# calculating a series

• Feb 24th 2013, 02:48 AM
hasti
calculating a series
Hi,

I'm trying to calculate this series!

Attachment 27214

a and b are two constants.

is there any answer for this series??

hasti
• Feb 24th 2013, 03:26 AM
Bashyboy
Re: calculating a series
Well, seeing as b is a constant, that implies that as m goes through the different values, from one to infinity, b is unaffected. So, what can we do with it?

EDIT: If it isn't immediately evident why b remains unaffected as m goes through its different values, and what we can do with it, let's calculate a partial sum of this series. Let's find the sum for just $m=1,2,~and~3$

$\frac{e^{-a \cdot (1)^2}}{b \cdot (1)^2} + \frac{e^{-a \cdot (2)^2}}{b \cdot (2)^2} + \frac{e^{-a \cdot (3)^2}}{b \cdot (3)^2}$

What do you notice about "a" and "b," as m goes from 1 to 3?
• Feb 24th 2013, 03:41 AM
Bashyboy
Re: calculating a series
I edited my post, hasti.

EDIT: Actually, I don't know how helpful it will be. I misread what sort of series you have.
• Feb 24th 2013, 03:48 AM
hasti
Re: calculating a series
Yes, It's correct. Then now, We should calculate this series when b is eliminated!

But i don't know how to do it!

Attachment 27223
• Feb 24th 2013, 03:56 AM
Bashyboy
Re: calculating a series
Well, b isn't eliminated, it's just factored out of the series. Are you sure there isn't suppose to be an addition sign between a and m^2?
• Feb 24th 2013, 04:10 AM
hasti
Re: calculating a series
let me to explain with an example!
we know that
Attachment 27225
now i should find an equivalent for
Attachment 27226

thanks
• Feb 24th 2013, 04:38 AM
hasti
Re: calculating a series
we don't need any extra sign between a and m^2. a is a coefficient for m^2 and (a*m^2) is a power for e.