# Sum of sequence

1. Jul 25, 2009

### songoku

1. The problem statement, all variables and given/known data
Let {a_n} be the sequence defined by
$$a_{n} = \frac{n^{2}+8n+10}{n+9}$$

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

2. Relevant equations

3. 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

Last edited: Jul 25, 2009
2. Jul 25, 2009

### tiny-tim

Hi songoku!

(you mean (n + 9) )

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

3. Jul 25, 2009

### songoku

Hi tiny-tim :)

Oh yes i mean n+9

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

more hint?

thx

4. Jul 25, 2009

### tiny-tim

Hi songoku!

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

5. Jul 25, 2009

### songoku

Hi tiny-tim :tongue2:

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...

6. Jul 25, 2009

### tiny-tim

Hi songoku!

i] nooo …

ii] try putting x = 1

7. Jul 25, 2009

### songoku

Hi tiny-tim !:rofl:

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

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

i'm still trying to catch the hint

Last edited: Jul 25, 2009
8. Jul 25, 2009

### tiny-tim

Hi songoku!

Sorry … I've just realised 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)

9. Jul 25, 2009

### songoku

Hi tiny-tim !

OMG psi function....

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

thx a lot tiny-tim ^^