Question

If $S_n=\frac{1}{1^3}+\frac{1+2}{1^3+2^3}+...+\frac{1+2+3+...+n}{1^3+2^3+3^3...+n^3},n=1,2,3...$ Then $S_n$ is not greater than

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

Answer 1

The correct option is C $2$

$S_r=\frac{1+2+3+4+.....+r}{1^3+2^3+.....+r^3}=\frac{\frac{r(r+1)}{2}}{\frac{r^2{(r+1)}^2}{4}}$
$S_r=\frac{2}{r(r+1)}$
$\sum _{r=1}^n{S}_r=S_n=\sum _{r=1}^n\frac{2}{r(r+1)}$
$=2\sum _{r=1}^n\frac{1}{r}-\frac{1}{r+1}$
$=2\left(\begin{array}{cc}1/1& -1/2\\ 1/2& -1/3\\ .& .\\ .& .\\ 1/n& -1/n+1\end{array}\right)$
$=2(1-\frac{1}{n+1})=\frac{2n}{n+1}$
So this value not greater than $2$ when, $n⟶\infty$
$\underset{n\to \infty }{lim}\frac{2}{1+\frac{1}{n}}=2$