$
Now, use $p$-series to make conclusion that it diverges.
answered Dec 12 '18 at 5:40
Siong Thye GohSiong Thye Goh
102k1466118
answered Dec 12 '18 at 5:40
Siong Thye GohSiong Thye Goh
102k1466118
answered Dec 12 '18 at 5:40
Siong Thye GohSiong Thye Goh
102k1466118
answered Dec 12 '18 at 5:40
Siong Thye GohSiong Thye Goh
102k1466118
102k1466118
add a comment |
add a comment |
$begingroup$
We have that
$$frac{n}{sqrt[3]{8n^5-1}}sim frac1{2n^{2/3}}$$
therefore we can conclude that the given series diverges by limit comparison test with $sum frac1{n^{2/3}}$.
answered Dec 12 '18 at 7:16
gimusigimusi
92.9k84494
$endgroup$
add a comment |
$begingroup$
We have that
$$frac{n}{sqrt[3]{8n^5-1}}sim frac1{2n^{2/3}}$$
therefore we can conclude that the given series diverges by limit comparison test with $sum frac1{n^{2/3}}$.
answered Dec 12 '18 at 7:16
gimusigimusi
92.9k84494
$endgroup$
add a comment |
$begingroup$
We have that
$$frac{n}{sqrt[3]{8n^5-1}}sim frac1{2n^{2/3}}$$
therefore we can conclude that the given series diverges by limit comparison test with $sum frac1{n^{2/3}}$.
answered Dec 12 '18 at 7:16
gimusigimusi
92.9k84494
$endgroup$
We have that
$$frac{n}{sqrt[3]{8n^5-1}}sim frac1{2n^{2/3}}$$
therefore we can conclude that the given series diverges by limit comparison test with $sum frac1{n^{2/3}}$.
answered Dec 12 '18 at 7:16
gimusigimusi
92.9k84494
answered Dec 12 '18 at 7:16
gimusigimusi
92.9k84494
answered Dec 12 '18 at 7:16
gimusigimusi
92.9k84494
answered Dec 12 '18 at 7:16
gimusigimusi
92.9k84494
92.9k84494
add a comment |
add a comment |
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$begingroup$
it's same as $frac{1}{n^{2/3}}$, so no, it doesn't converge
$endgroup$
– mathworker21
Dec 12 '18 at 5:36