Question

For $a$ $\in R$ (the set of all real numbers), $a\ne -1,\underset{n\to \infty }{lim}\frac{(1^a+2^a+\dots +n^a)}{(n+1{)}^{a-1}[(na+1)+(na+2)+\dots +(na+n)]}=\frac{1}{60}$. Then $a=$

• A. $5$
• B. $7$
• C. $\frac{-15}{2}$
• D. $\frac{-17}{2}$

Answer 1

The correct option is B $7$

Denominator :
${(n+1)}^{a-1}\left[\right(na+1)+(na+2)++(na+n\left)\right]$
= ${(n+1)}^{a-1}\left[\right(n^{2}a+n(n+1)/2]$
= { ${(n+1)}^{a-1}${(2a+1)$n^2$+ n}}/2
Numerator :
${lim}_{n\to \infty }(1^a+2^a+3^a+-----+n^a)$ =${lim}_{n\to \infty }{\int }_0^n{x}^{a}dx$
= $\frac{1}{a+1}{lim}_{n\to \infty }{n}^{a+1}$
Now as a whole :

$L=\underset{n\to \infty }{lim}\frac{(1^a+2^a+\dots +n^a)}{(n+1{)}^{a-1}[(na+1)+(na+2)+\dots +(na+n)]}$
$=\frac{2}{a+1}{lim}_{n\to \infty }\frac{n^{a+1}}{{(n+1)}^{a-1}\left[\right(2a+1)n^2+n]}$

Dividing Numerator and Denominator by${(n+1)}^{a+1}$, we get
$L=\frac{2}{a+1}{lim}_{n\to \infty }\frac{{\left(\frac{n}{n+1}\right)}^{a+1}}{(2a+1){\left(\frac{n}{n+1}\right)}^2+\frac{n}{{(n+1)}^2}}$
As $\underset{n\to \infty }{lim}{\left(\frac{n}{n+1}\right)}^r=\underset{n\to \infty }{lim}{\left(\frac{1}{1+\frac{1}{n}}\right)}^r=1$ and ${lim}_{n\to \infty }\frac{n}{(n+1{)}^r}=0ifr>1$

We get,
$\frac{2}{(a+1)(2a+1)}=\frac{1}{60}$
$\therefore (a+1)(2a+1)=120$
By solving the above quadratic equation we get two values of $a$
$a=7$ or $(-17/2)$

For a<0 numerator diverges

$\therefore a=7$