## anonymous 3 years ago Use partial sums to determine if the following series diverge or converge. If the series converges, determine its sum. Σ (from n=0 to infinity) (2/(n^2 + 4n + 3)

1. anonymous

it definitely converges because the degree of the denominator is 2 and the degree of the numerator is 0 and the difference is greater than 1

2. anonymous

to find the sum, i think the easiest thing to do would be partial fractions do you know how to do that?

3. anonymous

yes thank you!

4. anonymous

yw

5. anonymous

once you get the partial fraction, write out the first few terms and you will see that the sum "telescopes" that will not only give you a formula for the partial sums, it will also tell you what the infinite sum is

6. anonymous

ok, thanks. Going to try it now.

7. anonymous

good luck!

8. anonymous

write back if you get stuck, i will try it too

9. anonymous

For the partial sums I got this

10. anonymous

$\frac{ 1 }{ 2 }+\frac{ 1 }{ 3 }-\frac{ 1 }{n+2 }-\frac{ 1 }{ n+3 }$

11. anonymous

if you are starting at $$n=0$$ isn't the first term 1 ?

12. anonymous

Oh yes you are correct. My mistake.

13. anonymous

a small quibble, but it does make a difference i suppose

14. anonymous

i get $$1+\frac{1}{2}$$ and everything after that gets killed off

15. anonymous

also i think ( i may be wrong about this) that for the partial sum you get $\frac{3}{2}-\frac{1}{n+3}$

16. anonymous

yes, I got this as well. Thanks for your help!

17. anonymous

yw

18. anonymous

Can you help me again please. How do I go about doing this kind: $\sum_{n=1}^{\infty}5(\frac{ 2 }{ 3 })^{n-1}$ Use partial sums to determine if that series converges or diverges and if it converges, determine its sum.

19. anonymous

geometric series for this one pull out the 5

20. anonymous

ohh

21. anonymous

$5\sum_{k=0}^{\infty}(\frac{2}{3})^k$ use $\frac{1}{1-r}$

22. anonymous

you can almost do it in your head $1-\frac{2}{3}=\frac{1}{3}$ the reciprocal of $$\frac{1}{3}$$ is 3 and you get 15

23. anonymous

Thanks you so much.

24. anonymous

yw again partial sums are a pain, but not too bad, it is just $\frac{1-r^n}{1-r}$ in general

25. anonymous

yeah, I'll try to remember that from now on :)