Does $(1+frac12-frac13) + (frac14+frac15-frac16)+(frac17+frac18-frac19)+cdots$ converge?
Does the series $$S=left(1+frac{1}{2}-frac{1}{3} right) + left(frac{1}{4}+frac{1}{5}-frac{1}{6} right)+left(frac{1}{7}+frac{1}{8}-frac{1}{9}right)+cdots$$ converge?
Here's my attempt at a solution: $$S = sum_{n=1}^{infty}frac{1}{n}-2sum_{n=1}^{infty}frac{1}{3n}=sum_{n=1}^{infty}frac{1}{3n}=frac{1}{3}sum_{n=1}^{infty}frac{1}{n}$$
As we can "rewrite" this series as one third of the harmonical series (that diverges), we conclude the divergence of $S$.
Is this right? Which other convergence tests could be used?
calculus sequences-and-series proof-verification convergence divergent-series
New contributor
add a comment |
Does the series $$S=left(1+frac{1}{2}-frac{1}{3} right) + left(frac{1}{4}+frac{1}{5}-frac{1}{6} right)+left(frac{1}{7}+frac{1}{8}-frac{1}{9}right)+cdots$$ converge?
Here's my attempt at a solution: $$S = sum_{n=1}^{infty}frac{1}{n}-2sum_{n=1}^{infty}frac{1}{3n}=sum_{n=1}^{infty}frac{1}{3n}=frac{1}{3}sum_{n=1}^{infty}frac{1}{n}$$
As we can "rewrite" this series as one third of the harmonical series (that diverges), we conclude the divergence of $S$.
Is this right? Which other convergence tests could be used?
calculus sequences-and-series proof-verification convergence divergent-series
New contributor
2
your solution is not correct. You cannot change the order of summation of a series that doesn't converge absolutely, neither make (standard) arithmetic with divergent series
– Masacroso
yesterday
@Masacroso Is it not fine to do something along the lines of " Suppose the following series converges. Then in particular, it converges absolutely, so we can reorder the series in any way. One particular ordering gives a scalar multiple of the harmonic series which diverges, so the given series cannot converge".
– Saad
yesterday
4
@Saad That is not a valid argument because convergence doesn't imply absolute convergence (you've got it the wrong way round).
– AlephNull
22 hours ago
add a comment |
Does the series $$S=left(1+frac{1}{2}-frac{1}{3} right) + left(frac{1}{4}+frac{1}{5}-frac{1}{6} right)+left(frac{1}{7}+frac{1}{8}-frac{1}{9}right)+cdots$$ converge?
Here's my attempt at a solution: $$S = sum_{n=1}^{infty}frac{1}{n}-2sum_{n=1}^{infty}frac{1}{3n}=sum_{n=1}^{infty}frac{1}{3n}=frac{1}{3}sum_{n=1}^{infty}frac{1}{n}$$
As we can "rewrite" this series as one third of the harmonical series (that diverges), we conclude the divergence of $S$.
Is this right? Which other convergence tests could be used?
calculus sequences-and-series proof-verification convergence divergent-series
New contributor
Does the series $$S=left(1+frac{1}{2}-frac{1}{3} right) + left(frac{1}{4}+frac{1}{5}-frac{1}{6} right)+left(frac{1}{7}+frac{1}{8}-frac{1}{9}right)+cdots$$ converge?
Here's my attempt at a solution: $$S = sum_{n=1}^{infty}frac{1}{n}-2sum_{n=1}^{infty}frac{1}{3n}=sum_{n=1}^{infty}frac{1}{3n}=frac{1}{3}sum_{n=1}^{infty}frac{1}{n}$$
As we can "rewrite" this series as one third of the harmonical series (that diverges), we conclude the divergence of $S$.
Is this right? Which other convergence tests could be used?
calculus sequences-and-series proof-verification convergence divergent-series
calculus sequences-and-series proof-verification convergence divergent-series
New contributor
New contributor
edited 16 hours ago
Did
246k23221455
246k23221455
New contributor
asked yesterday
Raúl Astete
496
496
New contributor
New contributor
2
your solution is not correct. You cannot change the order of summation of a series that doesn't converge absolutely, neither make (standard) arithmetic with divergent series
– Masacroso
yesterday
@Masacroso Is it not fine to do something along the lines of " Suppose the following series converges. Then in particular, it converges absolutely, so we can reorder the series in any way. One particular ordering gives a scalar multiple of the harmonic series which diverges, so the given series cannot converge".
– Saad
yesterday
4
@Saad That is not a valid argument because convergence doesn't imply absolute convergence (you've got it the wrong way round).
– AlephNull
22 hours ago
add a comment |
2
your solution is not correct. You cannot change the order of summation of a series that doesn't converge absolutely, neither make (standard) arithmetic with divergent series
– Masacroso
yesterday
@Masacroso Is it not fine to do something along the lines of " Suppose the following series converges. Then in particular, it converges absolutely, so we can reorder the series in any way. One particular ordering gives a scalar multiple of the harmonic series which diverges, so the given series cannot converge".
– Saad
yesterday
4
@Saad That is not a valid argument because convergence doesn't imply absolute convergence (you've got it the wrong way round).
– AlephNull
22 hours ago
2
2
your solution is not correct. You cannot change the order of summation of a series that doesn't converge absolutely, neither make (standard) arithmetic with divergent series
– Masacroso
yesterday
your solution is not correct. You cannot change the order of summation of a series that doesn't converge absolutely, neither make (standard) arithmetic with divergent series
– Masacroso
yesterday
@Masacroso Is it not fine to do something along the lines of " Suppose the following series converges. Then in particular, it converges absolutely, so we can reorder the series in any way. One particular ordering gives a scalar multiple of the harmonic series which diverges, so the given series cannot converge".
– Saad
yesterday
@Masacroso Is it not fine to do something along the lines of " Suppose the following series converges. Then in particular, it converges absolutely, so we can reorder the series in any way. One particular ordering gives a scalar multiple of the harmonic series which diverges, so the given series cannot converge".
– Saad
yesterday
4
4
@Saad That is not a valid argument because convergence doesn't imply absolute convergence (you've got it the wrong way round).
– AlephNull
22 hours ago
@Saad That is not a valid argument because convergence doesn't imply absolute convergence (you've got it the wrong way round).
– AlephNull
22 hours ago
add a comment |
2 Answers
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Your answer is correct, but your reasoning is not. Order matters. You should write the sum as follows:
$$sum_{n=0}^inftyleft(frac{1}{3n+1}+frac{1}{3n+2}-frac{1}{3n+3}right)$$
Simplify what is in the parentheses and then evaluate in the usual way.
Oh, and it is true that order does not matter if all the terms are positive. But when some terms are positive and others negative, you have to be more careful.
2
In this case you can simply say $frac 1{3n+2}-frac 1{3n+3}>0$ thus the series is $>sumfrac 1{3n+1}$ which is a divergent series. I mean we do not even need to calculate that the parenthesis is equivalent to $frac {9n^2}{27n^3}$.
– zwim
yesterday
@zwim quite true, that is a nice little shortcut
– Ben W
20 hours ago
Simplify and it behaves like $1/n$, so it diverges.
– ncmathsadist
16 hours ago
add a comment |
The idea is correct, but you cannot write like that, because the series involved are both divergent and generally it is moot to do arithmetic operations of divergent series.
Let $H_n = sum_1^n 1/k$, i.e. the $n$-th partial sum of the harmonic series, then the
$$
S_n = H_{3n} - frac 23 H_n,
$$
then use the asymptotic expression $H_n = log n + gamma + varepsilon_n$ where $varepsilon_n to 0 [n to infty]$, we have
$$
S_n = log (3n) -frac 23log n + frac 13 gamma + varepsilon_{3n} - frac 23 varepsilon_n = log(3n^{1/3}) + frac 13 gamma + alpha_n xrightarrow{n to infty} +infty.
$$
add a comment |
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2 Answers
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Your answer is correct, but your reasoning is not. Order matters. You should write the sum as follows:
$$sum_{n=0}^inftyleft(frac{1}{3n+1}+frac{1}{3n+2}-frac{1}{3n+3}right)$$
Simplify what is in the parentheses and then evaluate in the usual way.
Oh, and it is true that order does not matter if all the terms are positive. But when some terms are positive and others negative, you have to be more careful.
2
In this case you can simply say $frac 1{3n+2}-frac 1{3n+3}>0$ thus the series is $>sumfrac 1{3n+1}$ which is a divergent series. I mean we do not even need to calculate that the parenthesis is equivalent to $frac {9n^2}{27n^3}$.
– zwim
yesterday
@zwim quite true, that is a nice little shortcut
– Ben W
20 hours ago
Simplify and it behaves like $1/n$, so it diverges.
– ncmathsadist
16 hours ago
add a comment |
Your answer is correct, but your reasoning is not. Order matters. You should write the sum as follows:
$$sum_{n=0}^inftyleft(frac{1}{3n+1}+frac{1}{3n+2}-frac{1}{3n+3}right)$$
Simplify what is in the parentheses and then evaluate in the usual way.
Oh, and it is true that order does not matter if all the terms are positive. But when some terms are positive and others negative, you have to be more careful.
2
In this case you can simply say $frac 1{3n+2}-frac 1{3n+3}>0$ thus the series is $>sumfrac 1{3n+1}$ which is a divergent series. I mean we do not even need to calculate that the parenthesis is equivalent to $frac {9n^2}{27n^3}$.
– zwim
yesterday
@zwim quite true, that is a nice little shortcut
– Ben W
20 hours ago
Simplify and it behaves like $1/n$, so it diverges.
– ncmathsadist
16 hours ago
add a comment |
Your answer is correct, but your reasoning is not. Order matters. You should write the sum as follows:
$$sum_{n=0}^inftyleft(frac{1}{3n+1}+frac{1}{3n+2}-frac{1}{3n+3}right)$$
Simplify what is in the parentheses and then evaluate in the usual way.
Oh, and it is true that order does not matter if all the terms are positive. But when some terms are positive and others negative, you have to be more careful.
Your answer is correct, but your reasoning is not. Order matters. You should write the sum as follows:
$$sum_{n=0}^inftyleft(frac{1}{3n+1}+frac{1}{3n+2}-frac{1}{3n+3}right)$$
Simplify what is in the parentheses and then evaluate in the usual way.
Oh, and it is true that order does not matter if all the terms are positive. But when some terms are positive and others negative, you have to be more careful.
edited yesterday
answered yesterday
Ben W
1,995615
1,995615
2
In this case you can simply say $frac 1{3n+2}-frac 1{3n+3}>0$ thus the series is $>sumfrac 1{3n+1}$ which is a divergent series. I mean we do not even need to calculate that the parenthesis is equivalent to $frac {9n^2}{27n^3}$.
– zwim
yesterday
@zwim quite true, that is a nice little shortcut
– Ben W
20 hours ago
Simplify and it behaves like $1/n$, so it diverges.
– ncmathsadist
16 hours ago
add a comment |
2
In this case you can simply say $frac 1{3n+2}-frac 1{3n+3}>0$ thus the series is $>sumfrac 1{3n+1}$ which is a divergent series. I mean we do not even need to calculate that the parenthesis is equivalent to $frac {9n^2}{27n^3}$.
– zwim
yesterday
@zwim quite true, that is a nice little shortcut
– Ben W
20 hours ago
Simplify and it behaves like $1/n$, so it diverges.
– ncmathsadist
16 hours ago
2
2
In this case you can simply say $frac 1{3n+2}-frac 1{3n+3}>0$ thus the series is $>sumfrac 1{3n+1}$ which is a divergent series. I mean we do not even need to calculate that the parenthesis is equivalent to $frac {9n^2}{27n^3}$.
– zwim
yesterday
In this case you can simply say $frac 1{3n+2}-frac 1{3n+3}>0$ thus the series is $>sumfrac 1{3n+1}$ which is a divergent series. I mean we do not even need to calculate that the parenthesis is equivalent to $frac {9n^2}{27n^3}$.
– zwim
yesterday
@zwim quite true, that is a nice little shortcut
– Ben W
20 hours ago
@zwim quite true, that is a nice little shortcut
– Ben W
20 hours ago
Simplify and it behaves like $1/n$, so it diverges.
– ncmathsadist
16 hours ago
Simplify and it behaves like $1/n$, so it diverges.
– ncmathsadist
16 hours ago
add a comment |
The idea is correct, but you cannot write like that, because the series involved are both divergent and generally it is moot to do arithmetic operations of divergent series.
Let $H_n = sum_1^n 1/k$, i.e. the $n$-th partial sum of the harmonic series, then the
$$
S_n = H_{3n} - frac 23 H_n,
$$
then use the asymptotic expression $H_n = log n + gamma + varepsilon_n$ where $varepsilon_n to 0 [n to infty]$, we have
$$
S_n = log (3n) -frac 23log n + frac 13 gamma + varepsilon_{3n} - frac 23 varepsilon_n = log(3n^{1/3}) + frac 13 gamma + alpha_n xrightarrow{n to infty} +infty.
$$
add a comment |
The idea is correct, but you cannot write like that, because the series involved are both divergent and generally it is moot to do arithmetic operations of divergent series.
Let $H_n = sum_1^n 1/k$, i.e. the $n$-th partial sum of the harmonic series, then the
$$
S_n = H_{3n} - frac 23 H_n,
$$
then use the asymptotic expression $H_n = log n + gamma + varepsilon_n$ where $varepsilon_n to 0 [n to infty]$, we have
$$
S_n = log (3n) -frac 23log n + frac 13 gamma + varepsilon_{3n} - frac 23 varepsilon_n = log(3n^{1/3}) + frac 13 gamma + alpha_n xrightarrow{n to infty} +infty.
$$
add a comment |
The idea is correct, but you cannot write like that, because the series involved are both divergent and generally it is moot to do arithmetic operations of divergent series.
Let $H_n = sum_1^n 1/k$, i.e. the $n$-th partial sum of the harmonic series, then the
$$
S_n = H_{3n} - frac 23 H_n,
$$
then use the asymptotic expression $H_n = log n + gamma + varepsilon_n$ where $varepsilon_n to 0 [n to infty]$, we have
$$
S_n = log (3n) -frac 23log n + frac 13 gamma + varepsilon_{3n} - frac 23 varepsilon_n = log(3n^{1/3}) + frac 13 gamma + alpha_n xrightarrow{n to infty} +infty.
$$
The idea is correct, but you cannot write like that, because the series involved are both divergent and generally it is moot to do arithmetic operations of divergent series.
Let $H_n = sum_1^n 1/k$, i.e. the $n$-th partial sum of the harmonic series, then the
$$
S_n = H_{3n} - frac 23 H_n,
$$
then use the asymptotic expression $H_n = log n + gamma + varepsilon_n$ where $varepsilon_n to 0 [n to infty]$, we have
$$
S_n = log (3n) -frac 23log n + frac 13 gamma + varepsilon_{3n} - frac 23 varepsilon_n = log(3n^{1/3}) + frac 13 gamma + alpha_n xrightarrow{n to infty} +infty.
$$
answered yesterday
xbh
5,7651522
5,7651522
add a comment |
add a comment |
Raúl Astete is a new contributor. Be nice, and check out our Code of Conduct.
Raúl Astete is a new contributor. Be nice, and check out our Code of Conduct.
Raúl Astete is a new contributor. Be nice, and check out our Code of Conduct.
Raúl Astete is a new contributor. Be nice, and check out our Code of Conduct.
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2
your solution is not correct. You cannot change the order of summation of a series that doesn't converge absolutely, neither make (standard) arithmetic with divergent series
– Masacroso
yesterday
@Masacroso Is it not fine to do something along the lines of " Suppose the following series converges. Then in particular, it converges absolutely, so we can reorder the series in any way. One particular ordering gives a scalar multiple of the harmonic series which diverges, so the given series cannot converge".
– Saad
yesterday
4
@Saad That is not a valid argument because convergence doesn't imply absolute convergence (you've got it the wrong way round).
– AlephNull
22 hours ago