1. anonymous

can anyone help????

2. anonymous

maybe partial fractions, but I need help on that part too...

3. anonymous

First, you should break up the 5/(n(n + 9)) into two separate fractions. A/n + B/(n + 9) = 5/(n(n+9)) --> A(n + 9) + Bn = 5 Let n = 0 to find A = 5/9 and n = -9 to find B = -5/9. You should find that 5/(n(n + 9)) = 5/(9n) - 5(9(n + 9)). You should be wanting to integrate this: $\lim_{b \rightarrow \infty}\int\limits_{1}^{b}[5/(9n)-5/(9(n + 9))]dn$

4. anonymous

Awww yeah, thats how it's done

5. anonymous

Just use some u substitution and your done!

6. anonymous

Other than the limit, just use log rules to combine them and it should work out

7. anonymous

Factor out a 5/9 to make the calculations somewhat easier: $(5/9)\int\limits_{1}^{b}[1/n-1/(n+9)]dn$ $(5/9)[\ln |n|-\ln |n+9|]_{1}^{b}$ $(5/9)[\ln |b|-\ln |b+9|-\ln1+\ln10]$ $(5/9)\ln[10|b/(b+9)|]$

8. anonymous

Take the limit as b approaches infinity of the last thing, I got (5/9)*ln10

9. anonymous

I think it's e^ of that answer. I don't think you can use L'hospitals rule inside logs, can you?

10. anonymous

I didn't use L'Hopital's rule. I divided the top and the bottom of the fraction by b.

11. anonymous

yeah... it's just (5/9)ln10

12. anonymous

thank you so much, that was so hard for me!

13. anonymous

Try wolframalpha.com for these things, if you just want a solution, by the way.

14. anonymous