Question

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)

Answer 1

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

Answer 2

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

Answer 3

yes thank you!

Answer 4

yw

Answer 5

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

Answer 6

ok, thanks. Going to try it now.

Answer 7

good luck!

Answer 8

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

Answer 9

For the partial sums I got this

Answer 10

\[\frac{ 1 }{ 2 }+\frac{ 1 }{ 3 }-\frac{ 1 }{n+2 }-\frac{ 1 }{ n+3 }\]

Answer 11

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

Answer 12

Oh yes you are correct.
My mistake.

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

yw

Answer 18

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.

Answer 19

geometric series for this one
pull out the 5

Answer 20

ohh

Answer 21

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

Answer 22

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

Answer 23

Thanks you so much.

Answer 24

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

Answer 25

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