Why is the Euler Sum Infinite for Zeta=1?

[overlook1977]

This may be a stupid question, but I do not understand why the Euler sum is infinite for zeta=1. Why is "1+1/2+1/3+1/4+... " infinite, but zeta=2 (1+1/4+1/9+...) not?

 

[mjsd]

1+1/2+1/3+... is $$\sum_n \frac{1}{n}$$ is a harmonic series and there are various tests to show that this diverges while
1+1/4+1/9+... is really $$\sum_n \frac{1}{n^2}$$ and this series converges

 

[Gib Z]

The first series is a p series where p is equal to 1. We know for convergence of p series that a condition is that p>1.

The second series converges to $\pi^2/6$ but is somewhat harder to show. Just to show it converges at all though, p=2 which is more than 1. Look up p series.

 

[mathman]

1+1/2+1/3+1/4+1/5+1/6+1/7+1/8+...>
1+1/2+1/4+1/4+1/8+1/8+1/8+1/8+...=
1+1/2+1/2+ 1/2+... which diverges.

 

[Werg22]

1+1/2+1/3+1/4+1/5+1/6+1/7+1/8+...>
1+1/2+1/4+1/4+1/8+1/8+1/8+1/8+...=
1+1/2+1/2+ 1/2+... which diverges.

Brilliant!

 

[robert Ihnot]

You can use the integral test: $$1+1/2+1/3+++1/n> \int_N\frac{1}{x}dx =ln (N )\rightarrow \infty.$$

And: $$1/4+1/9+1/16 +(1/(N+1)^2 < \int_1^N\frac{1}{x^2}dx =1-1/N.$$