Question

Let $a_n = (1^2 + 2^2 + \cdots + n^2)$ and $b_n = n^n (n!)$. Then

• A. an<bn ∀n
• B. an> bn ∀n
• C. an = bn for infinitely many n
• D. an<bn if n be even and an>bn if n be odd

Answer 1

The Correct Option is B

We are given two sequences: \[ a_n = 1^2 + 2^2 + \cdots + n^2 \] and \[ b_n = n^n (n!) \] We need to compare these sequences and determine the relationship between them for all \( n \).

Step 1: Understanding \( a_n \)
The sum of squares of the first \( n \) natural numbers is given by: \[ a_n = 1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6} \] This expression grows roughly as \( \frac{n^3}{3} \) for large \( n \), which means that \( a_n \) increases cubicly as \( n \) increases. 

Step 2: Understanding \( b_n \)
The sequence \( b_n \) is given by: \[ b_n = n^n \cdot (n!) \] This grows much faster than \( a_n \), since \( n^n \) grows exponentially and \( n! \) grows faster than any polynomial function. To get a rough idea of the growth rate of \( b_n \), recall that \( n! \) grows as \( \sim \sqrt{2\pi n} \left(\frac{n}{e}\right)^n \) (by Stirling's approximation), and multiplying this by \( n^n \) gives an even faster growth rate. 

Step 3: Comparing \( a_n \) and \( b_n \)
Clearly, \( b_n \) grows much faster than \( a_n \). For large \( n \), the exponential growth of \( b_n \) will always outpace the cubic growth of \( a_n \). To illustrate, let’s check the first few values of \( a_n \) and \( b_n \): - For \( n = 1 \): \[ a_1 = 1^2 = 1, \quad b_1 = 1^1 \cdot 1! = 1 \] So \( a_1 = b_1 \). - For \( n = 2 \): \[ a_2 = 1^2 + 2^2 = 1 + 4 = 5, \quad b_2 = 2^2 \cdot 2! = 4 \cdot 2 = 8 \] So \( a_2 < b_2 \). - For \( n = 3 \): \[ a_3 = 1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14, \quad b_3 = 3^3 \cdot 3! = 27 \cdot 6 = 162 \] So \( a_3 < b_3 \). - For \( n = 4 \): \[ a_4 = 1^2 + 2^2 + 3^2 + 4^2 = 1 + 4 + 9 + 16 = 30, \quad b_4 = 4^4 \cdot 4! = 256 \cdot 24 = 6144 \] So \( a_4 < b_4 \). As we can see, for every \( n \), \( b_n \) grows much faster than \( a_n \). 

Step 4: Conclusion
From the calculations and growth rates, we can conclude that: \[ a_n < b_n \quad \forall n \]

Answer:

\[ \boxed{a_n < b_n \, \forall n} \]