0
$\begingroup$

5(b) $$\frac{1}{(2k-1)(2k+1)(2k+3)}= \frac{A}{2k-1}+\frac{B}{2k+1}+\frac{C}{2k+3}$$ $$1=A(2k+1)(2k+3)+B(2k-1)(2k+3)+C(2k-1)(2k+1)$$ $$\left. \begin{array}{ll} k=\frac{1}{2}\Rightarrow& 1=8A\\ k=-\frac{1}{2}\Rightarrow& 1=-4B\\ k=-\frac{3}{2}\Rightarrow& 8C=1\\ \end{array} \right\}\Rightarrow \begin{array}{l} A=\frac18\\ B=-\frac14\\ C=\frac18\\ \end{array} $$ $$\frac18\sum\limits_{k=1}^n\frac{1}{2k-1} -\frac14\sum\limits_{k=1}^n\frac{1}{2k+1} +\frac18\sum\limits_{k=1}^n\frac{1}{2k+3}= $$

$$=\frac18\sum\limits_{k=1}^n\frac{1}{2k-1} -\frac14\sum\limits_{k=2}^{n+1}\frac{1}{2k-1} +\frac18\sum\limits_{k=3}^{n+2}\frac{1}{2k-1}= $$ $$=\frac18\left(1+\frac13\right) -\frac14\left(\frac13+\frac{1}{2n+1}\right) +\frac18\left(\frac{1}{2n+1}+\frac{1}{2n+3}\right)=\ldots$$

Sorry for photo, only thing i have. Appreciate any help

$\endgroup$
3
  • 1
    $\begingroup$ I think the technique used here is not to evaluate the sum $$ \sum \frac{1}{2k-1} $$ (it would be a bit tricky ... math.stackexchange.com/questions/141224/…) but to rather use the property $$ \sum_{k=1}^{n+1} = \sum_{k=1}^n + \sum_{k=n+1}^{n+1} $$ For example, the last term can be written as $$ \sum_{k=3}^{n+2} = \sum_{k=1}^n + \sum_{k=n+1}^{n+2} - \sum_{k=2}^3 $$ So just juggle with the indices like that ... $\endgroup$
    – Matti P.
    Commented Aug 5, 2020 at 11:00
  • $\begingroup$ Yup, there's no point evaluating the sum since it telescopes. $\endgroup$
    – Alexey Burdin
    Commented Aug 5, 2020 at 11:12
  • $\begingroup$ The problem is to find the infinite sum of $\sum_{n=1}^\infty\frac{1}{(2k-1)(2k+1)(2k+3)}$. It was applied partial fractions decomposition. And the last two sums were re-indexed in order to have same denominator. $\endgroup$
    – cgiovanardi
    Commented Aug 5, 2020 at 23:30

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .