Convergence of a series

May 2009
86
1
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
 
May 2009
959
362
\(\displaystyle \frac{1}{i^{2}-1} = \frac{1}{(i+1)(i-1)} \)


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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