Convergence of a series

**poorna** · May 23rd 2010, 07:12 AM

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,

But I dont understand how we get this formula. Can you please help me

**Random Variable** · May 23rd 2010, 07:56 AM

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

now use partial fractions

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

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

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

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

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

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

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

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

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

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