Mixing
Determine whether the series $sum_{n=1}^{infty} frac{ln(n)}{n^2 +1}$ converges or not.
2
$begingroup$
Determine whether the series $sum_{n=1}^{infty} frac{ln(n)}{n^2 +1}$ converges or not.
** My trial **
I tried dividing $frac{ln(n)}{n^2 +1}$ by $1/n^2$ and finding the limit which was $infty$ so I could not use the limit comparison test and this idea did not work.
Could anyone give me a hint for studying the convergence of this series?
calculus sequences-and-series
edited Jan 16 at 5:08
Aaron Quitta
328214
asked Jan 16 at 4:47
hopefullyhopefully
249114
$endgroup$
add a comment |
2
$begingroup$
Determine whether the series $sum_{n=1}^{infty} frac{ln(n)}{n^2 +1}$ converges or not.
** My trial **
I tried dividing $frac{ln(n)}{n^2 +1}$ by $1/n^2$ and finding the limit which was $infty$ so I could not use the limit comparison test and this idea did not work.
Could anyone give me a hint for studying the convergence of this series?
calculus sequences-and-series
edited Jan 16 at 5:08
Aaron Quitta
328214
asked Jan 16 at 4:47
hopefullyhopefully
249114
$endgroup$
$begingroup$
do you mean $$sum_{n=1}^{infty}frac{ln n}{n^2+1}$$?
$endgroup$
– clathratus
Jan 16 at 4:52
$begingroup$
Hint: Use the inequality $ln n < n$ in the following way $ln n = 2 ln sqrt{n} < 2 sqrt{n}$
$endgroup$
– RRL
Jan 16 at 4:54
$begingroup$
@clathratus yes sorry I corrected it.
$endgroup$
– hopefully
Jan 16 at 4:56
add a comment |
2
2
2
$begingroup$
Determine whether the series $sum_{n=1}^{infty} frac{ln(n)}{n^2 +1}$ converges or not.
** My trial **
I tried dividing $frac{ln(n)}{n^2 +1}$ by $1/n^2$ and finding the limit which was $infty$ so I could not use the limit comparison test and this idea did not work.
Could anyone give me a hint for studying the convergence of this series?
calculus sequences-and-series
edited Jan 16 at 5:08
Aaron Quitta
328214
asked Jan 16 at 4:47
hopefullyhopefully
249114
$endgroup$
Determine whether the series $sum_{n=1}^{infty} frac{ln(n)}{n^2 +1}$ converges or not.
** My trial **
I tried dividing $frac{ln(n)}{n^2 +1}$ by $1/n^2$ and finding the limit which was $infty$ so I could not use the limit comparison test and this idea did not work.
Could anyone give me a hint for studying the convergence of this series?
calculus sequences-and-series
calculus sequences-and-series
edited Jan 16 at 5:08
Aaron Quitta
328214
asked Jan 16 at 4:47
hopefullyhopefully
249114
edited Jan 16 at 5:08
Aaron Quitta
328214
asked Jan 16 at 4:47
hopefullyhopefully
249114
edited Jan 16 at 5:08
Aaron Quitta
328214
edited Jan 16 at 5:08
Aaron Quitta
328214
edited Jan 16 at 5:08
Aaron Quitta
328214
328214
asked Jan 16 at 4:47
hopefullyhopefully
249114
asked Jan 16 at 4:47
hopefullyhopefully
249114
asked Jan 16 at 4:47
hopefullyhopefully
249114
249114
$begingroup$
do you mean $$sum_{n=1}^{infty}frac{ln n}{n^2+1}$$?
$endgroup$
– clathratus
Jan 16 at 4:52
$begingroup$
Hint: Use the inequality $ln n < n$ in the following way $ln n = 2 ln sqrt{n} < 2 sqrt{n}$
$endgroup$
– RRL
Jan 16 at 4:54
$begingroup$
@clathratus yes sorry I corrected it.
$endgroup$
– hopefully
Jan 16 at 4:56
add a comment |
$begingroup$
do you mean $$sum_{n=1}^{infty}frac{ln n}{n^2+1}$$?
$endgroup$
– clathratus
Jan 16 at 4:52
$begingroup$
Hint: Use the inequality $ln n < n$ in the following way $ln n = 2 ln sqrt{n} < 2 sqrt{n}$
$endgroup$
– RRL
Jan 16 at 4:54
$begingroup$
@clathratus yes sorry I corrected it.
$endgroup$
– hopefully
Jan 16 at 4:56
$begingroup$
do you mean $$sum_{n=1}^{infty}frac{ln n}{n^2+1}$$?
$endgroup$
– clathratus
Jan 16 at 4:52
$begingroup$
do you mean $$sum_{n=1}^{infty}frac{ln n}{n^2+1}$$?
$endgroup$
– clathratus
Jan 16 at 4:52
$begingroup$
Hint: Use the inequality $ln n < n$ in the following way $ln n = 2 ln sqrt{n} < 2 sqrt{n}$
$endgroup$
– RRL
Jan 16 at 4:54
$begingroup$
Hint: Use the inequality $ln n < n$ in the following way $ln n = 2 ln sqrt{n} < 2 sqrt{n}$
$endgroup$
– RRL
Jan 16 at 4:54
$begingroup$
@clathratus yes sorry I corrected it.
$endgroup$
– hopefully
Jan 16 at 4:56
$begingroup$
@clathratus yes sorry I corrected it.
$endgroup$
– hopefully
Jan 16 at 4:56
add a comment |
1 Answer
1
active
oldest
votes
2
$begingroup$
Compare with $$sum_{n=1}^{infty}frac{1}{n^{1.5}}$$
answered Jan 16 at 4:54
alex.jordanalex.jordan
39.3k560121
$endgroup$
$begingroup$
you mean comparison not limit comparison?
$endgroup$
– hopefully
Jan 16 at 4:56
$begingroup$
Either way. This series converges. With direct comparison, the OP series has smaller terms (eventually). With a limit comparison, the ratio of the OP terms to the terms here converges to $0$. Either implies the OP series converges.
$endgroup$
– alex.jordan
Jan 16 at 5:06
$begingroup$
but if I use the direct comparison test how can I compare them?
$endgroup$
– hopefully
Jan 16 at 14:19
$begingroup$
You would need to know that (eventually) $ln(n)<n^{0.5}$. There are proofs out there that (eventually) $ln(n)<n^{varepsilon}$ for any positive $varepsilon$.
$endgroup$
– alex.jordan
Jan 16 at 15:41
1
$begingroup$
For example, $ln(n)/n^{0.5}$. What is the limit of this as $ntoinfty$? By L'Hospital, it is the same as the limit of $1/(0.5n^{0.5})$, which is $0$. So for large enough $n$, $ln(n)/n^{0.5}<1$. So for large enough $n$, $ln(n)<n^{0.5}$.
$endgroup$
– alex.jordan
Jan 16 at 16:39
|
show 1 more comment
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
$begingroup$
Compare with $$sum_{n=1}^{infty}frac{1}{n^{1.5}}$$
answered Jan 16 at 4:54
alex.jordanalex.jordan
39.3k560121
$endgroup$
$begingroup$
you mean comparison not limit comparison?
$endgroup$
– hopefully
Jan 16 at 4:56
$begingroup$
Either way. This series converges. With direct comparison, the OP series has smaller terms (eventually). With a limit comparison, the ratio of the OP terms to the terms here converges to $0$. Either implies the OP series converges.
$endgroup$
– alex.jordan
Jan 16 at 5:06
$begingroup$
but if I use the direct comparison test how can I compare them?
$endgroup$
– hopefully
Jan 16 at 14:19
$begingroup$
You would need to know that (eventually) $ln(n)<n^{0.5}$. There are proofs out there that (eventually) $ln(n)<n^{varepsilon}$ for any positive $varepsilon$.
$endgroup$
– alex.jordan
Jan 16 at 15:41
1
$begingroup$
For example, $ln(n)/n^{0.5}$. What is the limit of this as $ntoinfty$? By L'Hospital, it is the same as the limit of $1/(0.5n^{0.5})$, which is $0$. So for large enough $n$, $ln(n)/n^{0.5}<1$. So for large enough $n$, $ln(n)<n^{0.5}$.
$endgroup$
– alex.jordan
Jan 16 at 16:39
|
show 1 more comment
2
$begingroup$
Compare with $$sum_{n=1}^{infty}frac{1}{n^{1.5}}$$
answered Jan 16 at 4:54
alex.jordanalex.jordan
39.3k560121
$endgroup$
$begingroup$
you mean comparison not limit comparison?
$endgroup$
– hopefully
Jan 16 at 4:56
$begingroup$
Either way. This series converges. With direct comparison, the OP series has smaller terms (eventually). With a limit comparison, the ratio of the OP terms to the terms here converges to $0$. Either implies the OP series converges.
$endgroup$
– alex.jordan
Jan 16 at 5:06
$begingroup$
but if I use the direct comparison test how can I compare them?
$endgroup$
– hopefully
Jan 16 at 14:19
$begingroup$
You would need to know that (eventually) $ln(n)<n^{0.5}$. There are proofs out there that (eventually) $ln(n)<n^{varepsilon}$ for any positive $varepsilon$.
$endgroup$
– alex.jordan
Jan 16 at 15:41
1
$begingroup$
For example, $ln(n)/n^{0.5}$. What is the limit of this as $ntoinfty$? By L'Hospital, it is the same as the limit of $1/(0.5n^{0.5})$, which is $0$. So for large enough $n$, $ln(n)/n^{0.5}<1$. So for large enough $n$, $ln(n)<n^{0.5}$.
$endgroup$
– alex.jordan
Jan 16 at 16:39
|
show 1 more comment
2
2
2
$begingroup$
Compare with $$sum_{n=1}^{infty}frac{1}{n^{1.5}}$$
answered Jan 16 at 4:54
alex.jordanalex.jordan
39.3k560121
$endgroup$
Compare with $$sum_{n=1}^{infty}frac{1}{n^{1.5}}$$
answered Jan 16 at 4:54
alex.jordanalex.jordan
39.3k560121
answered Jan 16 at 4:54
alex.jordanalex.jordan
39.3k560121
answered Jan 16 at 4:54
alex.jordanalex.jordan
39.3k560121
answered Jan 16 at 4:54
alex.jordanalex.jordan
39.3k560121
39.3k560121
$begingroup$
you mean comparison not limit comparison?
$endgroup$
– hopefully
Jan 16 at 4:56
$begingroup$
Either way. This series converges. With direct comparison, the OP series has smaller terms (eventually). With a limit comparison, the ratio of the OP terms to the terms here converges to $0$. Either implies the OP series converges.
$endgroup$
– alex.jordan
Jan 16 at 5:06
$begingroup$
but if I use the direct comparison test how can I compare them?
$endgroup$
– hopefully
Jan 16 at 14:19
$begingroup$
You would need to know that (eventually) $ln(n)<n^{0.5}$. There are proofs out there that (eventually) $ln(n)<n^{varepsilon}$ for any positive $varepsilon$.
$endgroup$
– alex.jordan
Jan 16 at 15:41
1
$begingroup$
For example, $ln(n)/n^{0.5}$. What is the limit of this as $ntoinfty$? By L'Hospital, it is the same as the limit of $1/(0.5n^{0.5})$, which is $0$. So for large enough $n$, $ln(n)/n^{0.5}<1$. So for large enough $n$, $ln(n)<n^{0.5}$.
$endgroup$
– alex.jordan
Jan 16 at 16:39
|
show 1 more comment
$begingroup$
you mean comparison not limit comparison?
$endgroup$
– hopefully
Jan 16 at 4:56
$begingroup$
Either way. This series converges. With direct comparison, the OP series has smaller terms (eventually). With a limit comparison, the ratio of the OP terms to the terms here converges to $0$. Either implies the OP series converges.
$endgroup$
– alex.jordan
Jan 16 at 5:06
$begingroup$
but if I use the direct comparison test how can I compare them?
$endgroup$
– hopefully
Jan 16 at 14:19
$begingroup$
You would need to know that (eventually) $ln(n)<n^{0.5}$. There are proofs out there that (eventually) $ln(n)<n^{varepsilon}$ for any positive $varepsilon$.
$endgroup$
– alex.jordan
Jan 16 at 15:41
1
$begingroup$
For example, $ln(n)/n^{0.5}$. What is the limit of this as $ntoinfty$? By L'Hospital, it is the same as the limit of $1/(0.5n^{0.5})$, which is $0$. So for large enough $n$, $ln(n)/n^{0.5}<1$. So for large enough $n$, $ln(n)<n^{0.5}$.
$endgroup$
– alex.jordan
Jan 16 at 16:39
$begingroup$
you mean comparison not limit comparison?
$endgroup$
– hopefully
Jan 16 at 4:56
$begingroup$
you mean comparison not limit comparison?
$endgroup$
– hopefully
Jan 16 at 4:56
$begingroup$
Either way. This series converges. With direct comparison, the OP series has smaller terms (eventually). With a limit comparison, the ratio of the OP terms to the terms here converges to $0$. Either implies the OP series converges.
$endgroup$
– alex.jordan
Jan 16 at 5:06
$begingroup$
Either way. This series converges. With direct comparison, the OP series has smaller terms (eventually). With a limit comparison, the ratio of the OP terms to the terms here converges to $0$. Either implies the OP series converges.
$endgroup$
– alex.jordan
Jan 16 at 5:06
$begingroup$
but if I use the direct comparison test how can I compare them?
$endgroup$
– hopefully
Jan 16 at 14:19
$begingroup$
but if I use the direct comparison test how can I compare them?
$endgroup$
– hopefully
Jan 16 at 14:19
$begingroup$
You would need to know that (eventually) $ln(n)<n^{0.5}$. There are proofs out there that (eventually) $ln(n)<n^{varepsilon}$ for any positive $varepsilon$.
$endgroup$
– alex.jordan
Jan 16 at 15:41
$begingroup$
You would need to know that (eventually) $ln(n)<n^{0.5}$. There are proofs out there that (eventually) $ln(n)<n^{varepsilon}$ for any positive $varepsilon$.
$endgroup$
– alex.jordan
Jan 16 at 15:41
1
1
$begingroup$
For example, $ln(n)/n^{0.5}$. What is the limit of this as $ntoinfty$? By L'Hospital, it is the same as the limit of $1/(0.5n^{0.5})$, which is $0$. So for large enough $n$, $ln(n)/n^{0.5}<1$. So for large enough $n$, $ln(n)<n^{0.5}$.
$endgroup$
– alex.jordan
Jan 16 at 16:39
$begingroup$
For example, $ln(n)/n^{0.5}$. What is the limit of this as $ntoinfty$? By L'Hospital, it is the same as the limit of $1/(0.5n^{0.5})$, which is $0$. So for large enough $n$, $ln(n)/n^{0.5}<1$. So for large enough $n$, $ln(n)<n^{0.5}$.
$endgroup$
– alex.jordan
Jan 16 at 16:39
|
show 1 more comment
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$begingroup$
do you mean $$sum_{n=1}^{infty}frac{ln n}{n^2+1}$$?
$endgroup$
– clathratus
Jan 16 at 4:52
$begingroup$
Hint: Use the inequality $ln n < n$ in the following way $ln n = 2 ln sqrt{n} < 2 sqrt{n}$
$endgroup$
– RRL
Jan 16 at 4:54
$begingroup$
@clathratus yes sorry I corrected it.
$endgroup$
– hopefully
Jan 16 at 4:56