Limit of a sequence...

**chisigma** · June 3rd 2009, 01:31 AM

Does the sequence...

$a_{n} = \frac{1}{\sqrt{n}}\cdot \sum_{i=1}^{n-1} \frac{1}{\sqrt{i+\alpha}}$ , $0<\alpha<1$ (1)

... have finite limit or not?...

I will appreciate very much an answer from you!...

Kind regards

$\chi$ $\sigma$

**NonCommAlg** · June 3rd 2009, 02:29 AM

Originally Posted by chisigma

Does the sequence...

$a_{n} = \frac{1}{\sqrt{n}}\cdot \sum_{i=1}^{n-1} \frac{1}{\sqrt{i+\alpha}}$ , $0<\alpha<1$ (1)

... have finite limit or not?...

I will appreciate very much an answer from you!...

Kind regards

$\chi$ $\sigma$

$a_n=\frac{1}{n} \sum_{i=0}^{n-1}\frac{1}{\sqrt{\frac{i + \alpha}{n}}} - \frac{1}{\sqrt{\alpha n}}.$ therefore: $\lim_{n\to\infty} a_n = \lim_{n\to\infty}\frac{1}{n} \sum_{i=0}^{n-1}\frac{1}{\sqrt{\frac{i + \alpha}{n}}} = \int_0^1 \frac{1}{\sqrt{x}} \ dx= 2.$

**chisigma** · June 3rd 2009, 03:37 AM

Excellent!

...

Kind regards

$\chi$ $\sigma$

**chisigma** · June 16th 2009, 10:55 PM

Instead of opening a new thread I can use this and ask you for another limit...

$\lim_{x \rightarrow \infty} x\cdot e^{- c\cdot \sqrt{\ln x}}$

... where c>0 is a constant... Does it exist?... if yes, what is its value?...

Thank you very much!...

Kind regards

$\chi$ $\sigma$

**NonCommAlg** · June 17th 2009, 06:00 PM

Originally Posted by chisigma

Instead of opening a new thread I can use this and ask you for another limit...

$\lim_{x \rightarrow \infty} x\cdot e^{- c\cdot \sqrt{\ln x}}$

... where c>0 is a constant... Does it exist?... if yes, what is its value?...

Thank you very much!...

Kind regards

$\chi$ $\sigma$

let $\ln x = t^2.$ then $\lim_{x\to\infty}xe^{-c\sqrt{\ln x}}=\lim_{t\to\infty}e^{t^2-ct}=\infty.$

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