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kd123
 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)
kd123
 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)

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satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1it 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

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1to find the sum, i think the easiest thing to do would be partial fractions do you know how to do that?

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1once 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

kd123
 3 years ago
Best ResponseYou've already chosen the best response.0ok, thanks. Going to try it now.

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1write back if you get stuck, i will try it too

kd123
 3 years ago
Best ResponseYou've already chosen the best response.0For the partial sums I got this

kd123
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ 1 }{ 2 }+\frac{ 1 }{ 3 }\frac{ 1 }{n+2 }\frac{ 1 }{ n+3 }\]

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1if you are starting at \(n=0\) isn't the first term 1 ?

kd123
 3 years ago
Best ResponseYou've already chosen the best response.0Oh yes you are correct. My mistake.

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1a small quibble, but it does make a difference i suppose

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1i get \(1+\frac{1}{2}\) and everything after that gets killed off

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1also i think ( i may be wrong about this) that for the partial sum you get \[\frac{3}{2}\frac{1}{n+3}\]

kd123
 3 years ago
Best ResponseYou've already chosen the best response.0yes, I got this as well. Thanks for your help!

kd123
 3 years ago
Best ResponseYou've already chosen the best response.0Can you help me again please. How do I go about doing this kind: \[\sum_{n=1}^{\infty}5(\frac{ 2 }{ 3 })^{n1} \] Use partial sums to determine if that series converges or diverges and if it converges, determine its sum.

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1geometric series for this one pull out the 5

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1\[5\sum_{k=0}^{\infty}(\frac{2}{3})^k\] use \[\frac{1}{1r}\]

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1you 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

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1yw again partial sums are a pain, but not too bad, it is just \[\frac{1r^n}{1r}\] in general

kd123
 3 years ago
Best ResponseYou've already chosen the best response.0yeah, I'll try to remember that from now on :)
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