If $$\lim_{n \to \infty} \left(\frac{1^a + 2^a + \ldots + n^a}{(n+1)^{a-1}[(na+1) + (na+2) + \ldots + (na+n)]}\right) = \frac{1}{60}$$ for some positive real number $$a$$, then $$a$$ is equal to

Let us denote the required limit by $$L$$

$$L=\lim_{n\to\infty}\left(\frac{1^{a}+2^{a}+\ldots +n^{a}}{(n+1)^{\,a-1}\Big[(na+1)+(na+2)+\ldots +(na+n)\Big]}\right)$$ $$-(1)$$

First we examine the numerator. For large $$n$$ the sum of powers can be expanded by the Euler-Maclaurin (or simply the integral) approximation:

$$1^{a}+2^{a}+\ldots +n^{a}=\frac{n^{\,a+1}}{a+1}+\frac{n^{\,a}}{2}+O\!\left(n^{\,a-1}\right)$$ $$-(2)$$

Only the two leading terms are written because every later term will contain a lower power of $$n$$ and will vanish after division by the denominator which will also be of order $$n^{\,a+1}$$.

Next we handle the bracketed sum in the denominator. The terms inside form an arithmetic progression whose first term is $$na+1$$ and last term is $$na+n$$ with common difference $$1$$. For an arithmetic progression of $$n$$ terms, the sum equals half the number of terms multiplied by the sum of the extreme terms, therefore

$$\big(na+1\big)+\big(na+2\big)+\ldots +\big(na+n\big)=\frac{n}{2}\Big[\,(na+1)+(na+n)\Big]$$ $$-(3)$$

Simplifying the bracket inside (3):

$$(na+1)+(na+n)=2na+n+1=n(2a+1)+1$$ $$-(4)$$

Putting (4) into (3) we get

$$\big(na+1\big)+\ldots +\big(na+n\big)=\frac{n}{2}\Big[n(2a+1)+1\Big]$$

$$=\frac{n^{2}}{2}(2a+1)+\frac{n}{2}$$ $$-(5)$$

Now multiply this by the factor $$(n+1)^{\,a-1}$$ appearing in the denominator of (1). Write $$(n+1)^{\,a-1}$$ in powers of $$n$$ by the binomial expansion for large $$n$$:

$$(n+1)^{\,a-1}=n^{\,a-1}\Big(1+\frac{1}{n}\Big)^{a-1}=n^{\,a-1}\left[1+\frac{a-1}{n}+O\!\left(\frac{1}{n^{2}}\right)\right]$$ $$-(6)$$

Combining (5) and (6), the entire denominator becomes

$$\Big[(n+1)^{\,a-1}\Big]\Big[(na+1)+\ldots +(na+n)\Big]$$

$$=n^{\,a-1}\left[1+\frac{a-1}{n}+O\!\left(\frac{1}{n^{2}}\right)\right]\left[\frac{n^{2}}{2}(2a+1)+\frac{n}{2}\right]$$ $$-(7)$$

In (7) factor out the highest power of $$n$$, namely $$n^{\,a+1}$$, and collect the numerical coefficient:

$$=n^{\,a+1}\,\frac{2a+1}{2}\left[1+\frac{a-1}{n}+O\!\left(\frac{1}{n}\right)\right]\left[1+\frac{1}{n(2a+1)}\right]$$

$$=n^{\,a+1}\,\frac{2a+1}{2}\Big[1+O\!\left(\frac{1}{n}\right)\Big]$$ $$-(8)$$

Thus, to the leading order

$$\Big[(n+1)^{\,a-1}\Big]\Big[(na+1)+\ldots +(na+n)\Big]=\frac{2a+1}{2}\;n^{\,a+1}+O\!\left(n^{\,a}\right)$$ $$-(9)$$

Now we take the ratio appearing in (1). Using the leading terms from (2) and (9) we have

$$\frac{1^{a}+2^{a}+\ldots +n^{a}}{(n+1)^{\,a-1}\Big[(na+1)+\ldots +(na+n)\Big]}$$

$$=\frac{\displaystyle\frac{n^{\,a+1}}{a+1}+O\!\left(n^{\,a}\right)}{\displaystyle\frac{2a+1}{2}\;n^{\,a+1}+O\!\left(n^{\,a}\right)}$$ $$-(10)$$

Divide the numerator and denominator of the right-hand side of (10) by $$n^{\,a+1}$$ to obtain the finite limit:

$$=\frac{\dfrac{1}{a+1}+O\!\left(\dfrac{1}{n}\right)}{\dfrac{2a+1}{2}+O\!\left(\dfrac{1}{n}\right)}$$ $$-(11)$$

As $$n\rightarrow\infty$$ all the $$O\!\left(\dfrac{1}{n}\right)$$ terms vanish, leaving

$$L=\frac{\dfrac{1}{a+1}}{\dfrac{2a+1}{2}}=\frac{2}{(a+1)(2a+1)}$$ $$-(12)$$

The problem statement tells us that this limit equals $$\dfrac{1}{60}$$, so equate (12) to that value:

$$\frac{2}{(a+1)(2a+1)}=\frac{1}{60}$$ $$-(13)$$

Cross-multiplying in (13):

$$2\cdot 60=(a+1)(2a+1)$$

$$120=(a+1)(2a+1)$$ $$-(14)$$

Expand the product in (14):

$$(a+1)(2a+1)=2a^{2}+a+2a+1=2a^{2}+3a+1$$ $$-(15)$$

Hence (14) becomes the quadratic equation

$$2a^{2}+3a+1-120=0$$

$$2a^{2}+3a-119=0$$ $$-(16)$$

Use the quadratic formula for $$ax^{2}+bx+c=0$$, where $$a=2$$, $$b=3$$, $$c=-119$$:

$$a=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{-3\pm\sqrt{3^{2}-4\cdot 2\cdot(-119)}}{2\cdot 2}$$ $$-(17)$$

The discriminant evaluates to

$$3^{2}-4\cdot 2\cdot(-119)=9+952=961$$ $$-(18)$$

Since $$\sqrt{961}=31$$, substitute back in (17):

$$a=\frac{-3\pm 31}{4}$$ $$-(19)$$

This gives two possible values:

$$a=\frac{-3+31}{4}=\frac{28}{4}=7\quad\text{or}\quad a=\frac{-3-31}{4}=\frac{-34}{4}=-8.5$$ $$-(20)$$

The question specifies that $$a$$ is a positive real number, therefore we discard the negative root and accept

$$a=7$$ $$-(21)$$

Looking at the options, Option 3 corresponds to the value $$7$$.

Hence, the correct answer is Option 3.