Question

If $\underset{n\to \infty }{lim}\frac{1^a+2^a+3^a+...+n^a}{n^{a+1}}=\frac{1}{5}$ $(wherea>-1)$ then the value of $a$ is

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is C $4$

$\underset{n\to \infty }{lim}\frac{1^a+2^a+3^a+......+n^a}{n^{a+1}}=\frac{1}{5}$

$a>-1$

$a=2$ of $2$

Subsituting $a=2$

$\underset{n\to \infty }{lim}\frac{1^2+2^2+3^2+...n^2}{n^3}$

$\underset{n\to \infty }{lim}\frac{n(n+1)(2n+1)}{6n^3}=\underset{n\to \infty }{lim}\frac{\left(1\right)(1+\frac{1}{x})(2+\frac{1}{x})}{6}$

$=\frac{2}{6}=\frac{1}{3}\ne \frac{1}{5}$

of $3$

Subsituting

$\underset{x\to \infty }{lim}=\frac{1^3+2^3+3^3+........n^3}{n^4}$

$\underset{n\to \infty }{lim}\frac{n(n+1{)}^2}{4\times n^2}\frac{{(1+\frac{1}{x})}^2}{4}=\frac{1}{4}$

of $4$

$\underset{n\to \infty }{lim}\frac{1^4+2^4+3^4+......n^4}{n^5}$

$\underset{x\to \infty }{lim}\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30n^5}$

$\underset{n\to \infty }{lim}\frac{1(1+\frac{1}{x})(2+\frac{1}{x})(3+\frac{3}{n}-\frac{1}{n^2})}{30}$

$=\frac{6}{30}=\frac{1}{6}\therefore$