Question

Can anyone help with evaluating a series? I first answered it via proof by induction but I'm sure there must be a way to evaluate it.

Answer 1

Here's the question:

Show that the sum S(N) of the first N terms of the series
\[\frac{ 1 }{ 1\times2\times3 } + \frac{ 3 }{ 2\times3\times4 } + \frac{ 5 }{ 3\times4\times5 } + ... + \frac{ 2n-1 }{n(n+1)(n+2)} + ...\]
is
\[\frac{ 1 }{ 2 }(\frac{ 3 }{ 2 } + \frac{ 1 }{ N+1 } - \frac{5}{N+2})\]

Answer 2

I'm not exactly certain of the Solution. Though I did manage to find a simplified form of Nth term.

T(N) = \[\frac{ 2n-1 }{ (n)(n+1)(n+2) } = \frac{ A }{ n } + \frac{ B }{( n+1) } + \frac{ C }{ n+2 }\]

I'm trusting you can do the simplification
This implies 2n-1 = A(n+1)(n+2)+B(n)(n+2) +C (n)(n+1)
=(A+B+C)n^{2} + (3A+ 2B+ C)n + 2A

Which gives us, A+B+C=0
3A+2B+C=2 which implies 2A +B=2
2A=-1
This means A=-1/2, B=3, C= 5/2

Thus the n'th Term becomes

\[T(n) = \frac{ -1 }{ 2n } + \frac{ 3 }{ n+1 } + \frac{ 5 }{ 2(n+2) }\]

Can't really get beyond this, though.

Answer 3

And here's a correction, it's -5, not +5... -_-"

Answer 4

If I remember correctly this type of series can be solved by the method of differences
I'd have to look it up before I am sure. if I have time I'll get back to you today.

Answer 5

Thanks for the help - I did try to simplify the nth term and got what you did, but I have no idea how to sum the terms. I'll see if I can find anything about the method of differences with a quick google. :P