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

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

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(D) We observe that the numerator can be written as $(k+1)^3 - k^3 = k^3 + 3k^2 + 3k + 1 - k^3 = 3k^2 + 3k + 1$. Thus,the general term is $\frac{(k+1)^3 - k^3}{k^3(k+1)^3} = \frac{(k+1)^3}{k^3(k+1)^3} - \frac{k^3}{k^3(k+1)^3} = \frac{1}{k^3} - \frac{1}{(k+1)^3}$. This is a telescoping series. The sum is $\sum_{k=1}^n \left( \frac{1}{k^3} - \frac{1}{(k+1)^3} ight) = \left( 1 - \frac{1}{2^3} ight) + \left( \frac{1}{2^3} - \frac{1}{3^3} ight) + \dots + \left( \frac{1}{n^3} - \frac{1}{(n+1)^3} ight)$. As $n 	o \infty$,the sum converges to $1 - 0 = 1$.

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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