Question

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

• A. $\frac{1}{8}$
• B. $\frac{1}{4}$
• C. $\frac{1}{2}$
• D. $1$

Answer 1

The correct option is D $1$

Apply distributing rule

${(a+b)}^3=a^3+3a^{2}b+3a{b}^2+b^3$

$a=k^2,b=k$

$={\left(k^2\right)}^3+3{\left(k^2\right)}^{2}k+3k^2{k}^2+k^3$

Simplify $=k^6+3k^5+3k^4+k^3$

$={\sum }_{k=1}^{\infty }\frac{3k^2+3k+1}{k^6+3k^5+3k^4+k^3}$

For an infinite upper boundary, if $a_n\to 0,then{\sum }_{n=k}^{\infty }(a_{n+1}-a_n)=-a_k$

$\frac{3k^2+3k+1}{k^6+3k^5+3k^4+k^3}=-\left(\frac{1}{{(k+1)}^3}\right)-(-\frac{1}{k^3})$

$a_{n+1}=-\left(\frac{1}{{(k+1)}^3}\right)a_n=-\frac{1}{k^3}{a}_k=a_1=-\frac{1}{1^3}$

${lim}_{k\to \infty }(-\frac{1}{k^3})$

${lim}_{x\to a}[c\cdot f\left(x\right)]=c\cdot {lim}_{x\to a}f\left(x\right)$

$=-{lim}_{k\to \infty }\left(\frac{1}{k^3}\right)$

${lim}_{x\to a}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{{lim}_{x\to a}f\left(x\right)}{{lim}_{x\to a}g\left(x\right)},{lim}_{x\to a}g\left(x\right)\ne 0$

With the exception of indeterminate form

$=-\frac{{lim}_{k\to \infty }\left(1\right)}{{lim}_{k\to \infty }\left(k^3\right)}$

${lim}_{k\to \infty }\left(1\right)$

${lim}_{x\to a}c=c$

$=1$

${lim}_{k\to \infty }\left(k^3\right)$

{Apply Infinity Property}:

${lim}_{x\to \infty }(a{x}^n+\cdots +bx+c)=\infty ,a>0,nisodd$

$=1,n=3$

$=\infty$

$=-\frac{1}{\infty }$

$=0$

By the telescoping series test : $\sum {}_{k=1}^{\infty }\frac{3k^2+3k+1}{k^6+3k^5+3k^4+k^3}=-a_k$

$=-(-\frac{1}{1^3})$

$=1$

Option D is correct answer