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Divergence of this series and further investigation
Ways to make a series diverge “faster” to show divergenceHelp with proving that this series divergesProving the convergence/divergence of a seemingly oscillating seriesConvergence and Divergence and Using Various MethodsSeries Divergence ProofProving divergence of a sequenceNewton-Raphson method for the functionThe choice of $epsilon$ in the proofs of divergenceDivergence of power seriesA Sequence converges or diverges
$begingroup$
I was wondering if $1 + frac12 - frac13 +frac14 + frac15 - frac16+...$ diverges? I suspect that it does. I found the general term as $$frac2n^2 -1 - 3floor(frac2n^23)n,$$ which, to me, strongly suggested divergence, but I can't quite formalise the argument from there.
As a further investigation, I'm curious if there exists a function such that $sum |f(n)|$ diverges, but
$$f(1) - f(2) + f(3) ...,$$
$$f(1)+f(2)-f(3)+f(4)+f(5)-f(6)+...,$$
$$f(1)+f(2)+f(3)-f(4)+...$$
And so on, all converge.
I apologise for the bad formatting and notation, I don't know a better way to express the question, but I can clarify anything need be. I know this is technically two questions, but I wanted to also show that $f(n) = frac1n$ is not such a function, which made me curious if there IS such a function satisfying the above; if so, what are the necessary conditions a function must have for this to be true?
calculus sequences-and-series
asked Mar 21 at 5:38
Ryan GouldenRyan Goulden
500310
$endgroup$
add a comment |
$begingroup$
I was wondering if $1 + frac12 - frac13 +frac14 + frac15 - frac16+...$ diverges? I suspect that it does. I found the general term as $$frac2n^2 -1 - 3floor(frac2n^23)n,$$ which, to me, strongly suggested divergence, but I can't quite formalise the argument from there.
As a further investigation, I'm curious if there exists a function such that $sum |f(n)|$ diverges, but
$$f(1) - f(2) + f(3) ...,$$
$$f(1)+f(2)-f(3)+f(4)+f(5)-f(6)+...,$$
$$f(1)+f(2)+f(3)-f(4)+...$$
And so on, all converge.
I apologise for the bad formatting and notation, I don't know a better way to express the question, but I can clarify anything need be. I know this is technically two questions, but I wanted to also show that $f(n) = frac1n$ is not such a function, which made me curious if there IS such a function satisfying the above; if so, what are the necessary conditions a function must have for this to be true?
calculus sequences-and-series
asked Mar 21 at 5:38
Ryan GouldenRyan Goulden
500310
$endgroup$
$begingroup$
So its two terms added, then 1 term subtracted?
$endgroup$
– Don Thousand
Mar 21 at 5:49
$begingroup$
@DonThousand correct. Similar to an alternating series, where every second term is negative, this has it so that every third term is negative, the rest being positive.
$endgroup$
– Ryan Goulden
Mar 21 at 5:50
$begingroup$
@DonThousand the numerator in the general term I found for the sequence resolves to (1,1,-1,1,1,-1,...)
$endgroup$
– Ryan Goulden
Mar 21 at 5:51
$begingroup$
@SujitBhattacharyya that's the sum for the alternating harmonic series, this isn't quite that.
$endgroup$
– Ryan Goulden
Mar 21 at 5:52
add a comment |
$begingroup$
I was wondering if $1 + frac12 - frac13 +frac14 + frac15 - frac16+...$ diverges? I suspect that it does. I found the general term as $$frac2n^2 -1 - 3floor(frac2n^23)n,$$ which, to me, strongly suggested divergence, but I can't quite formalise the argument from there.
As a further investigation, I'm curious if there exists a function such that $sum |f(n)|$ diverges, but
$$f(1) - f(2) + f(3) ...,$$
$$f(1)+f(2)-f(3)+f(4)+f(5)-f(6)+...,$$
$$f(1)+f(2)+f(3)-f(4)+...$$
And so on, all converge.
I apologise for the bad formatting and notation, I don't know a better way to express the question, but I can clarify anything need be. I know this is technically two questions, but I wanted to also show that $f(n) = frac1n$ is not such a function, which made me curious if there IS such a function satisfying the above; if so, what are the necessary conditions a function must have for this to be true?
calculus sequences-and-series
asked Mar 21 at 5:38
Ryan GouldenRyan Goulden
500310
$endgroup$
I was wondering if $1 + frac12 - frac13 +frac14 + frac15 - frac16+...$ diverges? I suspect that it does. I found the general term as $$frac2n^2 -1 - 3floor(frac2n^23)n,$$ which, to me, strongly suggested divergence, but I can't quite formalise the argument from there.
As a further investigation, I'm curious if there exists a function such that $sum |f(n)|$ diverges, but
$$f(1) - f(2) + f(3) ...,$$
$$f(1)+f(2)-f(3)+f(4)+f(5)-f(6)+...,$$
$$f(1)+f(2)+f(3)-f(4)+...$$
And so on, all converge.
I apologise for the bad formatting and notation, I don't know a better way to express the question, but I can clarify anything need be. I know this is technically two questions, but I wanted to also show that $f(n) = frac1n$ is not such a function, which made me curious if there IS such a function satisfying the above; if so, what are the necessary conditions a function must have for this to be true?
calculus sequences-and-series
calculus sequences-and-series
asked Mar 21 at 5:38
Ryan GouldenRyan Goulden
500310
asked Mar 21 at 5:38
Ryan GouldenRyan Goulden
500310
edited Mar 21 at 5:54
Ryan Goulden
asked Mar 21 at 5:38
Ryan GouldenRyan Goulden
500310
asked Mar 21 at 5:38
Ryan GouldenRyan Goulden
500310
asked Mar 21 at 5:38
Ryan GouldenRyan Goulden
500310
500310
$begingroup$
So its two terms added, then 1 term subtracted?
$endgroup$
– Don Thousand
Mar 21 at 5:49
$begingroup$
@DonThousand correct. Similar to an alternating series, where every second term is negative, this has it so that every third term is negative, the rest being positive.
$endgroup$
– Ryan Goulden
Mar 21 at 5:50
$begingroup$
@DonThousand the numerator in the general term I found for the sequence resolves to (1,1,-1,1,1,-1,...)
$endgroup$
– Ryan Goulden
Mar 21 at 5:51
$begingroup$
@SujitBhattacharyya that's the sum for the alternating harmonic series, this isn't quite that.
$endgroup$
– Ryan Goulden
Mar 21 at 5:52
add a comment |
$begingroup$
So its two terms added, then 1 term subtracted?
$endgroup$
– Don Thousand
Mar 21 at 5:49
$begingroup$
@DonThousand correct. Similar to an alternating series, where every second term is negative, this has it so that every third term is negative, the rest being positive.
$endgroup$
– Ryan Goulden
Mar 21 at 5:50
$begingroup$
@DonThousand the numerator in the general term I found for the sequence resolves to (1,1,-1,1,1,-1,...)
$endgroup$
– Ryan Goulden
Mar 21 at 5:51
$begingroup$
@SujitBhattacharyya that's the sum for the alternating harmonic series, this isn't quite that.
$endgroup$
– Ryan Goulden
Mar 21 at 5:52
$begingroup$
So its two terms added, then 1 term subtracted?
$endgroup$
– Don Thousand
Mar 21 at 5:49
$begingroup$
So its two terms added, then 1 term subtracted?
$endgroup$
– Don Thousand
Mar 21 at 5:49
$begingroup$
@DonThousand correct. Similar to an alternating series, where every second term is negative, this has it so that every third term is negative, the rest being positive.
$endgroup$
– Ryan Goulden
Mar 21 at 5:50
$begingroup$
@DonThousand correct. Similar to an alternating series, where every second term is negative, this has it so that every third term is negative, the rest being positive.
$endgroup$
– Ryan Goulden
Mar 21 at 5:50
$begingroup$
@DonThousand the numerator in the general term I found for the sequence resolves to (1,1,-1,1,1,-1,...)
$endgroup$
– Ryan Goulden
Mar 21 at 5:51
$begingroup$
@DonThousand the numerator in the general term I found for the sequence resolves to (1,1,-1,1,1,-1,...)
$endgroup$
– Ryan Goulden
Mar 21 at 5:51
$begingroup$
@SujitBhattacharyya that's the sum for the alternating harmonic series, this isn't quite that.
$endgroup$
– Ryan Goulden
Mar 21 at 5:52
$begingroup$
@SujitBhattacharyya that's the sum for the alternating harmonic series, this isn't quite that.
$endgroup$
– Ryan Goulden
Mar 21 at 5:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: $1+frac12-frac13+frac14+frac15-frac16+cdots+frac13k-2+frac13k-1-frac13k > 1+frac14+cdots+frac13k-2$.
answered Mar 21 at 5:55
Greg MartinGreg Martin
36.5k23565
$endgroup$
$begingroup$
Wow, that's so simple I feel utterly dumb for not immediately thinking that...
$endgroup$
– Don Thousand
Mar 21 at 5:57
$begingroup$
Everything is simpler after we see the idea! No need to feel dumb. Indeed, a better answer (though less simple) is: the 1 (mod 3) terms up to $x$ add up to roughly $frac13log x$ (makes sense, given how the harmonic series itself grows); the 2 (mod 3) terms up to $x$ add up to roughly $frac13log x$ as well; and the 0 (mod 3) terms up to $x$ add up to roughly $-frac13log x$. You get the actual asymptotic rate of divergence this way; and it generalizes to any periodic sequence of signs (indeed of general coefficients): it converges if and only if the coefficients sum to $0$ in each period.
$endgroup$
– Greg Martin
Mar 21 at 6:02
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
Hint: $1+frac12-frac13+frac14+frac15-frac16+cdots+frac13k-2+frac13k-1-frac13k > 1+frac14+cdots+frac13k-2$.
answered Mar 21 at 5:55
Greg MartinGreg Martin
36.5k23565
$endgroup$
$begingroup$
Wow, that's so simple I feel utterly dumb for not immediately thinking that...
$endgroup$
– Don Thousand
Mar 21 at 5:57
$begingroup$
Everything is simpler after we see the idea! No need to feel dumb. Indeed, a better answer (though less simple) is: the 1 (mod 3) terms up to $x$ add up to roughly $frac13log x$ (makes sense, given how the harmonic series itself grows); the 2 (mod 3) terms up to $x$ add up to roughly $frac13log x$ as well; and the 0 (mod 3) terms up to $x$ add up to roughly $-frac13log x$. You get the actual asymptotic rate of divergence this way; and it generalizes to any periodic sequence of signs (indeed of general coefficients): it converges if and only if the coefficients sum to $0$ in each period.
$endgroup$
– Greg Martin
Mar 21 at 6:02
add a comment |
$begingroup$
Hint: $1+frac12-frac13+frac14+frac15-frac16+cdots+frac13k-2+frac13k-1-frac13k > 1+frac14+cdots+frac13k-2$.
answered Mar 21 at 5:55
Greg MartinGreg Martin
36.5k23565
$endgroup$
$begingroup$
Wow, that's so simple I feel utterly dumb for not immediately thinking that...
$endgroup$
– Don Thousand
Mar 21 at 5:57
$begingroup$
Everything is simpler after we see the idea! No need to feel dumb. Indeed, a better answer (though less simple) is: the 1 (mod 3) terms up to $x$ add up to roughly $frac13log x$ (makes sense, given how the harmonic series itself grows); the 2 (mod 3) terms up to $x$ add up to roughly $frac13log x$ as well; and the 0 (mod 3) terms up to $x$ add up to roughly $-frac13log x$. You get the actual asymptotic rate of divergence this way; and it generalizes to any periodic sequence of signs (indeed of general coefficients): it converges if and only if the coefficients sum to $0$ in each period.
$endgroup$
– Greg Martin
Mar 21 at 6:02
add a comment |
$begingroup$
Hint: $1+frac12-frac13+frac14+frac15-frac16+cdots+frac13k-2+frac13k-1-frac13k > 1+frac14+cdots+frac13k-2$.
answered Mar 21 at 5:55
Greg MartinGreg Martin
36.5k23565
$endgroup$
Hint: $1+frac12-frac13+frac14+frac15-frac16+cdots+frac13k-2+frac13k-1-frac13k > 1+frac14+cdots+frac13k-2$.
answered Mar 21 at 5:55
Greg MartinGreg Martin
36.5k23565
answered Mar 21 at 5:55
Greg MartinGreg Martin
36.5k23565
answered Mar 21 at 5:55
Greg MartinGreg Martin
36.5k23565
answered Mar 21 at 5:55
Greg MartinGreg Martin
36.5k23565
36.5k23565
$begingroup$
Wow, that's so simple I feel utterly dumb for not immediately thinking that...
$endgroup$
– Don Thousand
Mar 21 at 5:57
$begingroup$
Everything is simpler after we see the idea! No need to feel dumb. Indeed, a better answer (though less simple) is: the 1 (mod 3) terms up to $x$ add up to roughly $frac13log x$ (makes sense, given how the harmonic series itself grows); the 2 (mod 3) terms up to $x$ add up to roughly $frac13log x$ as well; and the 0 (mod 3) terms up to $x$ add up to roughly $-frac13log x$. You get the actual asymptotic rate of divergence this way; and it generalizes to any periodic sequence of signs (indeed of general coefficients): it converges if and only if the coefficients sum to $0$ in each period.
$endgroup$
– Greg Martin
Mar 21 at 6:02
add a comment |
$begingroup$
Wow, that's so simple I feel utterly dumb for not immediately thinking that...
$endgroup$
– Don Thousand
Mar 21 at 5:57
$begingroup$
Everything is simpler after we see the idea! No need to feel dumb. Indeed, a better answer (though less simple) is: the 1 (mod 3) terms up to $x$ add up to roughly $frac13log x$ (makes sense, given how the harmonic series itself grows); the 2 (mod 3) terms up to $x$ add up to roughly $frac13log x$ as well; and the 0 (mod 3) terms up to $x$ add up to roughly $-frac13log x$. You get the actual asymptotic rate of divergence this way; and it generalizes to any periodic sequence of signs (indeed of general coefficients): it converges if and only if the coefficients sum to $0$ in each period.
$endgroup$
– Greg Martin
Mar 21 at 6:02
$begingroup$
Wow, that's so simple I feel utterly dumb for not immediately thinking that...
$endgroup$
– Don Thousand
Mar 21 at 5:57
$begingroup$
Wow, that's so simple I feel utterly dumb for not immediately thinking that...
$endgroup$
– Don Thousand
Mar 21 at 5:57
$begingroup$
Everything is simpler after we see the idea! No need to feel dumb. Indeed, a better answer (though less simple) is: the 1 (mod 3) terms up to $x$ add up to roughly $frac13log x$ (makes sense, given how the harmonic series itself grows); the 2 (mod 3) terms up to $x$ add up to roughly $frac13log x$ as well; and the 0 (mod 3) terms up to $x$ add up to roughly $-frac13log x$. You get the actual asymptotic rate of divergence this way; and it generalizes to any periodic sequence of signs (indeed of general coefficients): it converges if and only if the coefficients sum to $0$ in each period.
$endgroup$
– Greg Martin
Mar 21 at 6:02
$begingroup$
Everything is simpler after we see the idea! No need to feel dumb. Indeed, a better answer (though less simple) is: the 1 (mod 3) terms up to $x$ add up to roughly $frac13log x$ (makes sense, given how the harmonic series itself grows); the 2 (mod 3) terms up to $x$ add up to roughly $frac13log x$ as well; and the 0 (mod 3) terms up to $x$ add up to roughly $-frac13log x$. You get the actual asymptotic rate of divergence this way; and it generalizes to any periodic sequence of signs (indeed of general coefficients): it converges if and only if the coefficients sum to $0$ in each period.
$endgroup$
– Greg Martin
Mar 21 at 6:02
add a comment |
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$begingroup$
So its two terms added, then 1 term subtracted?
$endgroup$
– Don Thousand
Mar 21 at 5:49
$begingroup$
@DonThousand correct. Similar to an alternating series, where every second term is negative, this has it so that every third term is negative, the rest being positive.
$endgroup$
– Ryan Goulden
Mar 21 at 5:50
$begingroup$
@DonThousand the numerator in the general term I found for the sequence resolves to (1,1,-1,1,1,-1,...)
$endgroup$
– Ryan Goulden
Mar 21 at 5:51
$begingroup$
@SujitBhattacharyya that's the sum for the alternating harmonic series, this isn't quite that.
$endgroup$
– Ryan Goulden
Mar 21 at 5:52