Convergence and sum of an infinite series: $\sum_{i=1}^{\infty}\frac{6}{24 i-4 i^2-35}$

Question

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.

Answer 1

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.