# Thread: covergense root question mmn11 2c

1. ## 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?

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

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

Telescoping Series.

3. by what test its absolut value converging?

4. 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

$\chi$ $\sigma$

5. 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

$\chi$ $\sigma$

6. 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?

7. 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