# Thread: Sum of infinite series

1. ## Sum of infinite series

Hello, I'm not sure what topic this comes under so apologies for putting it in the wrong place.

I have a series of numbers, see below. I want to basically continue this number to infinite or it get's very small and then add them all up.

My method using Excel and what I think wikipedia is telling me to do is;
1) Calculate the log of y where x is 1 to 20.
2) Using Excel's Linest function to calculate the slope and intercept of this curve
3) Then I calculate Mu by Intercept + Slope * x
3) Calculate variance by Residual Sum of Squared / Degrees if Freedom (output from Linest array formula)
4) Calculate the expected y using exp(Mu + Variance / 2 )

This formula like magic enables me to predict using the same curve as my actual data what it would be going forward. I can then copy this formula down and draw a new line until it gets very small. When I plot this line against my actual data it's close enough so I'm happy I've done it correctly. Again like magic if I add a trend line on my actual data it follows my predicted line perfectly the equation being y= 0.0412e-0.142x
I now want to add up all those predicted terms, happy to not use real data at this point. If I predict this number forward by 20 x's I get a sum of 0.272, if I go forward 40 terms I get 0.274 and for my purpose either number will do.

I'm hoping there's another magical way to calculate this is in a simple equation.

Anyone know?

 0.045532 0.036191 0.030401 0.026179 0.020096 0.017043 0.014503 0.012486 0.009496 0.008054 0.006973 0.006197 0.005379 0.004996 0.004649 0.004347 0.00386 0.00361 0.003418 0.003233

2. ## Re: Sum of infinite series

Alternatively:

Can I sum y where x = 1 to n

y= 0.0412e-0.142x

3. ## Re: Sum of infinite series

Originally Posted by RedundantDummy
Alternatively:

Can I sum y where x = 1 to n

y= 0.0412e-0.142x
$\displaystyle{\sum_{x=1}^n}~\alpha^x = \dfrac{1-\alpha^{n+1}}{1-\alpha} - 1 = \alpha \dfrac{1-\alpha^n}{1-\alpha}$

$e^{-bx} = \left(e^{-b}\right)^x$

$\alpha = e^{-b}$

$\displaystyle{\sum_{x=1}^n}~c e^{-bx} = c \displaystyle{\sum_{x=1}^n}~ \left(e^{-b}\right)^x = c e^{-b} \dfrac{1-e^{-n b}}{1-e^{-b}} = c\dfrac{1-e^{-nb}}{e^b - 1}$

here

$c = 0.0412$

$b = 0.142$

I leave you to plug in the constants.