The sum of the series 1 + frac{1}{1 + 2} + fr | vedclass

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(C) The $r$-th term of the series is given by $T_r = \frac{1}{1 + 2 + 3 + \dots + r}$.Using the sum formula for the first $r$ natural numbers,$1 + 2 +

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$\frac{20}{11}$

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(C) The $r$-th term of the series is given by $T_r = \frac{1}{1 + 2 + 3 + \dots + r}$.Using the sum formula for the first $r$ natural numbers,$1 + 2 + 3 + \dots + r = \frac{r(r + 1)}{2}$.Thus,$T_r = \frac{2}{r(r + 1)} = 2 \left( \frac{1}{r} - \frac{1}{r + 1} \right)$.The sum of the first $10$ terms is $S_{10} = \sum_{r=1}^{10} T_r = 2 \sum_{r=1}^{10} \left( \frac{1}{r} - \frac{1}{r + 1} \right)$.Expanding the summation,we get $S_{10} = 2 \left[ (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{10} - \frac{1}{11}) \right]$.This is a telescoping series where intermediate terms cancel out,leaving $S_{10} = 2 \left( 1 - \frac{1}{11} \right)$.$S_{10} = 2 \left( \frac{10}{11} \right) = \frac{20}{11}$.

The sum of the series $1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + \dots$ up to $10$ terms is:

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