Why Does 1/n Diverge & 1/n2 Converge? Explanation Beyond Formulas

[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

 

[Office_Shredder]

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?

 

[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.

 

[atomqwerty]

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?

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

 

[atomqwerty]

...

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

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