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Mathematics: This Will Blow Your Mind

Don't use the internet.

Estimate the value of these sums:

a) 1+2+3+4+...

b)1^2+2^2+3^2+...

The solution involves the Riemann zeta function but tell me what you think.

_________________

"God may not play dice with the universe, but something strange is going on with prime numbers."

-Paul Erdos

"There are two types of cryptography in this world: cryptography that will stop your kid sister from looking at your files, and cryptography that will stop major governments from reading your files."

-Bruce Schneider

You can add, and add, and add and still add even more - it never ends.

Seems correct however it's miraculously not.

1+2+3+4+...=-1/12 and even more miraculously 1^2+2^2+3^2+...=0. It has to do with a formula for Riemann zeta function when the variable is negative. when s<0. When s is negative and even the sum is 0.

BTW: 1+2+3+...=ζ(-1) , 1^2+2^2+3^2+...=ζ(-2).

_________________

"God may not play dice with the universe, but something strange is going on with prime numbers."

-Paul Erdos

"There are two types of cryptography in this world: cryptography that will stop your kid sister from looking at your files, and cryptography that will stop major governments from reading your files."

-Bruce Schneider

Nice write up with satisfying points of clarification at SciAm:

http://blogs.scientificamerican.com/roo ... equal-112/

**Quote:**

The first time I saw this, I thought it was going to be one of those proofs where a false result is proven with the help of division by zero. Looks like I was wrong. Even after completing a math degree it's amazing how much mathematics there still is to learn.

I hope people followed the link to the Scientific American article. It's well worth reading, not least for this:

http://blogs.scientificamerican.com/roo ... equal-112/

Estimate the value of these sums:

a) 1+2+3+4+...

b)1^2+2^2+3^2+...

The solution involves the Riemann zeta function but tell me what you think.

Neither series converges. If a series converges then its n-th term approaches 0 as n goes to infinity. Clearly this does not happen with either series, hence they do not converge.

Estimate the value of these sums:

a) 1+2+3+4+...

b)1^2+2^2+3^2+...

The solution involves the Riemann zeta function but tell me what you think.

Neither series converges. If a series converges then its n-th term approaches 0 as n goes to infinity. Clearly this does not happen with either series, hence they do not converge.

That is true, however they do converge due to the Riemann zeta functional equation.

Any convergence or divergence test will make it clear these two series diverge, however they don't.

_________________

"God may not play dice with the universe, but something strange is going on with prime numbers."

-Paul Erdos

"There are two types of cryptography in this world: cryptography that will stop your kid sister from looking at your files, and cryptography that will stop major governments from reading your files."

-Bruce Schneider

Estimate the value of these sums:

a) 1+2+3+4+...

b)1^2+2^2+3^2+...

The solution involves the Riemann zeta function but tell me what you think.

Neither series converges. If a series converges then its n-th term approaches 0 as n goes to infinity. Clearly this does not happen with either series, hence they do not converge.

That is true, however they do converge due to the Riemann zeta functional equation.

Any convergence or divergence test will make it clear these two series diverge, however they don't.

No these series cannot converge as firstly, they are infinite, and secondly, the Riemann zeta function requires a complex value operator in order to compute. If you use the ratio-limit test, the series is innately divergent.

_________________

Sebastian

"Don't forget to floss." - Darkwing Duck

Estimate the value of these sums:

a) 1+2+3+4+...

b)1^2+2^2+3^2+...

The solution involves the Riemann zeta function but tell me what you think.

Neither series converges. If a series converges then its n-th term approaches 0 as n goes to infinity. Clearly this does not happen with either series, hence they do not converge.

That is true, however they do converge due to the Riemann zeta functional equation.

Any convergence or divergence test will make it clear these two series diverge, however they don't.

No these series cannot converge as firstly, they are infinite, and secondly, the Riemann zeta function requires a complex value operator in order to compute. If you use the ratio-limit test, the series is innately divergent.

They do converge though. One can prove it mathematically.

_________________

"God may not play dice with the universe, but something strange is going on with prime numbers."

-Paul Erdos

"There are two types of cryptography in this world: cryptography that will stop your kid sister from looking at your files, and cryptography that will stop major governments from reading your files."

-Bruce Schneider

Estimate the value of these sums:

a) 1+2+3+4+...

b)1^2+2^2+3^2+...

The solution involves the Riemann zeta function but tell me what you think.

Neither series converges. If a series converges then its n-th term approaches 0 as n goes to infinity. Clearly this does not happen with either series, hence they do not converge.

That is true, however they do converge due to the Riemann zeta functional equation.

Any convergence or divergence test will make it clear these two series diverge, however they don't.

No these series cannot converge as firstly, they are infinite, and secondly, the Riemann zeta function requires a complex value operator in order to compute. If you use the ratio-limit test, the series is innately divergent.

They do converge though. One can prove it mathematically.

Your original question is incorrectly expressed. As the convergence/divergence determination is relative to gamma, and you did not mention a complex value operator in your original question, that still postulates a divergent solution in every sense in that word.

**One thing I learned when I worked for my Honors degree in Math that I have learned is that assume the fewest restrictions as a questions allows and interpret a proposition or explanation in the simplest ideals possible, that is called Occam's Razor**.

As long we are not given gamma, the solution will and always will be a divergent series as y>0, XeR, unless otherwise specified of a complex value operator.

I hope this will seal the issue of this thread.

_________________

Sebastian

"Don't forget to floss." - Darkwing Duck

Estimate the value of these sums:

a) 1+2+3+4+...

b)1^2+2^2+3^2+...

The solution involves the Riemann zeta function but tell me what you think.

Neither series converges. If a series converges then its n-th term approaches 0 as n goes to infinity. Clearly this does not happen with either series, hence they do not converge.

That is true, however they do converge due to the Riemann zeta functional equation.

Any convergence or divergence test will make it clear these two series diverge, however they don't.

No these series cannot converge as firstly, they are infinite, and secondly, the Riemann zeta function requires a complex value operator in order to compute. If you use the ratio-limit test, the series is innately divergent.

They do converge though. One can prove it mathematically.

At what point do they converge?

At what point do they stop diverging?

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