I Infinite Series - Divergence of 1/n question

[Ascendant0]

I understand mathematically several ways to test whether an infinite series converges or diverges. However, I came across one particular equation that is stumping me, $\sum_{n=1}^{\infty} 1/n$. I understand how to mathematically apply series tests to show it diverges. But intuitively, I thought it would surely converge, as the larger n becomes, the smaller the value. When (by the integral test) I came to find out it actually diverges (and yes, I am aware it's the correct answer), I'm now trying to fully wrap my brain around this, and hoping someone might be able to give me a different perspective to make sense of it. This is how I currently perceive it...

I understand the notion that as you increase the value of n, there is always going to be a value of n higher than that, which will have at least some amount of area under it, no matter how minuscule. But, the same goes for $\sum_{n=1}^{\infty} 1/(n^2)$ , yet it converges? No matter how small of a value you plug in, there is still always going to be another higher value you can put in there that will have at least some amount of area under it. So long as it’s not directly on “0,” which it can never reach, there is always additional area under it through infinity. It never reaches 0, and continues indefinitely. So, based on the explanation given in the book as for why $1/n$ diverges, I fail to understand how a similar equation, just one that decreases much faster, converges?

 

[fresh_42]

The answer is quite simple: $1/n$ isn't fast enough. It adds up quicker than it tends to zero. But $1/n^1$ is the tipping point, i.e. $\displaystyle{\sum_{n=1}^\infty \dfrac{1}{n^{1+\varepsilon}}}$ converges for $\varepsilon>0$ just not for $\varepsilon =0.$

I find Riemann's series theorem far more disturbing: if a series converges but not absolutely, then it can be rearranged so that it converges at an arbitrary given number.

Gabriel's horn is another strange example: It has a finite volume but an infinitely large surface. You can fill it with a gallon of paint but you cannot paint it with a gallon of paint.

 

[anuttarasammyak]

Taking a look at a case of integration of continuous variables
\int_1^\infty \frac{1}{x^{1+\epsilon}}dx = \frac{1}{-\epsilon}[x^{-\epsilon}]_1^\infty
, we see $\epsilon \rightarrow +0$ is critical of convergence/divergence.

 

[Ascendant0]

The answer is quite simple: $1/n$ isn't fast enough. It adds up quicker than it tends to zero. But $1/n^1$ is the tipping point, i.e. $\displaystyle{\sum_{n=1}^\infty \dfrac{1}{n^{1+\varepsilon}}}$ converges for $\varepsilon>0$ just not for $\varepsilon =0.$

I find Riemann's series theorem far more disturbing: if a series converges but not absolutely, then it can be rearranged so that it converges at an arbitrary given number.

Gabriel's horn is another strange example: It has a finite volume but an infinitely large surface. You can fill it with a gallon of paint but you cannot paint it with a gallon of paint.

Ok, I got you, so it's basically a matter of whether it adds up faster or tends to zero faster. That makes sense of it, thank you.

 

[WWGD]

For contrast, notice $\Sigma \frac{1}{n!} =e$~$2.718...$, where $\frac{1}{n!}\frac{}{}<< \frac{1}{n^2}$

 

[mathwonk]

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 +.......