frac{1}{1 cdot 2 cdot 3} + frac{1}{3 cdot 4 c | vedclass

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(B) The general term of the series is $T_n = \frac{1}{(2n - 1)(2n)(2n + 1)}$.Using partial fractions,we have $T_n = \frac{1}{2} \left[ \frac{1}{2n - 1

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$\log_{e} 2 - \frac{1}{2}$

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(B) The general term of the series is $T_n = \frac{1}{(2n - 1)(2n)(2n + 1)}$.Using partial fractions,we have $T_n = \frac{1}{2} \left[ \frac{1}{2n - 1} - \frac{2}{2n} + \frac{1}{2n + 1} \right] = \frac{1}{2} \left[ \left( \frac{1}{2n - 1} - \frac{1}{2n} \right) - \left( \frac{1}{2n} - \frac{1}{2n + 1} \right) \right]$.Summing from $n=1$ to $\infty$:$S = \frac{1}{2} \sum_{n=1}^{\infty} \left( \frac{1}{2n - 1} - \frac{1}{2n} \right) - \frac{1}{2} \sum_{n=1}^{\infty} \left( \frac{1}{2n} - \frac{1}{2n + 1} \right)$.We know the logarithmic series expansion $\log_{e} 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots = \sum_{n=1}^{\infty} \left( \frac{1}{2n - 1} - \frac{1}{2n} \right)$.Also,$\sum_{n=1}^{\infty} \left( \frac{1}{2n} - \frac{1}{2n + 1} \right) = \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \dots = 1 - \log_{e} 2$.Substituting these values:$S = \frac{1}{2} (\log_{e} 2) - \frac{1}{2} (1 - \log_{e} 2) = \frac{1}{2} \log_{e} 2 - \frac{1}{2} + \frac{1}{2} \log_{e} 2 = \log_{e} 2 - \frac{1}{2}$.

$\frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{3 \cdot 4 \cdot 5} + \frac{1}{5 \cdot 6 \cdot 7} + \dots \infty = $

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