Summing up a_n: Finding \sum_{n=1}^{30}\ a_n

[songoku]

Homework Statement

Let {a_n} be the sequence defined by
a_{n} = \frac{n^{2}+8n+10}{n+9}

Find the value of \sum_{n=1}^{30}\ a_n

Homework Equations

The Attempt at a Solution

\sum_{n=1}^{30}\ a_n

= \sum_{n=1}^{30}\ n -1+\frac{19}{n+9}

= \sum_{n=1}^{30}\ n - \sum_{n=1}^{30}\ 1 +\sum_{n=1}^{30}\frac{19}{n+9}

= 1/2 (30)(31) - 30 + \sum_{n=1}^{30}\frac{19}{n+9}

I'm stuck here...

thx

 

[tiny-tim]

Hi songoku!

(you mean (n + 9) )

Hint: can you do \sum_{n=1}^{39}\frac{1}{n} ?

 

[songoku]

Hi tiny-tim :)

Oh yes i mean n+9

Sorry i don't know how to do that...

more hint?

thx

 

[tiny-tim]

Hi songoku!

ok … new hint … can you sum \sum_{n=1}^{39}x^n and then integrate it?

 

[songoku]

Hi tiny-tim

Are you trying to say that :

\sum_{n=1}^{30}\frac{19}{n+9} = \int_{0}^{30}\frac{19}{n+9} dn ?

I did try your last hint and got 1/40 x^40 + 1/39 x^39 + ... + 1/2x^2 after the integration...

 

[tiny-tim]

Hi songoku!

i] nooo …

ii] try putting x = 1

 

[songoku]

Hi tiny-tim !

i] nooo …

I laughed a lot when i read this. I'm pretty sure that you won't state that \sum_{n=1}^{30}\frac{19}{n+9} = \int_{0}^{30}\frac{19}{n+9} dn

but i just gave it a shot and posted it, lol

ii] try putting x = 1

1/2 + 1/3 + ... + 1/40 ?

i'm still trying to catch the hint

 

[tiny-tim]

Hi songoku!

Sorry … I've just realized my method doesn't work for reciprocals

(I was thinking that the sum is (1 - xⁿ)/(1 - x), which I could then integrate, and put x = 1, but there isn't an easy way to integrate it )

In fact, I don't think there is any "short-cut" solution for this …

I think you have to use the "psi function" (see http://en.wikipedia.org/wiki/Digamma_function#Recurrence_formula)

 

[songoku]

Hi tiny-tim !

OMG psi function...

I think maybe i'll do it manually...

thx a lot tiny-tim ^^