Hey, will appreciate help with the following Question;
Prove That,
$\displaystyle \lim \frac{1}{\sqrt{n}} (1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}) = 0$
Thanks in advance for any help !
Alright I think I got it this time:
$\displaystyle 0\leq a_n= \frac{1}{\sqrt{n} } \sum_{k=1}^{n} \frac{1}{k} = \frac{1}{n} \sum_{k=1}^{n} \frac{\sqrt{n} }{k} \rightarrow 0$ by Cesaro's mean (your sequence is $\displaystyle b_n=\frac{\sqrt{n} }{n}=\frac{1}{\sqrt{n} } \rightarrow 0$), so $\displaystyle a_n \rightarrow 0$
I remember that
$\displaystyle \lim_{n\to\infty} \sum_{k=1}^n \frac{a_k}{n} = \lim_{n\to\infty} a_{n} $
this case $\displaystyle a_k = \frac{1}{k} $
so your series
$\displaystyle \lim_{n\to\infty} \frac{1}{\sqrt{n}} \sum_{k=1}^{n} \frac{1}{k}$
$\displaystyle = \lim_{n\to\infty} \frac{\sqrt{n}}{n} \sum_{k=1}^{n} \frac{1}{k} $
$\displaystyle = \lim_{n\to\infty} \sqrt{n} \frac{1}{n} $
$\displaystyle = \lim_{n\to\infty} \frac{1}{\sqrt{n}} = 0 $
I think this theorem is what Jose27 mentioned ...
Hey, thanks for the answer, there is still something unclear to me,
in order to use cesaro's mean, you assume that
$\displaystyle
\lim_{n\to\infty} \sum_{k=1}^{n} \frac{1}{k}
$ exists ?
at least by Cauchy Criterion For Convergence (to my knowledge) i find that this sequence does not have a limit.
i'd appreciate further explanations (also if i got the Cauchy criterion wrong... )
Thanks and have a nice weekend
I am not sure what the cesaro's mean is ,
But talking about the theorem I made use , the series should not have a limit ( diverges to infinty ) , otherwise , the result is to be zero .
$\displaystyle \lim_{n\to\infty} \sum_{k=1}^n \frac{a_k}{n} = \lim_{n\to\infty} \frac{E}{n} = 0 $
which is meaningless .