Mixing
Determine whether the series $sum_{n =1}^{infty} frac{n + sqrt{n}}{2n^3 -1}$ converges?
0
$begingroup$
Determine whether the series $sum_{n =1}^{infty} frac{n + sqrt{n}}{2n^3 -1}$ converges?
My answer:
For large n, the given series is smaller than $sum_{n =1}^{infty}frac {2n}{2n^3}$ which is equal to $sum_{n =1}^{infty}frac {1}{n^2}$, but we know that the later is convergent by the p-series test, then by comparison test the given series is convergent.
Is my answer correct?
calculus sequences-and-series
asked Jan 15 at 5:36
hopefullyhopefully
238114
$endgroup$
|
show 2 more comments
0
$begingroup$
Determine whether the series $sum_{n =1}^{infty} frac{n + sqrt{n}}{2n^3 -1}$ converges?
My answer:
For large n, the given series is smaller than $sum_{n =1}^{infty}frac {2n}{2n^3}$ which is equal to $sum_{n =1}^{infty}frac {1}{n^2}$, but we know that the later is convergent by the p-series test, then by comparison test the given series is convergent.
Is my answer correct?
calculus sequences-and-series
asked Jan 15 at 5:36
hopefullyhopefully
238114
$endgroup$
1
$begingroup$
It is correct, however you glossed over the exact details of how we know $frac{n+sqrt{n}}{2n^3-1}$ is smaller than $frac{2n}{2n^3}$ for large enough $n$. It would be best to show this more explicitly.
$endgroup$
– JMoravitz
Jan 15 at 5:39
1
$begingroup$
It is correct. In fact the inequality holds for all $n$. Just verify it carefully to make the argument complete.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 5:39
$begingroup$
@KaviRamaMurthy I think it does not hold for $n=1$
$endgroup$
– hopefully
Jan 15 at 5:42
1
$begingroup$
@hopefully Right. It is true for all $n>1$.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 5:45
1
$begingroup$
Also, because the Comparison Test applies only to series with nonnegative terms, you should mention that $frac{n + sqrt{n}}{2n^3 -1}>0$.
$endgroup$
– Taladris
Jan 15 at 5:49
|
show 2 more comments
0
0
0
$begingroup$
Determine whether the series $sum_{n =1}^{infty} frac{n + sqrt{n}}{2n^3 -1}$ converges?
My answer:
For large n, the given series is smaller than $sum_{n =1}^{infty}frac {2n}{2n^3}$ which is equal to $sum_{n =1}^{infty}frac {1}{n^2}$, but we know that the later is convergent by the p-series test, then by comparison test the given series is convergent.
Is my answer correct?
calculus sequences-and-series
asked Jan 15 at 5:36
hopefullyhopefully
238114
$endgroup$
Determine whether the series $sum_{n =1}^{infty} frac{n + sqrt{n}}{2n^3 -1}$ converges?
My answer:
For large n, the given series is smaller than $sum_{n =1}^{infty}frac {2n}{2n^3}$ which is equal to $sum_{n =1}^{infty}frac {1}{n^2}$, but we know that the later is convergent by the p-series test, then by comparison test the given series is convergent.
Is my answer correct?
calculus sequences-and-series
calculus sequences-and-series
asked Jan 15 at 5:36
hopefullyhopefully
238114
asked Jan 15 at 5:36
hopefullyhopefully
238114
asked Jan 15 at 5:36
hopefullyhopefully
238114
asked Jan 15 at 5:36
hopefullyhopefully
238114
asked Jan 15 at 5:36
hopefullyhopefully
238114
238114
1
$begingroup$
It is correct, however you glossed over the exact details of how we know $frac{n+sqrt{n}}{2n^3-1}$ is smaller than $frac{2n}{2n^3}$ for large enough $n$. It would be best to show this more explicitly.
$endgroup$
– JMoravitz
Jan 15 at 5:39
1
$begingroup$
It is correct. In fact the inequality holds for all $n$. Just verify it carefully to make the argument complete.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 5:39
$begingroup$
@KaviRamaMurthy I think it does not hold for $n=1$
$endgroup$
– hopefully
Jan 15 at 5:42
1
$begingroup$
@hopefully Right. It is true for all $n>1$.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 5:45
1
$begingroup$
Also, because the Comparison Test applies only to series with nonnegative terms, you should mention that $frac{n + sqrt{n}}{2n^3 -1}>0$.
$endgroup$
– Taladris
Jan 15 at 5:49
|
show 2 more comments
1
$begingroup$
It is correct, however you glossed over the exact details of how we know $frac{n+sqrt{n}}{2n^3-1}$ is smaller than $frac{2n}{2n^3}$ for large enough $n$. It would be best to show this more explicitly.
$endgroup$
– JMoravitz
Jan 15 at 5:39
1
$begingroup$
It is correct. In fact the inequality holds for all $n$. Just verify it carefully to make the argument complete.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 5:39
$begingroup$
@KaviRamaMurthy I think it does not hold for $n=1$
$endgroup$
– hopefully
Jan 15 at 5:42
1
$begingroup$
@hopefully Right. It is true for all $n>1$.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 5:45
1
$begingroup$
Also, because the Comparison Test applies only to series with nonnegative terms, you should mention that $frac{n + sqrt{n}}{2n^3 -1}>0$.
$endgroup$
– Taladris
Jan 15 at 5:49
1
1
$begingroup$
It is correct, however you glossed over the exact details of how we know $frac{n+sqrt{n}}{2n^3-1}$ is smaller than $frac{2n}{2n^3}$ for large enough $n$. It would be best to show this more explicitly.
$endgroup$
– JMoravitz
Jan 15 at 5:39
$begingroup$
It is correct, however you glossed over the exact details of how we know $frac{n+sqrt{n}}{2n^3-1}$ is smaller than $frac{2n}{2n^3}$ for large enough $n$. It would be best to show this more explicitly.
$endgroup$
– JMoravitz
Jan 15 at 5:39
1
1
$begingroup$
It is correct. In fact the inequality holds for all $n$. Just verify it carefully to make the argument complete.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 5:39
$begingroup$
It is correct. In fact the inequality holds for all $n$. Just verify it carefully to make the argument complete.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 5:39
$begingroup$
@KaviRamaMurthy I think it does not hold for $n=1$
$endgroup$
– hopefully
Jan 15 at 5:42
$begingroup$
@KaviRamaMurthy I think it does not hold for $n=1$
$endgroup$
– hopefully
Jan 15 at 5:42
1
1
$begingroup$
@hopefully Right. It is true for all $n>1$.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 5:45
$begingroup$
@hopefully Right. It is true for all $n>1$.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 5:45
1
1
$begingroup$
Also, because the Comparison Test applies only to series with nonnegative terms, you should mention that $frac{n + sqrt{n}}{2n^3 -1}>0$.
$endgroup$
– Taladris
Jan 15 at 5:49
$begingroup$
Also, because the Comparison Test applies only to series with nonnegative terms, you should mention that $frac{n + sqrt{n}}{2n^3 -1}>0$.
$endgroup$
– Taladris
Jan 15 at 5:49
|
show 2 more comments
1 Answer
1
active
oldest
votes
3
$begingroup$
Almost correct.
You have to make the denominator smaller, not larger.
So yse $n^3$ and everything is fine.
answered Jan 15 at 5:44
marty cohenmarty cohen
73.7k549128
$endgroup$
$begingroup$
could not understand what do you mean?
$endgroup$
– hopefully
Jan 15 at 5:47
$begingroup$
I think for large n increasing the denominator by 1 would not matter, because I have increased the numerator also.
$endgroup$
– hopefully
Jan 15 at 5:49
$begingroup$
yeah I got it ....great!
$endgroup$
– hopefully
Jan 15 at 6:05
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
3
$begingroup$
Almost correct.
You have to make the denominator smaller, not larger.
So yse $n^3$ and everything is fine.
answered Jan 15 at 5:44
marty cohenmarty cohen
73.7k549128
$endgroup$
$begingroup$
could not understand what do you mean?
$endgroup$
– hopefully
Jan 15 at 5:47
$begingroup$
I think for large n increasing the denominator by 1 would not matter, because I have increased the numerator also.
$endgroup$
– hopefully
Jan 15 at 5:49
$begingroup$
yeah I got it ....great!
$endgroup$
– hopefully
Jan 15 at 6:05
add a comment |
3
$begingroup$
Almost correct.
You have to make the denominator smaller, not larger.
So yse $n^3$ and everything is fine.
answered Jan 15 at 5:44
marty cohenmarty cohen
73.7k549128
$endgroup$
$begingroup$
could not understand what do you mean?
$endgroup$
– hopefully
Jan 15 at 5:47
$begingroup$
I think for large n increasing the denominator by 1 would not matter, because I have increased the numerator also.
$endgroup$
– hopefully
Jan 15 at 5:49
$begingroup$
yeah I got it ....great!
$endgroup$
– hopefully
Jan 15 at 6:05
add a comment |
3
3
3
$begingroup$
Almost correct.
You have to make the denominator smaller, not larger.
So yse $n^3$ and everything is fine.
answered Jan 15 at 5:44
marty cohenmarty cohen
73.7k549128
$endgroup$
Almost correct.
You have to make the denominator smaller, not larger.
So yse $n^3$ and everything is fine.
answered Jan 15 at 5:44
marty cohenmarty cohen
73.7k549128
answered Jan 15 at 5:44
marty cohenmarty cohen
73.7k549128
answered Jan 15 at 5:44
marty cohenmarty cohen
73.7k549128
answered Jan 15 at 5:44
marty cohenmarty cohen
73.7k549128
73.7k549128
$begingroup$
could not understand what do you mean?
$endgroup$
– hopefully
Jan 15 at 5:47
$begingroup$
I think for large n increasing the denominator by 1 would not matter, because I have increased the numerator also.
$endgroup$
– hopefully
Jan 15 at 5:49
$begingroup$
yeah I got it ....great!
$endgroup$
– hopefully
Jan 15 at 6:05
add a comment |
$begingroup$
could not understand what do you mean?
$endgroup$
– hopefully
Jan 15 at 5:47
$begingroup$
I think for large n increasing the denominator by 1 would not matter, because I have increased the numerator also.
$endgroup$
– hopefully
Jan 15 at 5:49
$begingroup$
yeah I got it ....great!
$endgroup$
– hopefully
Jan 15 at 6:05
$begingroup$
could not understand what do you mean?
$endgroup$
– hopefully
Jan 15 at 5:47
$begingroup$
could not understand what do you mean?
$endgroup$
– hopefully
Jan 15 at 5:47
$begingroup$
I think for large n increasing the denominator by 1 would not matter, because I have increased the numerator also.
$endgroup$
– hopefully
Jan 15 at 5:49
$begingroup$
I think for large n increasing the denominator by 1 would not matter, because I have increased the numerator also.
$endgroup$
– hopefully
Jan 15 at 5:49
$begingroup$
yeah I got it ....great!
$endgroup$
– hopefully
Jan 15 at 6:05
$begingroup$
yeah I got it ....great!
$endgroup$
– hopefully
Jan 15 at 6:05
add a comment |
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1
$begingroup$
It is correct, however you glossed over the exact details of how we know $frac{n+sqrt{n}}{2n^3-1}$ is smaller than $frac{2n}{2n^3}$ for large enough $n$. It would be best to show this more explicitly.
$endgroup$
– JMoravitz
Jan 15 at 5:39
1
$begingroup$
It is correct. In fact the inequality holds for all $n$. Just verify it carefully to make the argument complete.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 5:39
$begingroup$
@KaviRamaMurthy I think it does not hold for $n=1$
$endgroup$
– hopefully
Jan 15 at 5:42
1
$begingroup$
@hopefully Right. It is true for all $n>1$.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 5:45
1
$begingroup$
Also, because the Comparison Test applies only to series with nonnegative terms, you should mention that $frac{n + sqrt{n}}{2n^3 -1}>0$.
$endgroup$
– Taladris
Jan 15 at 5:49