## kd123 Group Title 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) one year ago one year ago

1. satellite73

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. satellite73

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

3. kd123

yes thank you!

4. satellite73

yw

5. satellite73

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. kd123

ok, thanks. Going to try it now.

7. satellite73

good luck!

8. satellite73

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

9. kd123

For the partial sums I got this

10. kd123

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

11. satellite73

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

12. kd123

Oh yes you are correct. My mistake.

13. satellite73

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

14. satellite73

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

15. satellite73

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

16. kd123

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

17. satellite73

yw

18. kd123

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. satellite73

geometric series for this one pull out the 5

20. kd123

ohh

21. satellite73

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

22. satellite73

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. kd123

Thanks you so much.

24. satellite73

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

25. kd123

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