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
edited 2 days ago
Did
246k23221456
asked Jan 6 at 2:43
Raúl AsteteRaúl Astete
496
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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
edited 2 days ago
Did
246k23221456
asked Jan 6 at 2:43
Raúl AsteteRaúl Astete
496
New contributor
Raúl Astete is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
Jan 6 at 3:15
@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
Jan 6 at 3:21
5
@Saad That is not a valid argument because convergence doesn't imply absolute convergence (you've got it the wrong way round).
– AlephNull
2 days ago
Of course, you're correct. For whatever reason, I thought each term was positive..
– Saad
yesterday
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
edited 2 days ago
Did
246k23221456
asked Jan 6 at 2:43
Raúl AsteteRaúl Astete
496
New contributor
Raúl Astete is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
edited 2 days ago
Did
246k23221456
asked Jan 6 at 2:43
Raúl AsteteRaúl Astete
496
New contributor
Raúl Astete is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
Did
246k23221456
asked Jan 6 at 2:43
Raúl AsteteRaúl Astete
496
New contributor
Raúl Astete is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
Did
246k23221456
edited 2 days ago
Did
246k23221456
edited 2 days ago
Did
246k23221456
246k23221456
asked Jan 6 at 2:43
Raúl AsteteRaúl Astete
496
New contributor
Raúl Astete is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 6 at 2:43
Raúl AsteteRaúl Astete
496
asked Jan 6 at 2:43
Raúl AsteteRaúl Astete
496
496
New contributor
Raúl Astete is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Raúl Astete is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Raúl Astete is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
Jan 6 at 3:15
@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
Jan 6 at 3:21
5
@Saad That is not a valid argument because convergence doesn't imply absolute convergence (you've got it the wrong way round).
– AlephNull
2 days ago
Of course, you're correct. For whatever reason, I thought each term was positive..
– Saad
yesterday
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
Jan 6 at 3:15
@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
Jan 6 at 3:21
5
@Saad That is not a valid argument because convergence doesn't imply absolute convergence (you've got it the wrong way round).
– AlephNull
2 days ago
Of course, you're correct. For whatever reason, I thought each term was positive..
– Saad
yesterday
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
Jan 6 at 3:15
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
Jan 6 at 3:15
@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
Jan 6 at 3:21
@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
Jan 6 at 3:21
5
5
@Saad That is not a valid argument because convergence doesn't imply absolute convergence (you've got it the wrong way round).
– AlephNull
2 days ago
@Saad That is not a valid argument because convergence doesn't imply absolute convergence (you've got it the wrong way round).
– AlephNull
2 days ago
Of course, you're correct. For whatever reason, I thought each term was positive..
– Saad
yesterday
Of course, you're correct. For whatever reason, I thought each term was positive..
– Saad
yesterday
add a comment |
2 Answers
2
active
oldest
votes
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.
answered Jan 6 at 3:03
Ben WBen W
2,007615
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
Jan 6 at 4:32
@zwim quite true, that is a nice little shortcut
– Ben W
2 days ago
Simplify and it behaves like $1/n$, so it diverges.
– ncmathsadist
2 days 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
$$
S_n = log (3n) -frac 23log n + frac 13 gamma + varepsilon_{3n} - frac 23 varepsilon_n = log(3n^{1/3}) + frac 13 gamma + alpha_n xrightarrow{n to infty} +infty.
$$
answered Jan 6 at 3:13
xbhxbh
5,7651522
add a comment |
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2 Answers
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active
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2 Answers
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oldest
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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.
answered Jan 6 at 3:03
Ben WBen W
2,007615
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
Jan 6 at 4:32
@zwim quite true, that is a nice little shortcut
– Ben W
2 days ago
Simplify and it behaves like $1/n$, so it diverges.
– ncmathsadist
2 days 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.
answered Jan 6 at 3:03
Ben WBen W
2,007615
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
Jan 6 at 4:32
@zwim quite true, that is a nice little shortcut
– Ben W
2 days ago
Simplify and it behaves like $1/n$, so it diverges.
– ncmathsadist
2 days 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.
answered Jan 6 at 3:03
Ben WBen W
2,007615
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.
answered Jan 6 at 3:03
Ben WBen W
2,007615
edited Jan 6 at 3:09
answered Jan 6 at 3:03
Ben WBen W
2,007615
answered Jan 6 at 3:03
Ben WBen W
2,007615
answered Jan 6 at 3:03
Ben WBen W
2,007615
2,007615
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
Jan 6 at 4:32
@zwim quite true, that is a nice little shortcut
– Ben W
2 days ago
Simplify and it behaves like $1/n$, so it diverges.
– ncmathsadist
2 days 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
Jan 6 at 4:32
@zwim quite true, that is a nice little shortcut
– Ben W
2 days ago
Simplify and it behaves like $1/n$, so it diverges.
– ncmathsadist
2 days 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
Jan 6 at 4:32
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
Jan 6 at 4:32
@zwim quite true, that is a nice little shortcut
– Ben W
2 days ago
@zwim quite true, that is a nice little shortcut
– Ben W
2 days ago
Simplify and it behaves like $1/n$, so it diverges.
– ncmathsadist
2 days ago
Simplify and it behaves like $1/n$, so it diverges.
– ncmathsadist
2 days 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
$$
S_n = log (3n) -frac 23log n + frac 13 gamma + varepsilon_{3n} - frac 23 varepsilon_n = log(3n^{1/3}) + frac 13 gamma + alpha_n xrightarrow{n to infty} +infty.
$$
answered Jan 6 at 3:13
xbhxbh
5,7651522
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
$$
S_n = log (3n) -frac 23log n + frac 13 gamma + varepsilon_{3n} - frac 23 varepsilon_n = log(3n^{1/3}) + frac 13 gamma + alpha_n xrightarrow{n to infty} +infty.
$$
answered Jan 6 at 3:13
xbhxbh
5,7651522
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
$$
S_n = log (3n) -frac 23log n + frac 13 gamma + varepsilon_{3n} - frac 23 varepsilon_n = log(3n^{1/3}) + frac 13 gamma + alpha_n xrightarrow{n to infty} +infty.
$$
answered Jan 6 at 3:13
xbhxbh
5,7651522
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 Jan 6 at 3:13
xbhxbh
5,7651522
answered Jan 6 at 3:13
xbhxbh
5,7651522
answered Jan 6 at 3:13
xbhxbh
5,7651522
answered Jan 6 at 3:13
xbhxbh
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
Jan 6 at 3:15
@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
Jan 6 at 3:21
5
@Saad That is not a valid argument because convergence doesn't imply absolute convergence (you've got it the wrong way round).
– AlephNull
2 days ago
Of course, you're correct. For whatever reason, I thought each term was positive..
– Saad
yesterday