Prove that the limit as p approaches infinity of the summation from k=1 to infinity of 1/k^p =1 and the limit as p approaches 1+ of the summation from k=1 to infinity of 1/k^p= infinity.
Thanks!
Lets call $\displaystyle \zeta(p)$ the function...
$\displaystyle \zeta (p) = \lim_{n \rightarrow \infty} \sum_{i=1}^{n} \frac{1}{i^{p}}$ (1)
First we prove that the limit (1) is $\displaystyle \infty$ if $\displaystyle p=1$. At this scope we compute the partial sum for $\displaystyle n=2^{k}$ of (1) for $\displaystyle p=1$...
$\displaystyle \sum_{i=1}^{2^{k}} \frac{1}{i}= 1 + \frac{1}{2} + \{\frac{1}{3} + \frac{1}{4}\} + \{\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}\} + \dots +\{\frac{1}{2^{k-1}+1} + \frac{1}{2^{k-1}+2} +\dots + \frac{1}{2^{k}-1} +\frac{1}{2^{k}} \}$ (2)
... and it is easy to see that is...
$\displaystyle \sum_{i=1}^{2^{k}} \frac{1}{i} > 1 + \frac {k}{2} \rightarrow \lim_{k \rightarrow \infty} \sum_{i=1}^{2^{k}} \frac{1}{i} = \infty $ (3)
Lets now compute the partial sum for $\displaystyle n=2^{k}$ of (1) for $\displaystyle p>1$...
$\displaystyle \sum_{i=1}^{2^{k}} \frac{1}{i^{p}}= 1 + \frac{1}{2^{p}} + \{\frac{1}{3^{p}} + \frac{1}{4^{p}}\} + \{\frac{1}{5^{p}} + \frac{1}{6^{p}} + \frac{1}{7^{p}} + \frac{1}{8^{p}}\} + \dots$
$\displaystyle \dots +\{\frac{1}{(2^{k-1)}+1)^{p}} + \frac{1}{(2^{k-1}+2)^{p}} +\dots + \frac{1}{(2^{k}-1)^{p}} +\frac{1}{2^{kp}} \}$ (4)
... and it is easy to see that is...
$\displaystyle \sum_{i=1}^{2^{k}} \frac{1}{i^{p}} < 1 + (\frac{1}{2})^{p} + (\frac{1}{2})^{2p} + \dots + (\frac{1}{2})^{(k-1)p} \rightarrow \sum_{i=1}^{\infty} \frac{1}{i^{p}} < \infty $ (5)
The results obatined in (3), (4) and (5) allow us to conclude that is...
$\displaystyle \lim_{p \rightarrow 1+} \sum_{i=1}^{\infty} \frac{1}{i^{p}}= \infty$
$\displaystyle \lim_{p \rightarrow \infty} \sum_{i=1}^{\infty} \frac{1}{i^{p}}=1$ (6)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$