Neater Method for Solving Series?

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SUMMARY

The discussion centers on solving the series S = ∑_{i=1}^{n} (i^4)/((2i - 1)(2i + 1)). Vivek describes his method involving polynomial long division and partial fractions to simplify the expression into a more manageable form. He identifies constants A, B, C, and D to break down the sum into standard forms, specifically utilizing the sums of squares and linear sequences. The conversation invites alternative methods, emphasizing the search for a neater solution.

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  • Understanding of polynomial long division
  • Familiarity with partial fraction decomposition
  • Knowledge of summation formulas for series
  • Basic algebraic manipulation skills
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  • Research polynomial long division techniques in algebra
  • Study partial fraction decomposition methods in calculus
  • Explore summation formulas for series, particularly for squares and linear terms
  • Investigate alternative methods for solving series in mathematical analysis
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Mathematicians, educators, and students interested in advanced series solutions and algebraic techniques will benefit from this discussion.

maverick280857
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Another "Solvable" Series

Hi

This is another series

[tex]S = \sum_{i = 1}^{n} \frac{i^4}{(2i - 1)(2i + 1)}[/tex]

I did it by dividing the numerator by the denominator (and doing the division once again) and finally using partial fractions. Is there a neater way? (I will post my solution to the problem in a few hours...going out right now).

Thanks a lot,

Cheers
Vivek
 
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I've started by saying that:

[tex]\frac{i^4}{(2i-1)(2i+1)} = Ai^2 + Bi + \frac{C}{2i-1} + \frac{D}{2i + 1}[/tex]

Once solved for A, B, C and D the sum becomes:

[tex]S = \left( A\sum_{i = 1}^{n} i^2 \right) + \left( B\sum_{i = 1}^{n} i \right) + \left( \sum_{i = 1}^{n} \frac{C}{2i-1} + \frac{D}{2i + 1} \right)[/tex]

The former 2 sums are standard form, the latter one is a quite simple as it takes the form [itex]f(i) = g(i) - h(i)[/itex]

That's the simplest way I see of doing it, but someone else might come up with a better method.
 
Hi Zurtex

That's EXACTLY what I did :biggrin:

Thanks anyway for clarifying my method. Yeah, let's see if someone comes up with something different.

Cheers
Vivek
 

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