Determine whether the following series is convergent or divergent. If convergent, find the sum. $$\sum_{i=1}^{\infty}\frac{6}{24 i-4 i^2-35}$$

Since the limit of the series is zero, I know that it is not divergent (divergence test).

How do i prove that the series is convergent, and futhermore, find the sum?

I rewrote the series (using partial fraction decomposition) as:

$$\sum_{i=1}^{\infty}\frac{6}{24 i-4 i^2-35}=\sum_{i=1}^{\infty}\frac1{1/4i-10(-(1/4 i-7))}$$

But I don't know what to do from here.

  • $\begingroup$ Why do you think this kind of math formatting is acceptable? $\endgroup$ – zhw. Oct 2 '15 at 23:41
  • 1
    $\begingroup$ Sorry, I am new to stack exchange and I am having a hard time figuring out how to use the appropriate formatting. $\endgroup$ – Hvb123 Oct 2 '15 at 23:43

Hint. You may use a partial fraction decomposition: $$ \frac{6}{24 i-4 i^2-35}=\frac{3}{2 i-5}-\frac{3}{2 (i-1)-5} $$ then you may observe that the series is a telescoping one:

$$ \sum_{i=1}^n\frac{6}{24 i-4 i^2-35}=\frac{3}{2 n-5}+\frac35,\qquad n\geq1, $$

then the conclusion is direct.

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