Question

Let $S_n,n=1,2,3,.........,$ be the sum of infinite geometric series whose first term is $n$ and the common ratio is $\frac{1}{n+1}$. Evaluate
${lim}_{n\to \infty }\frac{S_1{S}_n+S_2{S}_{n-1}+S_3{S}_{n-2}+...+S_n{S}_1}{S_1^2+S_2^2+.....+S_n^2}$

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

Answer 1

Answer: (C)

$\frac{1}{2}$

$S_n=\frac{n}{1-\frac{1}{n+1}}$

$S_n=n+1$

$\therefore S_1{S}_n+S_2{S}_{n-1}+S_3{S}_{n-2}+...+S_n{S}_1$

$=2.(n+1)+3.n+4.(n-1)+...+(n+1).2$

$={\sum }_{r=1}^n(r+1)(n-r+2)$

$={\sum }_{r=1}^n(nr-r^2+2r+n-r+2)$

$={\sum }_{r=1}^n\left\{\right(n+1)r-r^2+(n+2\left)\right\}$

$=(n+1){\sum }_{r=1}^{n}r-{\sum }_{r=1}^n{r}^2+(n+2){\sum }_{r=1}^{n}1$

$=(n+1)\sum n-\sum n^2+(n+2).n$

$=\frac{(n+1)n(n+1)}{2}-\frac{n(n+1)(2n+1)}{6}+n(n+2)$

$=\frac{n}{6}\left\{3\right(n^2+2n+1)-(2n^2+3n+1)+6n+12\}$

$=\frac{n}{6}\{n^2+9n+14\}\cdots \left(1\right)$

and $S_1^2+S_2^2+.....+S_n^2=2^2+3^2+...+(n+1{)}^2$

$=\frac{(n+1)(n+2)(2n+3)}{6}-1$

$=\frac{n(2n^2+9n+13)}{6}\cdots \left(2\right)$

from $\left(1\right)$ and $\left(2\right)$, we get

$\frac{S_1{S}_n+S_2{S}_{n-1}+S_3{S}_{n-2}+...+S_n{S}_1}{S_1^2+S_2^2+.....+S_n^2}=\frac{n^2+9n+14}{2n^2+9n+13}$

$=\frac{1+\frac{9}{n}+\frac{14}{n^2}}{2+\frac{9}{n}+\frac{13}{n^2}}$

$\therefore \underset{n\to \infty }{lim}\frac{S_1{S}_n+S_2{S}_{n-1}+S_3{S}_{n-2}+...+S_n{S}_1}{S_1^2+S_2^2+.....+S_n^2}$

$\therefore \underset{n\to \infty }{lim}\frac{n^2+9n+14}{2n^2+9n+13}=\frac{1+0+0}{2+0+0}=\frac{1}{2}$