Question

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

Answer 1

As we know${\int }_a^{b}f\left(x\right)dx=\underset{h\to 0}{lim}f(a+rh)$

More generally ${\int }_0^{k}f\left(x\right)dx=\underset{h\to 0}{lim}\frac{1}{n}\sum _{r=1}^{kn}f(\frac{r}{n})$ So, from given question

And we also know the formula of sum of infinte G.P

which is $\frac{a}{1+r}$ (where a and r used from usual notation)

so writing given question in general form $\underset{n\to \infty }{lim}\frac{{\sum }_{r=1}^n{(S}_{n-r+1})\left(S_r\right)}{{\sum }_{r=1}^n{\{S_r}^2\}}$ $where{S}_r=\frac{\left(r\right)(r+1)}{(r+2)}$= $\underset{n\to \infty }{lim}\frac{{\sum }_{r=1}^n\left\{\frac{(n-r+1)(n-r+2)}{n-r+3}\right\}\left\{\frac{\left(r\right)(r+1)}{(r+2)}\right\}}{{\sum }_{r=1}^n{\{\frac{\left(r\right)(r+1)}{(r+2)}}^2\}}$

Then we divide and multiply by $n^5$,and splitting each n into each bracket for making the given expression as limit as sum and we write $\frac{r}{n}=x$ and as tending to infinty $\frac{1}{n}=0$ so after that we got

$\frac{{\int }_0^1\left\{\frac{(1-x)(1-x)}{(1-x)}\right\}{\int }_0^1\left\{\frac{\left(x\right)\left(x\right)}{\left(x\right)}\right\}}{{\{{\int }_0^1\frac{\left(x\right)\left(x\right)}{\left(x\right)}\}}^2}=\frac{{\int }_0^1\{1-x\}\left\{x\right\}}{{\{{\int }_0^{1}x\}}^2}$

and after simple integration we got $\frac{\frac{1}{2}\times \frac{1}{2}}{{\left(\frac{1}{2}\right)}^2}=1$