Mixing
Prove convergence of $sum_{n=2}^infty frac{ln^5{(2n^7 + 13)} + 10sin{n}}{n ln^6{(n^{7/8} + sqrt{n} - 1)} cdot...
$begingroup$
Prove convergence of $sum_{n=2}^infty frac{ln^5{(2n^7 + 13)} + 10sin{n}}{n ln^6{(n^frac{7}{8} + sqrt{n} - 1)} cdot ln( ln (n + (-1)^n))}$. I get that most of the things in this sum doesn't really matter in analyzing its convergence, but I don't know how to solve it formally so as not to make any bad assumptions. Once I get rid of this fancy stuff I guess I should apply Cauchy condensetion.
sequences-and-series convergence
edited Jan 14 at 0:52
Henning Makholm
240k17306543
asked Jan 14 at 0:26
Quo Si ThanQuo Si Than
1467
$endgroup$
add a comment |
$begingroup$
Prove convergence of $sum_{n=2}^infty frac{ln^5{(2n^7 + 13)} + 10sin{n}}{n ln^6{(n^frac{7}{8} + sqrt{n} - 1)} cdot ln( ln (n + (-1)^n))}$. I get that most of the things in this sum doesn't really matter in analyzing its convergence, but I don't know how to solve it formally so as not to make any bad assumptions. Once I get rid of this fancy stuff I guess I should apply Cauchy condensetion.
sequences-and-series convergence
edited Jan 14 at 0:52
Henning Makholm
240k17306543
asked Jan 14 at 0:26
Quo Si ThanQuo Si Than
1467
$endgroup$
add a comment |
$begingroup$
Prove convergence of $sum_{n=2}^infty frac{ln^5{(2n^7 + 13)} + 10sin{n}}{n ln^6{(n^frac{7}{8} + sqrt{n} - 1)} cdot ln( ln (n + (-1)^n))}$. I get that most of the things in this sum doesn't really matter in analyzing its convergence, but I don't know how to solve it formally so as not to make any bad assumptions. Once I get rid of this fancy stuff I guess I should apply Cauchy condensetion.
sequences-and-series convergence
edited Jan 14 at 0:52
Henning Makholm
240k17306543
asked Jan 14 at 0:26
Quo Si ThanQuo Si Than
1467
$endgroup$
Prove convergence of $sum_{n=2}^infty frac{ln^5{(2n^7 + 13)} + 10sin{n}}{n ln^6{(n^frac{7}{8} + sqrt{n} - 1)} cdot ln( ln (n + (-1)^n))}$. I get that most of the things in this sum doesn't really matter in analyzing its convergence, but I don't know how to solve it formally so as not to make any bad assumptions. Once I get rid of this fancy stuff I guess I should apply Cauchy condensetion.
sequences-and-series convergence
sequences-and-series convergence
edited Jan 14 at 0:52
Henning Makholm
240k17306543
asked Jan 14 at 0:26
Quo Si ThanQuo Si Than
1467
edited Jan 14 at 0:52
Henning Makholm
240k17306543
asked Jan 14 at 0:26
Quo Si ThanQuo Si Than
1467
edited Jan 14 at 0:52
Henning Makholm
240k17306543
edited Jan 14 at 0:52
Henning Makholm
240k17306543
edited Jan 14 at 0:52
Henning Makholm
240k17306543
240k17306543
asked Jan 14 at 0:26
Quo Si ThanQuo Si Than
1467
asked Jan 14 at 0:26
Quo Si ThanQuo Si Than
1467
asked Jan 14 at 0:26
Quo Si ThanQuo Si Than
1467
1467
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This kind of series is designed typically for you to try and understand which parts matter.
So the solution follows two steps: first parse the general term (which is always positive and has a rather nice factorized expression with no same-order terms difference) and rewrite/approximate everything in the simplest way possible (best example is: if you have a sum that is not nested in an diverging exponential — or a hyperbolic cosine, you can usually forget all about the non-dominant term; constant factors in logarithms usually contribute little too).
This part is informal and just has to help you « get a grasp of what is going on ».
Here:
1) in the log of the numerator, we can forget about the constant term and the $2$ factor so it becomes $ln^5(n^7)$.
2) the sine term is bounded while the log goes to infinity so the latter can be forgotten.
3) the numerator can thus be re-written as $7^5ln^5(n)$.
4) in the first log of the denominator, $sqrt{n}-1$ is negligible before $n^{7/8}$ thus can be forgotten and this becomes $(7/8)^6ln^6(n)$.
5) Similarly in the second log (iterated), you can « forget » the $(-1)^n$ part.
6) so your denominator is approximately equal to some constant times $nln(n)ln(ln(n))$.
7) As explained in the other answer, the general expression is « roughly equal to » $frac{1}{nln{n}ln{ln{n}}}$.
Now, to the second step: the simplified series is always positive and diverges, as explained in the other answer.
So now you know what to prove and which estimates you have to do: you need a simpler lower bound on the numerator and a simple higher bound on the denominator such that the resulting series diverges.
The idea is to retake the steps you took for the first part and see how you make them rigorous.
1) applies as is.
2) turns the sine part into a $-10$, which is least than half the term in $1)$ for all sufficiently large $n$ (the only ones that matter).
3) so the numerator is not lower than some $cln^6(n)$, where $c>0$ is a numeric constant, for all large enough $n$.
4) $sqrt{n}-1 leq n^{7/8}$, so the log is not lower than $(7/8)^6ln^6(2^{8/7}n) leq (7/8)^6(ln(n)+1)^6 leq (7/4)^6ln^6(n)$ for all large enough $n$.
5) Similarly, the log log part is lower than $ln{ln{2n}} leq ln{ln{n}+ln{2}} leq ln{2ln{n}} leq ln{ln{n}}+ln{2} leq 2ln{ln{n}}$ for all large enough $n$.
6) so your term, for all large enough $n$ is no lower than some positive constant times $frac{1}{nln(n)ln(ln(n))}$, hence the series diverges.
answered Jan 14 at 1:58
MindlackMindlack
4,500210
$endgroup$
add a comment |
$begingroup$
With enough effort you can compare this to a constant times
$$ frac{1}{x log x log log x} $$
which is the derivative of
$$ log log log x $$
which goes to infinity "with great dignity." So your sum diverges.
answered Jan 14 at 0:46
Will JagyWill Jagy
103k5102200
$endgroup$
$begingroup$
Nice reference. (+1)
$endgroup$
– Markus Scheuer
Jan 14 at 20:05
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
This kind of series is designed typically for you to try and understand which parts matter.
So the solution follows two steps: first parse the general term (which is always positive and has a rather nice factorized expression with no same-order terms difference) and rewrite/approximate everything in the simplest way possible (best example is: if you have a sum that is not nested in an diverging exponential — or a hyperbolic cosine, you can usually forget all about the non-dominant term; constant factors in logarithms usually contribute little too).
This part is informal and just has to help you « get a grasp of what is going on ».
Here:
1) in the log of the numerator, we can forget about the constant term and the $2$ factor so it becomes $ln^5(n^7)$.
2) the sine term is bounded while the log goes to infinity so the latter can be forgotten.
3) the numerator can thus be re-written as $7^5ln^5(n)$.
4) in the first log of the denominator, $sqrt{n}-1$ is negligible before $n^{7/8}$ thus can be forgotten and this becomes $(7/8)^6ln^6(n)$.
5) Similarly in the second log (iterated), you can « forget » the $(-1)^n$ part.
6) so your denominator is approximately equal to some constant times $nln(n)ln(ln(n))$.
7) As explained in the other answer, the general expression is « roughly equal to » $frac{1}{nln{n}ln{ln{n}}}$.
Now, to the second step: the simplified series is always positive and diverges, as explained in the other answer.
So now you know what to prove and which estimates you have to do: you need a simpler lower bound on the numerator and a simple higher bound on the denominator such that the resulting series diverges.
The idea is to retake the steps you took for the first part and see how you make them rigorous.
1) applies as is.
2) turns the sine part into a $-10$, which is least than half the term in $1)$ for all sufficiently large $n$ (the only ones that matter).
3) so the numerator is not lower than some $cln^6(n)$, where $c>0$ is a numeric constant, for all large enough $n$.
4) $sqrt{n}-1 leq n^{7/8}$, so the log is not lower than $(7/8)^6ln^6(2^{8/7}n) leq (7/8)^6(ln(n)+1)^6 leq (7/4)^6ln^6(n)$ for all large enough $n$.
5) Similarly, the log log part is lower than $ln{ln{2n}} leq ln{ln{n}+ln{2}} leq ln{2ln{n}} leq ln{ln{n}}+ln{2} leq 2ln{ln{n}}$ for all large enough $n$.
6) so your term, for all large enough $n$ is no lower than some positive constant times $frac{1}{nln(n)ln(ln(n))}$, hence the series diverges.
answered Jan 14 at 1:58
MindlackMindlack
4,500210
$endgroup$
add a comment |
$begingroup$
This kind of series is designed typically for you to try and understand which parts matter.
So the solution follows two steps: first parse the general term (which is always positive and has a rather nice factorized expression with no same-order terms difference) and rewrite/approximate everything in the simplest way possible (best example is: if you have a sum that is not nested in an diverging exponential — or a hyperbolic cosine, you can usually forget all about the non-dominant term; constant factors in logarithms usually contribute little too).
This part is informal and just has to help you « get a grasp of what is going on ».
Here:
1) in the log of the numerator, we can forget about the constant term and the $2$ factor so it becomes $ln^5(n^7)$.
2) the sine term is bounded while the log goes to infinity so the latter can be forgotten.
3) the numerator can thus be re-written as $7^5ln^5(n)$.
4) in the first log of the denominator, $sqrt{n}-1$ is negligible before $n^{7/8}$ thus can be forgotten and this becomes $(7/8)^6ln^6(n)$.
5) Similarly in the second log (iterated), you can « forget » the $(-1)^n$ part.
6) so your denominator is approximately equal to some constant times $nln(n)ln(ln(n))$.
7) As explained in the other answer, the general expression is « roughly equal to » $frac{1}{nln{n}ln{ln{n}}}$.
Now, to the second step: the simplified series is always positive and diverges, as explained in the other answer.
So now you know what to prove and which estimates you have to do: you need a simpler lower bound on the numerator and a simple higher bound on the denominator such that the resulting series diverges.
The idea is to retake the steps you took for the first part and see how you make them rigorous.
1) applies as is.
2) turns the sine part into a $-10$, which is least than half the term in $1)$ for all sufficiently large $n$ (the only ones that matter).
3) so the numerator is not lower than some $cln^6(n)$, where $c>0$ is a numeric constant, for all large enough $n$.
4) $sqrt{n}-1 leq n^{7/8}$, so the log is not lower than $(7/8)^6ln^6(2^{8/7}n) leq (7/8)^6(ln(n)+1)^6 leq (7/4)^6ln^6(n)$ for all large enough $n$.
5) Similarly, the log log part is lower than $ln{ln{2n}} leq ln{ln{n}+ln{2}} leq ln{2ln{n}} leq ln{ln{n}}+ln{2} leq 2ln{ln{n}}$ for all large enough $n$.
6) so your term, for all large enough $n$ is no lower than some positive constant times $frac{1}{nln(n)ln(ln(n))}$, hence the series diverges.
answered Jan 14 at 1:58
MindlackMindlack
4,500210
$endgroup$
add a comment |
$begingroup$
This kind of series is designed typically for you to try and understand which parts matter.
So the solution follows two steps: first parse the general term (which is always positive and has a rather nice factorized expression with no same-order terms difference) and rewrite/approximate everything in the simplest way possible (best example is: if you have a sum that is not nested in an diverging exponential — or a hyperbolic cosine, you can usually forget all about the non-dominant term; constant factors in logarithms usually contribute little too).
This part is informal and just has to help you « get a grasp of what is going on ».
Here:
1) in the log of the numerator, we can forget about the constant term and the $2$ factor so it becomes $ln^5(n^7)$.
2) the sine term is bounded while the log goes to infinity so the latter can be forgotten.
3) the numerator can thus be re-written as $7^5ln^5(n)$.
4) in the first log of the denominator, $sqrt{n}-1$ is negligible before $n^{7/8}$ thus can be forgotten and this becomes $(7/8)^6ln^6(n)$.
5) Similarly in the second log (iterated), you can « forget » the $(-1)^n$ part.
6) so your denominator is approximately equal to some constant times $nln(n)ln(ln(n))$.
7) As explained in the other answer, the general expression is « roughly equal to » $frac{1}{nln{n}ln{ln{n}}}$.
Now, to the second step: the simplified series is always positive and diverges, as explained in the other answer.
So now you know what to prove and which estimates you have to do: you need a simpler lower bound on the numerator and a simple higher bound on the denominator such that the resulting series diverges.
The idea is to retake the steps you took for the first part and see how you make them rigorous.
1) applies as is.
2) turns the sine part into a $-10$, which is least than half the term in $1)$ for all sufficiently large $n$ (the only ones that matter).
3) so the numerator is not lower than some $cln^6(n)$, where $c>0$ is a numeric constant, for all large enough $n$.
4) $sqrt{n}-1 leq n^{7/8}$, so the log is not lower than $(7/8)^6ln^6(2^{8/7}n) leq (7/8)^6(ln(n)+1)^6 leq (7/4)^6ln^6(n)$ for all large enough $n$.
5) Similarly, the log log part is lower than $ln{ln{2n}} leq ln{ln{n}+ln{2}} leq ln{2ln{n}} leq ln{ln{n}}+ln{2} leq 2ln{ln{n}}$ for all large enough $n$.
6) so your term, for all large enough $n$ is no lower than some positive constant times $frac{1}{nln(n)ln(ln(n))}$, hence the series diverges.
answered Jan 14 at 1:58
MindlackMindlack
4,500210
$endgroup$
This kind of series is designed typically for you to try and understand which parts matter.
So the solution follows two steps: first parse the general term (which is always positive and has a rather nice factorized expression with no same-order terms difference) and rewrite/approximate everything in the simplest way possible (best example is: if you have a sum that is not nested in an diverging exponential — or a hyperbolic cosine, you can usually forget all about the non-dominant term; constant factors in logarithms usually contribute little too).
This part is informal and just has to help you « get a grasp of what is going on ».
Here:
1) in the log of the numerator, we can forget about the constant term and the $2$ factor so it becomes $ln^5(n^7)$.
2) the sine term is bounded while the log goes to infinity so the latter can be forgotten.
3) the numerator can thus be re-written as $7^5ln^5(n)$.
4) in the first log of the denominator, $sqrt{n}-1$ is negligible before $n^{7/8}$ thus can be forgotten and this becomes $(7/8)^6ln^6(n)$.
5) Similarly in the second log (iterated), you can « forget » the $(-1)^n$ part.
6) so your denominator is approximately equal to some constant times $nln(n)ln(ln(n))$.
7) As explained in the other answer, the general expression is « roughly equal to » $frac{1}{nln{n}ln{ln{n}}}$.
Now, to the second step: the simplified series is always positive and diverges, as explained in the other answer.
So now you know what to prove and which estimates you have to do: you need a simpler lower bound on the numerator and a simple higher bound on the denominator such that the resulting series diverges.
The idea is to retake the steps you took for the first part and see how you make them rigorous.
1) applies as is.
2) turns the sine part into a $-10$, which is least than half the term in $1)$ for all sufficiently large $n$ (the only ones that matter).
3) so the numerator is not lower than some $cln^6(n)$, where $c>0$ is a numeric constant, for all large enough $n$.
4) $sqrt{n}-1 leq n^{7/8}$, so the log is not lower than $(7/8)^6ln^6(2^{8/7}n) leq (7/8)^6(ln(n)+1)^6 leq (7/4)^6ln^6(n)$ for all large enough $n$.
5) Similarly, the log log part is lower than $ln{ln{2n}} leq ln{ln{n}+ln{2}} leq ln{2ln{n}} leq ln{ln{n}}+ln{2} leq 2ln{ln{n}}$ for all large enough $n$.
6) so your term, for all large enough $n$ is no lower than some positive constant times $frac{1}{nln(n)ln(ln(n))}$, hence the series diverges.
answered Jan 14 at 1:58
MindlackMindlack
4,500210
answered Jan 14 at 1:58
MindlackMindlack
4,500210
answered Jan 14 at 1:58
MindlackMindlack
4,500210
answered Jan 14 at 1:58
MindlackMindlack
4,500210
4,500210
add a comment |
add a comment |
$begingroup$
With enough effort you can compare this to a constant times
$$ frac{1}{x log x log log x} $$
which is the derivative of
$$ log log log x $$
which goes to infinity "with great dignity." So your sum diverges.
answered Jan 14 at 0:46
Will JagyWill Jagy
103k5102200
$endgroup$
$begingroup$
Nice reference. (+1)
$endgroup$
– Markus Scheuer
Jan 14 at 20:05
add a comment |
$begingroup$
With enough effort you can compare this to a constant times
$$ frac{1}{x log x log log x} $$
which is the derivative of
$$ log log log x $$
which goes to infinity "with great dignity." So your sum diverges.
answered Jan 14 at 0:46
Will JagyWill Jagy
103k5102200
$endgroup$
$begingroup$
Nice reference. (+1)
$endgroup$
– Markus Scheuer
Jan 14 at 20:05
add a comment |
$begingroup$
With enough effort you can compare this to a constant times
$$ frac{1}{x log x log log x} $$
which is the derivative of
$$ log log log x $$
which goes to infinity "with great dignity." So your sum diverges.
answered Jan 14 at 0:46
Will JagyWill Jagy
103k5102200
$endgroup$
With enough effort you can compare this to a constant times
$$ frac{1}{x log x log log x} $$
which is the derivative of
$$ log log log x $$
which goes to infinity "with great dignity." So your sum diverges.
answered Jan 14 at 0:46
Will JagyWill Jagy
103k5102200
answered Jan 14 at 0:46
Will JagyWill Jagy
103k5102200
answered Jan 14 at 0:46
Will JagyWill Jagy
103k5102200
answered Jan 14 at 0:46
Will JagyWill Jagy
103k5102200
103k5102200
$begingroup$
Nice reference. (+1)
$endgroup$
– Markus Scheuer
Jan 14 at 20:05
add a comment |
$begingroup$
Nice reference. (+1)
$endgroup$
– Markus Scheuer
Jan 14 at 20:05
$begingroup$
Nice reference. (+1)
$endgroup$
– Markus Scheuer
Jan 14 at 20:05
$begingroup$
Nice reference. (+1)
$endgroup$
– Markus Scheuer
Jan 14 at 20:05
add a comment |
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