This is a result of the Riemann zeta function where: \[\zeta\ (-1) = \displaystyle \sum_{i=1}^{\infty} n = -\frac{1}{12}\]
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Curtis Clement
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1 year, 8 months ago

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but how?

i do have a very interesting case

1/(1+x) = 1-x+x^2-x^3+x^4 ...

putting x=1

1/2 = 1-1+1-1+1-1+1-1...... (which is awesome because at any finite point if u stop the series, u get 1 or -1 or 0, but at infinity u get 1/2, this is weird because it is not even the limiting case that some calculator can check)
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Mvs Saketh
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1 year, 8 months ago

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@Mvs Saketh
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First of all, the following series: \(\dfrac{1}{1+x}=1-x+x^2-x^3+x^4-\ldots \infty\) is the Maclaurin series of \(f(x)=\dfrac{1}{1+x}\) and is valid (converges) iff \(|x|\lt 1\).

When \(x=1\), then the infinite series becomes equivalent with the Grandi's series which is divergent and thus technically has no value. But, one can use Caesaro Summation to provide a value to this divergent sum by checking what value does the partial sums approach.

It's not hard to see that the partial sums approach to \(\dfrac{1}{2}\), and as such, the Caesaro summation value of this divergent sum is \(\dfrac{1}{2}\), although the series has no value in the technical sense.
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Prasun Biswas
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1 year, 8 months ago

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@Prasun Biswas
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Yes bro i have a doubt regarding it, when we prove the taylor series, where exactly do we assume that the series must be convergent for the series to work, ?

Definitely i understand that it is necessary for it to be convergent and after we derive the taylor series, we check that it is convergent, but we dont need to assume that to prove taylor series,
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Mvs Saketh
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1 year, 8 months ago

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@Mvs Saketh
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In this case though, note that the expansion on RHS is an infinite geometric progression with \(a=1\) and \(r=(-x)\). As such, we know that an infinite GP converges iff \(|r|\lt 1\). So, we must have,

What do you want to say ? Are you thinking of a fantasy world of Mathematics ? @Mayank Singh
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Sandeep Bhardwaj
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1 year, 8 months ago

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@Sandeep Bhardwaj
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It is just a logical extension of the Riemann zeta function in calculus. In string theory the result of (is thought to be)\[\large {\sum_{1}^{\infty} n = -\dfrac {1}{12}}\]

P.s. It must be remembered that this is just an analytic contiuation. The Euler zeta function is actually only valid for \(x>1\). So it can be regaurded as a fallacy.
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Sualeh Asif
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1 year, 8 months ago

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@Sandeep Bhardwaj
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These formulas were given by Ramanujan for divergent series like the one mvs & mayank shared.
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Krishna Sharma
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1 year, 8 months ago

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If you want to sum that series, you need to restrict yourself a bit.

Adding down the columns, we arrive at \(0=1\)! The only conclusion is that, if we want to attribute this sum any finite value, we can't shift over the summands like that — either that, or we can't distribute, or we can't add down. In any case, if we were to sum this, we'd have to either give up some rules on how infinite sums behave, or get a contradiction.
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Akiva Weinberger
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1 year, 5 months ago

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TopNewestThis is a result of the Riemann zeta function where: \[\zeta\ (-1) = \displaystyle \sum_{i=1}^{\infty} n = -\frac{1}{12}\] – Curtis Clement · 1 year, 8 months ago

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but how?

i do have a very interesting case

1/(1+x) = 1-x+x^2-x^3+x^4 ...

putting x=1

1/2 = 1-1+1-1+1-1+1-1...... (which is awesome because at any finite point if u stop the series, u get 1 or -1 or 0, but at infinity u get 1/2, this is weird because it is not even the limiting case that some calculator can check) – Mvs Saketh · 1 year, 8 months ago

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When \(x=1\), then the infinite series becomes equivalent with the Grandi's series which is divergent and thus technically has no value. But, one can use Caesaro Summation to provide a value to this divergent sum by checking what value does the partial sums approach.

The partial sums for the series are:

\[\frac{1}{1},\frac{1}{2},\frac{2}{3},\frac{2}{4},\frac{3}{5},\frac{3}{6},\ldots\]

It's not hard to see that the partial sums approach to \(\dfrac{1}{2}\), and as such, the Caesaro summation value of this divergent sum is \(\dfrac{1}{2}\), although the series has no value in the technical sense. – Prasun Biswas · 1 year, 8 months ago

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Definitely i understand that it is necessary for it to be convergent and after we derive the taylor series, we check that it is convergent, but we dont need to assume that to prove taylor series, – Mvs Saketh · 1 year, 8 months ago

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\[|r|\lt 1\implies |(-x)|\lt 1\implies |x|\lt 1\] – Prasun Biswas · 1 year, 8 months ago

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What do you want to say ? Are you thinking of a fantasy world of Mathematics ? @Mayank Singh – Sandeep Bhardwaj · 1 year, 8 months ago

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Riemann zeta functionin calculus. In string theory the result of (is thought to be)\[\large {\sum_{1}^{\infty} n = -\dfrac {1}{12}}\]You might want to see this

P.s. It must be remembered that this is just an analytic contiuation. The

Euler zeta functionis actually only valid for \(x>1\). So it can be regaurded as a fallacy. – Sualeh Asif · 1 year, 8 months agoLog in to reply

– Krishna Sharma · 1 year, 8 months ago

These formulas were given by Ramanujan for divergent series like the one mvs & mayank shared.Log in to reply

If you want to sum that series, you need to restrict yourself a bit.

Call the sum \(S\). Thus:

\(\phantom{-2}S=1+2+3+4+5+\phantom16+\dotsb\)

\(-2S=\phantom1-2-4-6-8-10-\dotsb\)

\(\phantom{-2}S=\phantom{1+2+{}}1+2+3+\phantom14+\dotsb\)

Adding down the columns, we arrive at \(0=1\)! The only conclusion is that, if we want to attribute this sum any finite value, we can't shift over the summands like that — either that, or we can't distribute, or we can't add down. In any case, if we were to sum this, we'd have to either give up some rules on how infinite sums behave, or get a contradiction. – Akiva Weinberger · 1 year, 5 months ago

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