For a ∈ R (the set of all real numbers), a ≠ -1, lim (n→∞) (1ᵃ + 2ᵃ + ... + nᵃ)/((n+1)ᵃ⁻¹ [(na+1) +

Q: Why can we just ignore the $+1$ in $(n+1)^{a-1}$? Isn't that mathematically imprecise?

In the context of limits as $n \to \infty$, we are interested in the asymptotic behavior. The term $(n+1)^{a-1}$ can be written as $n^{a-1}(1 + \frac{1}{n})^{a-1}$. As $n \to \infty$, the term $(1 + \frac{1}{n})^{a-1}$ approaches $1^a = 1$. Thus, for large $n$, the expression is dominated by $n^{a-1}$. We aren't ignoring it; we are identifying the leading order term that dictates the limit's value.

Q: Why does the Riemann sum approach work here?

The Riemann sum definition states that $\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n f(\frac{k}{n}) = \int_0^1 f(x) dx$. In our numerator, we have $\sum k^a$. If we divide by $n^{a+1}$, we get $\frac{1}{n} \sum (\frac{k}{n})^a$, which perfectly matches the definition of the integral of $x^a$ from $0$ to $1$. It is the most elegant way to handle sums of powers.

Q: I got two values for $a$. How do I know which one is correct?

Both values are mathematically valid solutions to the quadratic equation derived from the limit. Unless the problem specifies a domain restriction (like $a > 0$), both $7$ and $-\frac{17}{2}$ are correct. Always check if your solutions violate any given constraints, such as the problem's statement that $a \neq -1$.

Question

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

Options:

(A) 5

(B) 7

(C) $-15/2$

(D) $-17/2$

Sigma Percentile · Accepted Answer

### The Symphony of Limits: Taming the Infinite Sum

Welcome, fellow traveler of the mathematical landscape. Today, we stand before a problem that looks like a chaotic mess of variables and summations.

At first glance, it is intimidating. But in the world of JEE Advanced, we do not fear complexity; we dissect it. We are going to break this limit down, piece by piece, until the answer reveals itself with absolute clarity.

### Phase 1: The Numerator's Hidden Geometry

Look at the numerator: $1^a + 2^a + \dots + n^a$. This is a sum of powers. When you see a sum of powers where $n$ approaches infinity, your mind should immediately jump to the **Riemann sum**.

We want to transform this discrete sum into a continuous integral. To do this, we need to create the structure $\frac{1}{n} \sum (\frac{k}{n})^a$. If we factor out $n^a$ from the sum, we get:

$$n^a \left( \left( \frac{1}{n} ight)^a + \left( \frac{2}{n} ight)^a + \dots + \left( \frac{n}{n} ight)^a ight)$$

This is almost there. If we multiply and divide by $n$, we get $n^{a+1} \cdot \frac{1}{n} \sum_{k=1}^n (\frac{k}{n})^a$.

As $n 	o \infty$, this sum converges to the definite integral:

$$\int_0^1 x^a \, dx = \left[ \frac{x^{a+1}}{a+1} ight]_0^1 = \frac{1}{a+1}$$

Thus, our numerator behaves asymptotically like $\frac{n^{a+1}}{a+1}$.

### Phase 2: The Denominator's Rhythmic Progression

Now, let us turn to the denominator. It consists of two parts: $(n+1)^{a-1}$ and the sum $S_2 = (na+1) + (na+2) + \dots + (na+n)$.

The second part is a classic Arithmetic Progression. The sum of an A.P. is $\frac{n}{2}(	ext{first} + 	ext{last})$. Here, the first term is $na+1$ and the last is $na+n$.

So, $S_2 = \frac{n}{2}(na+1 + na+n) = \frac{n}{2}(2na + n + 1) = \frac{n^2}{2}(2a+1) + \frac{n}{2}$.

As $n 	o \infty$, the term $\frac{n^2}{2}(2a+1)$ dominates, which we can rewrite as $n^2(a + \frac{1}{2})$.

Now, bring back the first part of the denominator, $(n+1)^{a-1}$, which behaves like $n^{a-1}$. Multiplying these together, the entire denominator behaves like:

$$n^{a-1} \cdot n^2 \left( a + \frac{1}{2} ight) = n^{a+1} \left( a + \frac{1}{2} ight)$$

### Phase 3: The Grand Convergence

We have successfully reduced the numerator and the denominator to their leading terms. Let us place them back into the limit:

$$\lim_{n 	o \infty} \frac{\frac{n^{a+1}}{a+1}}{n^{a+1}(a + \frac{1}{2})} = \frac{1}{60}$$

Notice the beauty here? The $n^{a+1}$ terms cancel out perfectly! We are left with a pure algebraic equation:

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

To make this cleaner, multiply the denominator by 2:

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

### Phase 4: The Quadratic Finale

We are in the home stretch. Cross-multiplying gives us $(a+1)(2a+1) = 120$.

Expanding the left side, we get $2a^2 + a + 2a + 1 = 120$, which simplifies to $2a^2 + 3a + 1 = 120$. Subtracting 120 from both sides yields the quadratic equation:

$$2a^2 + 3a - 119 = 0$$

Using the quadratic formula $a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we find:

$$a = \frac{-3 \pm \sqrt{9 - 4(2)(-119)}}{4} = \frac{-3 \pm \sqrt{9 + 952}}{4} = \frac{-3 \pm \sqrt{961}}{4}$$

Since $\sqrt{961} = 31$, our roots are:

$$a = \frac{-3 + 31}{4} = 7 \quad 	ext{and} \quad a = \frac{-3 - 31}{4} = -\frac{17}{2}$$

Both values are valid, and we have successfully tamed the beast. Remember, the path to the answer is just as important as the answer itself. Keep practicing, and keep that curiosity alive!

Animated Solution for Mathematics - Definite Integration: For $a \in R$ (the set of all real numbers), $a \neq = - 1, lim_{n \to \infty} \frac{( 1 ^{a} + 2 ^{a} + \dots + n ^{a} )}{( n + 1 ) ^{a - 1} [( na + 1 ) + ( na + 2 ) + \dots + ( na + n )]} = \frac{1}{60}$ . Then $a = .........$

Select Answer:

Visualized Solution

Analyzing the Limit Expression

Approximating the Numerator

Evaluating the Numerator Integral

Analyzing the Denominator Summation

Summing the Arithmetic Progression

Extracting the Leading Term of $S_{2}$

Leading Term of the Full Denominator

Substituting Back into the Limit

Simplifying the Limit Expression

Forming the Quadratic Equation

Standardizing the Quadratic Equation

Solving for $a$

Final Values of $a$

The Sigma Insight: Definite Integral as a Limit of a Sum

The Symphony of Limits

Phase 1

Phase 2

Phase 3

Phase 4

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