Question

Let $S_n={lim}_{n\to \infty }(\frac{1}{n^6}+\frac{32}{n^6}+...+\frac{1}{n})$ and $T_n={lim}_{n\to \infty }(\frac{1}{n^6}+\frac{32}{n^6}+...+\frac{(n-1{)}^5}{n^6})$, then which of the following is/are true?

• A. $S_n\to \frac{1^{+}}{6}$
  
• B. $(S_n+T_n)<\frac{1}{3}$
  
• C. $(S_n+T_n)>\frac{1}{3}$
  
• D. $T_n\to \frac{1^{-}}{6}$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $S_n\to \frac{1^{+}}{6}$

C $(S_n+T_n)>\frac{1}{3}$

D $T_n\to \frac{1^{-}}{6}$

$S_n={lim}_{n\to \infty }(\frac{1}{n^6}+\frac{32}{n^6}+\frac{243}{n^6}+...\frac{1}{n})T_n={lim}_{n\to \infty }(\frac{1}{n^6}+\frac{32}{n^6}+\frac{243}{n^6}+....\frac{(n-1{)}^5}{n^6})$
$S_n={lim}_{n\to \infty }{\sum }_{r=1}^n\frac{r^5}{n^6}={\int }_0^1{x}^{5}dx=\frac{x^6}{6}|\begin{array}{c}1\\ 0\end{array}=\frac{1}{6}$
$\Rightarrow S_n={\left(\frac{1}{6}\right)}^{+};T_n={\left(\frac{1}{6}\right)}^{-}$

But since $x^5$ is concave upward the area included in $S_n$ for any two consecutive values of r is more than area excluded in $T_n$ for same values of r
$\Rightarrow S_n-\frac{1}{6}>\frac{1}{6}-T_n\Rightarrow S_n+T_n>\frac{1}{3}$