Question
The $\sum\frac{19}{1 \cdot 2 \cdot 3}\frac{1}{4} + \frac{28}{2 \cdot 3 \cdot 4}\frac{1}{8} + \frac{39}{3 \cdot 4 \cdot 5}\frac{1}{16} + \frac{52}{4 \cdot 5 \cdot 6}\frac{1}{32} + \ldots\ldots. +$ upto infinite terms is equal to
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Solution
Sol. [Key Concept: Split each term by partial fractions; the product series telescopes to a finite infinite-sum.]
$$S_{n} = \sum_{r = 1}^{n}\mspace{2mu}\frac{\left( r^{2} + 6r + 12 \right)}{(r)(r + 1)(r + 2)}\frac{1}{2^{r + 1}} = \sum_{r = 1}^{n}\mspace{2mu}\frac{2(r + 2)(r + 3) - r(r + 4)}{(r)(r + 1)(r + 2)^{r + 1}}$$
$$\begin{matrix} & \ = \sum_{r = 1}^{n}\mspace{2mu}\mspace{2mu}\left( \frac{r + 3}{r(r + 1)2^{r}} - \frac{r + 4}{(r + 1)(r + 2)2^{r + 1}} \right) \\ & \ = 1 - \frac{n + 4}{(n + 1)(n + 2)2^{n + 1}} \end{matrix}$$
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