logarithmic decreasing serie

**oscaralive** · February 8th 2011, 12:34 PM

Hi all,

anyone knows how to compute the value of the following serie:

$alog(a) + (a-1)log(a-1)+....+(a-(a-1))log(a-(a-1))+(a-a)log(a-a)$

Many thanks in advance!

**topsquark** · February 8th 2011, 01:38 PM

I'll ignore your last term, which is undefined.
$a~log(a) + (a - 1)~log(a - 1) + (a-2)~log(a - 2) +~...$

$=~log(a^a) + log \left ( (a-1) ^{(a-1)} \right ) + log \left ( (a-2) ^{(a-2)} \right ) +~...$

$=~log \left ( a^a \cdot (a - 1)^{(a-1)} \cdot (a - 2)^{(a - 2)} \cdot ~...~ \right )$

Take it from here.

-Dan

**oscaralive** · February 9th 2011, 12:45 AM

Originally Posted by topsquark

I'll ignore your last term, which is undefined.
$a~log(a) + (a - 1)~log(a - 1) + (a-2)~log(a - 2) +~...$

$=~log(a^a) + log \left ( (a-1) ^{(a-1)} \right ) + log \left ( (a-2) ^{(a-2)} \right ) +~...$

$=~log \left ( a^a \cdot (a - 1)^{(a-1)} \cdot (a - 2)^{(a - 2)} \cdot ~...~ \right )$

Take it from here.

-Dan

Thanks Dan, but now

$\prod_{i=1}^{a}(a-i+1)^{(a-i+1)}$

how do you compute this?

Thanks again,

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