# Why 1/n converges

1. Aug 30, 2010

### atomqwerty

Why the serie $$\sum\frac{1}{n}$$ diverges and the serie $$\sum\frac{1}{n^{2}}$$ converges? I'd appreciate an explanation beyond the definition of geometric series (I know that the sum of a geometric serie is given by a formula).

I've found an explanation, that involves the creation of groups in the series so each of them result 1/2 (at least), so the sum diverges. Could I apply the same operation to the sum 1/n2?

thanks

2. Aug 30, 2010

### Office_Shredder

Staff Emeritus
These aren't geometric series so it will be hard to get an explanation from that direction. The integral test is a very natural way to understand why one converges and another doesn't; do you know what the integral test for convergence is?

3. Aug 30, 2010

### Petr Mugver

$$\sum_{n=1}^{2^{(N+1)}-1}\frac{1}{n^\alpha}=\frac{1}{1^\alpha}+\frac{1}{2^\alpha}+\frac{1}{3^\alpha}+\frac{1}{4^\alpha}+\frac{1}{5^\alpha}+\frac{1}{6^\alpha}+\frac{1}{7^\alpha}+\dots<\frac{1}{1^\alpha}+\frac{1}{2^\alpha}+\frac{1}{2^\alpha}+\frac{1}{4^\alpha}+\frac{1}{4^\alpha}+\frac{1}{4^\alpha}+\frac{1}{4^\alpha}+\dots=\sum_{n=0}^{N}2^{n(1-\alpha)}$$

this converges if $$\alpha> 1$$. In a similar way you show that the generalized harmonic series diverges for $$\alpha\leq 1$$.

4. Aug 30, 2010

### atomqwerty

If i'm not wrong, $$a_{n}$$ converges if the integral sum of the associated function f(x), $$\int f(x)$$ has a finite valor. I think I've heard about that way of see it. Now I've realized that this is not a geometric serie. thanks

5. Aug 30, 2010

### atomqwerty

Thanks for the explanation, I'll remember the alpha in the future.