Here's the question you clicked on:
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Proof Question
If \[ T_{n} = \frac{ n-1 }{ n }\], prove \[T_{n+1} - T _{n-1} = \frac{ 2 }{ n^2 -1 }\]
This is nice exercise for the hands
\[ T_{n+1} = \frac{n}{n+1}\]
So \[ T_{n+1} - T_{n-1} = \frac{ n }{ n+1 } - \frac{n-1}{n} \]
Sorry \[ T_{n+1} - T_{n-1} = \frac{ n }{ n+1 } - \frac{n-2}{n-1}\]
Just bring to the common denominator by multiplying the denominators and recall that \[ (n+1)(n-1) = ...^2 - 1^2 \]
it seems like i cant go any further