covergense root question mmn11 2c

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March 2nd 2011, 12:52 AM
transgalactic

covergense root question mmn11 2c

$\sum_{k=1}^{infinity}ln\frac{\sqrt{k}+1}{\sqrt{k}}$
is it converges absolutly ,or with condition(dont know the english term),
or is it diverges?
March 2nd 2011, 02:31 AM
Ted

$\dispalystyle ln \left( \frac{\sqrt{k}+1}{\sqrt{k}} \right) = ln(\sqrt{k}+1) - ln(\sqrt{k})$

so your series can be rewritten as $\displaystyle \sum_{k=1}^{\infty} \left( ln(\sqrt{k}+1) - ln(\sqrt{k}) \right)$

Telescoping Series.
March 3rd 2011, 05:28 AM
transgalactic

by what test its absolut value converging?
March 3rd 2011, 05:48 AM
chisigma

Quote:

Originally Posted by Ted

$\dispalystyle ln \left( \frac{\sqrt{k}+1}{\sqrt{k}} \right) = ln(\sqrt{k}+1) - ln(\sqrt{k})$

so your series can be rewritten as $\displaystyle \sum_{k=1}^{\infty} \left( ln(\sqrt{k}+1) - ln(\sqrt{k}) \right)$

Telescoping Series.

$\displaystyle \sum_{k=1}^{\infty} \ln \frac{1+\sqrt{k}}{\sqrt{k}}= \ln 2 - \ln 1 + \ln (1+\sqrt{2}}) - \ln {\sqrt{2}} + \ln (1+\sqrt{3}}) - \ln {\sqrt{3}})+ ...$ (1)

Is (1) 'telescopic'?...

Kind regards

$\chi$ $\sigma$
March 3rd 2011, 05:53 AM
chisigma

Quote:

Originally Posted by transgalactic

$\sum_{k=1}^{infinity}ln\frac{\sqrt{k}+1}{\sqrt{k}}$
is it converges absolutly ,or with condition(dont know the english term),
or is it diverges?

For k 'large enough' is...

$\displaystyle \ln \frac{1+\sqrt{k}}{\sqrt{k}} = \ln (1+\frac{1}{\sqrt{k}}) > \frac{1}{k}$

... so that...

Kind regards

$\chi$ $\sigma$
March 3rd 2011, 06:07 AM
transgalactic

why ln(1+x)>x
??

1/k diverges
so it cannot be absolutly convergent
only conditionaly
for this series to convers it 1. need to make a1>a2>...>ak i dont see it here

2.
and lim a_k =0

the limit is zero

how to prove 1?
April 1st 2011, 01:17 PM
transgalactic

Quote:

Originally Posted by chisigma

$\displaystyle \sum_{k=1}^{\infty} \ln \frac{1+\sqrt{k}}{\sqrt{k}}= \ln 2 - \ln 1 + \ln (1+\sqrt{2}}) - \ln {\sqrt{2}} + \ln (1+\sqrt{3}}) - \ln {\sqrt{3}})+ ...$ (1)

Is (1) 'telescopic'?...

Kind regards

$\chi$ $\sigma$

cant see how the numbers cancel out each other
cant see two similar members

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