Hi everyone! Need help with a series... sigma n=1 to infinity of 3(5)(7)times...(2n+1) divided by 5(14)(29)times...(3n^2+2) I posted a pic of my work...please click on 20.bmp to see a better written example of this problem... ultimately I need to understand why they wrote the series expansion they way they did, then how to construct a similar converging series that is larger so that I can prove that the above series indeed converges!...thanks!

Question

anonymous · Answer

attachment

anonymous · Answer

so, we have $$\sum_{n=1}^\infty\frac{2n+1}{3n^2+2}$$
this series DIVERGES but all the common tests for convergence fail!!

anonymous · Answer

check the limit test. it gives you "0"

anonymous · Answer

uhmm...the professor said this converges by the comparison test

anonymous · Answer

wait.. 
they are multiplying!!!
$$a_n\ne \frac{2n+1}{3n^2+2}$$

anonymous · Answer

yes...they are all multiplying...

anonymous · Answer

infact,
$$a_n={3\over5}\times\prod_{n=0}^\infty\left(2n+1\over3n^2+2\right)$$

anonymous · Answer

they wrote in x's to signify multiplication...those are not addition symbols

anonymous · Answer

that is why you cannot use any of the regular methods!!

anonymous · Answer

yeah...i'm kinda lost here...one sec while i type out a thought that may help...

anonymous · Answer

so, did you notice that $a_n$ is not a ratio? it is a product of another sequence.

anonymous · Answer

$$a_n=\prod_{k=0}^n\left(2k+1\over3k^2+2\right)$$

anonymous · Answer

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