The value of sumlimits_{k = 1}^infty {frac{{ | vedclass

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(D) We observe that the numerator $3k^2 + 3k + 1$ can be written as $(k+1)^3 - k^3$.Thus,the general term of the series is $T_k = \frac{(k+1)^3 - k^3}

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$1$

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(D) We observe that the numerator $3k^2 + 3k + 1$ can be written as $(k+1)^3 - k^3$.Thus,the general term of the series is $T_k = \frac{(k+1)^3 - k^3}{k^3(k+1)^3}$.This simplifies to $T_k = \frac{(k+1)^3}{k^3(k+1)^3} - \frac{k^3}{k^3(k+1)^3} = \frac{1}{k^3} - \frac{1}{(k+1)^3}$.This is a telescoping series.The sum $S_n = \sum_{k=1}^n \left( \frac{1}{k^3} - \frac{1}{(k+1)^3} ight) = \left( \frac{1}{1^3} - \frac{1}{2^3} ight) + \left( \frac{1}{2^3} - \frac{1}{3^3} ight) + \dots + \left( \frac{1}{n^3} - \frac{1}{(n+1)^3} ight)$.All intermediate terms cancel out,leaving $S_n = 1 - \frac{1}{(n+1)^3}$.Taking the limit as $n 	o \infty$,we get $S = \lim_{n 	o \infty} \left( 1 - \frac{1}{(n+1)^3} ight) = 1 - 0 = 1$.

The value of $\sum\limits_{k = 1}^\infty {\frac{{3{k^2} + 3k + 1}}{{{{\left( {{k^2} + k} \right)}^3}}}} $ is equal to

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