# Convergence of a series

#### poorna

I have a doubt in the following problem.

Determine if the following series converges or diverges. If it converges determine its sum.

It is given that the general formula for the partial sums is,

#### Random Variable

$$\displaystyle \frac{1}{i^{2}-1} = \frac{1}{(i+1)(i-1)}$$

now use partial fractions

$$\displaystyle \frac{1}{(i+1)(i-1)} = \frac{A}{i+1} + \frac{B}{i-1}$$

so $$\displaystyle A(i-1) + B(i+1) = 1$$

$$\displaystyle (A+B)i +(-A+B) = 1$$

therefore $$\displaystyle A+B = 0$$ and $$\displaystyle -A+B = 1$$

which means that $$\displaystyle A = -\frac{1}{2}$$ and $$\displaystyle B = \frac{1}{2}$$

so we have $$\displaystyle \sum^{n}_{i=2} \frac{1}{i^{2}-1} = \sum^{n}_{i=2} \Big(\frac{1}{2(i-1)} - \frac{1}{2(i+1)} \Big)$$

$$\displaystyle = \frac{1}{2} - \frac{1}{6} + \frac{1}{4} - \frac{1}{8} + \frac{1}{6} - \frac{1}{10} + ...+ \frac{1}{2(n-1)} - \frac{1}{2(n+1)}$$

so everything cancels except the first two terms of $$\displaystyle \frac{1}{2(i-1)}$$ and the last two terms of $$\displaystyle - \frac{1}{2(i+1)}$$

$$\displaystyle = \frac{1}{2} + \frac{1}{4} - \frac{1}{2\big(n-1+1\big)} - \frac{1}{2(n+1)} = \frac{3}{4} - \frac{1}{2n} - \frac{1}{2(n+1)}$$

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