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I have just studied p-series and I was taught that when p<=1 the series diverges and that when p>1 the series converges. It seems that both serieses converge only that when p>1 it happens faster.

For example: ∑(1/n)=1+1/2+1/3+1/4+1/5...

and now when p=2: ∑(1/n^2)=1+1/4+1/9+1/16...

Reboot_87
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  • The harmonic series $\sum \frac1n$ diverges slowly, but it diverges. – Daniel Fischer Jan 14 '14 at 22:16
  • You should google for Harmonic Series. The argument/proof that $\sum_{k=1}^{\infty} \frac{1}{n}$ diverges is in my opinion rather nice. – Nigel Overmars Jan 14 '14 at 22:16
  • $p$ is not the power of n, it is the ratio between terms. – Alec Teal Jan 14 '14 at 22:18
  • Part of what you say is true: If $1\lt p\lt q$, then $\sum\frac{1}{n^q}$ converges faster than $\sum \frac{1}{n^p}$. But we need $p\gt 1$ for convergence. – André Nicolas Jan 14 '14 at 22:18
  • See also http://math.stackexchange.com/questions/29450/self-contained-proof-that-sum-limits-n-1-infty-frac1np-converges-for – Harald Hanche-Olsen Jan 14 '14 at 22:19
  • Do you know what it means for a series to converge? I'm baffled as to how you can say a $p$-series converges when $0 < p \leq 1$. Related question: do you think the natural logarithm of $x$ converges as $x \rightarrow \infty$? – KCd Jan 14 '14 at 22:50

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Using the integral test, $$\sum_{n=1}^{\infty}{\frac{1}{n^p}}\text{ converges} \Leftrightarrow \int_{1}^{\infty}{\frac{1}{x^p}} \text{ converges}$$

Expanding the right hand side, $$\int_{1}^{\infty}{\frac{1}{x^p}}=\lim_{t\to\infty}{\left(\frac{t^{1-p}}{1-p}-\frac{1}{1-p}\right)}=\left\{\begin{align}\infty, \quad &p\le1 \\\frac{1}{p-1}, \quad &p>1 \end{align}\right.$$

We obtain that the series diverges for $p\ge 1$, and converges for $p>1$. This is one possible way of justifying the result you ask about.

GPerez
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