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Telescoping Series: find formula for nth term of sequence of partial sums Sn. Evaluate lim n approaches infinity Sn to obtain value of the series or state it diverges
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\[\sum_{k=1}^{\infty} \frac{ 1 }{ 4k^2+4k-3 }\]
got it too \[\frac{ 1 }{ (2k+3)(2k-1) }\]
did this series \[\frac{ 1 }{ 2 }\sum_{k=1}^{\infty}(\frac{ 1 }{ 2k-1 } - \frac{ 1}{ 2k+3 })\]
i got an answer of -1/2 but i dont think thats right
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