I Harmonic series Ʃ1/n diverges but p-series Ʃ(1/n)^p diverges?

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TL;DR Summary
Struggling with harmonic series
Maybe it's obvious why and I'm just dumb, but I hope you can help me understand this
-----------------------------+∞
I sorta get why the Ʃ(1/n) diverges to +∞
-------------------------------n=1
------------------+∞
But why the Ʃ(1/n)^p (p>1), converges?
------------------n=1
 
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Ok I think I get it. I consider the improper integral, if p is 1 I have ln(M) so the series diverges as the limit of M--> +∞ of the integer is ln(+∞)=+∞, if I have p > 1 I have the improper integral of 1/(x^p) or x^-p, so {[M^(1-p)]-1}/(1-p), for M--> +∞ I have a (+∞)^-k (as 1-p <0) so (zero -1)/(1-p) and It converges. For p<1 I have instead (+∞)^k (as 1-p >0) so (+∞-1)/(1-p) so it diverges, right?
 
##p=1## in ##s(p)=\displaystyle{\sum_{n=1}^\infty \dfrac{1}{n^p}}## is a hard boundary. We have known since the 14th century (possibly for longer) that ##s(1)## diverges. However, as soon as we add a bit, say we consider ##s(1+\varepsilon)## for ##\varepsilon>0## then it converges.

As noted, the integral test points to the reason for it: ##\displaystyle{\int \dfrac{dx}{x}=\log|x|}## involves a logarithmic function but ##\displaystyle{\int \dfrac{dx}{x^{1+\varepsilon}}=-\dfrac{1}{\varepsilon x^\varepsilon}} ## is a polynomial in ##1/x .## That is, we change from a logarithmic result to a polynomial result. This is a fundamental difference in behavior and finally, the reason why ##s(1)## diverges and ##s(1+\varepsilon)## converges.
 
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Dimwitted_UniStudent said:
TL;DR Summary: Struggling with harmonic series

Maybe it's obvious why and I'm just dumb, but I hope you can help me understand this
-----------------------------+∞
I sorta get why the Ʃ(1/n) diverges to +∞
-------------------------------n=1
------------------+∞
But why the Ʃ(1/n)^p (p>1), converges?
------------------n=1
Please learn and use latex when entering math expressions. It took me a minute or two to realize those dash lines were part of your summation.

We have a tutorial for latex whose link is listed below in this post.
 
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