Question

Let $S_n=\sum _{k=1}^n\frac{k}{(k-1{)}^{4/3}+(k^2-1{)}^{2/3}+(k+1{)}^{4/3}}$ and $\underset{n\to \infty }{lim}\frac{S_n}{n^{2/3}}=\frac{1}{p}.$ Then the value of $p$ is

Answer 1

Let $T_k=\frac{k}{(k-1{)}^{4/3}+(k^2-1{)}^{2/3}+(k+1{)}^{4/3}}$
Since $a^3-b^3=(a-b)(a^2+ab+b^2),$
$T_k=\frac{k\left[\right(k+1{)}^{2/3}-(k-1{)}^{2/3}]}{{\left[\right(k+1{)}^{2/3}]}^3-{\left[\right(k-1{)}^{2/3}]}^3}{T}_k=\frac{1}{4}\left[\right(k+1{)}^{2/3}-(k-1{)}^{2/3}]$

$T_1=\frac{1}{4}[2^{2/3}-0]T_2=\frac{1}{4}[3^{2/3}-1^{2/3}]T_3=\frac{1}{4}[4^{2/3}-2^{2/3}]\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot T_{(n-1)}=\frac{1}{4}\left[\right(n{)}^{2/3}-(n-2{)}^{2/3}]T_n=\frac{1}{4}\left[\right(n+1{)}^{2/3}-(n-1{)}^{2/3}]$

$S_n=T_1+T_2+T_3+\cdots +T_n{S}_n=\frac{1}{4}\left[\right(n+1{)}^{2/3}+n^{2/3}-1-0]=\frac{1}{4}\left[\right(n+1{)}^{2/3}+n^{2/3}-1]$

$\frac{S_n}{n^{2/3}}=\frac{1}{4}[{(1+\frac{1}{n})}^{2/3}+1-\frac{1}{n^{2/3}}]\Rightarrow \underset{n\to \infty }{lim}\frac{S_n}{n^{2/3}}=\frac{1}{2}$