Attempt on modified harmonic series
Does the series $left(1+frac{1}{2}-frac{1}{3} right) + left(frac{1}{4}+frac{1}{5}-frac{1}{6} right)+left(frac{1}{7}+frac{1}{8}-frac{1}{9} right)+ldots quad$ converge?
Here's my attempt at a solution: begin{eqnarray*}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)+ldots &= 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} end{eqnarray*}
As we can "rewrite" this series as one third of the harmonical series (that diverges), we conclude the divergence of the requiered series.
Is this right? Which other convergence tests could be used?
Thanks in advance.
calculus sequences-and-series proof-verification divergent-series
New contributor
add a comment |
Does the series $left(1+frac{1}{2}-frac{1}{3} right) + left(frac{1}{4}+frac{1}{5}-frac{1}{6} right)+left(frac{1}{7}+frac{1}{8}-frac{1}{9} right)+ldots quad$ converge?
Here's my attempt at a solution: begin{eqnarray*}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)+ldots &= 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} end{eqnarray*}
As we can "rewrite" this series as one third of the harmonical series (that diverges), we conclude the divergence of the requiered series.
Is this right? Which other convergence tests could be used?
Thanks in advance.
calculus sequences-and-series proof-verification divergent-series
New contributor
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
38 mins ago
@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
32 mins ago
add a comment |
Does the series $left(1+frac{1}{2}-frac{1}{3} right) + left(frac{1}{4}+frac{1}{5}-frac{1}{6} right)+left(frac{1}{7}+frac{1}{8}-frac{1}{9} right)+ldots quad$ converge?
Here's my attempt at a solution: begin{eqnarray*}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)+ldots &= 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} end{eqnarray*}
As we can "rewrite" this series as one third of the harmonical series (that diverges), we conclude the divergence of the requiered series.
Is this right? Which other convergence tests could be used?
Thanks in advance.
calculus sequences-and-series proof-verification divergent-series
New contributor
Does the series $left(1+frac{1}{2}-frac{1}{3} right) + left(frac{1}{4}+frac{1}{5}-frac{1}{6} right)+left(frac{1}{7}+frac{1}{8}-frac{1}{9} right)+ldots quad$ converge?
Here's my attempt at a solution: begin{eqnarray*}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)+ldots &= 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} end{eqnarray*}
As we can "rewrite" this series as one third of the harmonical series (that diverges), we conclude the divergence of the requiered series.
Is this right? Which other convergence tests could be used?
Thanks in advance.
calculus sequences-and-series proof-verification divergent-series
calculus sequences-and-series proof-verification divergent-series
New contributor
New contributor
New contributor
asked 1 hour ago
Raúl Astete
255
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New contributor
New contributor
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
38 mins ago
@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
32 mins ago
add a comment |
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
38 mins ago
@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
32 mins ago
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
38 mins ago
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
38 mins ago
@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
32 mins ago
@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
32 mins ago
add a comment |
3 Answers
3
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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.
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 |
Yes, your answer is correct.
By using Cauchy's Convergence test, for $sum_{n=1}^inftydfrac1n$
Taking $$S_{2n}-S_n=sum_{n=1}^{2n}dfrac1n-sum_{n=1}^ndfrac1n=dfrac{1}{n+1}+dfrac{1}{n+2}+....+dfrac{1}{2n}gedfrac{1}{2n}+....+dfrac{1}{2n}=dfrac{n}{2n}=dfrac{1}{2}$$
Since, there is no number $N$ that satisfies Cauchy's condition, the series diverges.
add a comment |
Your Answer
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3 Answers
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3 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.
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.
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 44 mins ago
answered 50 mins ago
Ben W
1,815514
1,815514
add a comment |
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 40 mins ago
xbh
5,6951522
5,6951522
add a comment |
add a comment |
Yes, your answer is correct.
By using Cauchy's Convergence test, for $sum_{n=1}^inftydfrac1n$
Taking $$S_{2n}-S_n=sum_{n=1}^{2n}dfrac1n-sum_{n=1}^ndfrac1n=dfrac{1}{n+1}+dfrac{1}{n+2}+....+dfrac{1}{2n}gedfrac{1}{2n}+....+dfrac{1}{2n}=dfrac{n}{2n}=dfrac{1}{2}$$
Since, there is no number $N$ that satisfies Cauchy's condition, the series diverges.
add a comment |
Yes, your answer is correct.
By using Cauchy's Convergence test, for $sum_{n=1}^inftydfrac1n$
Taking $$S_{2n}-S_n=sum_{n=1}^{2n}dfrac1n-sum_{n=1}^ndfrac1n=dfrac{1}{n+1}+dfrac{1}{n+2}+....+dfrac{1}{2n}gedfrac{1}{2n}+....+dfrac{1}{2n}=dfrac{n}{2n}=dfrac{1}{2}$$
Since, there is no number $N$ that satisfies Cauchy's condition, the series diverges.
add a comment |
Yes, your answer is correct.
By using Cauchy's Convergence test, for $sum_{n=1}^inftydfrac1n$
Taking $$S_{2n}-S_n=sum_{n=1}^{2n}dfrac1n-sum_{n=1}^ndfrac1n=dfrac{1}{n+1}+dfrac{1}{n+2}+....+dfrac{1}{2n}gedfrac{1}{2n}+....+dfrac{1}{2n}=dfrac{n}{2n}=dfrac{1}{2}$$
Since, there is no number $N$ that satisfies Cauchy's condition, the series diverges.
Yes, your answer is correct.
By using Cauchy's Convergence test, for $sum_{n=1}^inftydfrac1n$
Taking $$S_{2n}-S_n=sum_{n=1}^{2n}dfrac1n-sum_{n=1}^ndfrac1n=dfrac{1}{n+1}+dfrac{1}{n+2}+....+dfrac{1}{2n}gedfrac{1}{2n}+....+dfrac{1}{2n}=dfrac{n}{2n}=dfrac{1}{2}$$
Since, there is no number $N$ that satisfies Cauchy's condition, the series diverges.
answered 54 mins ago
Key Flex
7,56241232
7,56241232
add a comment |
add a comment |
Raúl Astete is a new contributor. Be nice, and check out our Code of Conduct.
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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
38 mins ago
@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
32 mins ago