## Euler sum

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?
 Recognitions: Homework Help 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
 Recognitions: Homework Help 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.

Recognitions:

## Euler sum

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.

 Quote by 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.
Brilliant!
 Recognitions: Gold Member You can use the integral test: $$1+1/2+1/3+++1/n> \int_N\frac{1}{x}dx =ln (N )\rightarrow \infty.$$ And: $$1/4+1/9+1/16 +(1/(N+1)^2 < \int_1^N\frac{1}{x^2}dx =1-1/N.$$