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

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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
ok, thanks. Going to try it now.
good luck!
write back if you get stuck, i will try it too
For the partial sums I got this
\[\frac{ 1 }{ 2 }+\frac{ 1 }{ 3 }-\frac{ 1 }{n+2 }-\frac{ 1 }{ n+3 }\]
if you are starting at \(n=0\) isn't the first term 1 ?
Oh yes you are correct. My mistake.
a small quibble, but it does make a difference i suppose
i get \(1+\frac{1}{2}\) and everything after that gets killed off
also i think ( i may be wrong about this) that for the partial sum you get \[\frac{3}{2}-\frac{1}{n+3}\]
yes, I got this as well. Thanks for your help!
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.
geometric series for this one pull out the 5
\[5\sum_{k=0}^{\infty}(\frac{2}{3})^k\] use \[\frac{1}{1-r}\]
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
Thanks you so much.
yw again partial sums are a pain, but not too bad, it is just \[\frac{1-r^n}{1-r}\] in general
yeah, I'll try to remember that from now on :)

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