Question

Series Expansion $ n^6 + \frac{1}{2} n^4 + \frac{1}{3} n^2 + \cdots + \frac{1}{n} C_n + 1 \quad n \to \infty $

• A. \( 9 \)
• B. \( 8 \)
• C. \( 10 \)
• D. 7

Hint:

For series expansions, analyze the highest-order terms and their behavior as \( n \) becomes large to identify the limit of the series.

Answer 1

The Correct Option is C

The given series is:

$$ n^6 + \frac{1}{2}n^4 + \frac{1}{3}n^2 + \cdots + \frac{1}{n} C_n + 1 $$

We need to find the asymptotic behavior of this series as \( n \to \infty \).

Let's consider the general term of the series:

$$ a_k = \frac{1}{k}n^{6 - 2(k-1)} $$

Here, the series general term is:

$$ a_1 = n^6, \, a_2 = \frac{1}{2}n^4, \, a_3 = \frac{1}{3}n^2, $$

Continuing this pattern, we have for term \( k \):

$$ a_k = \frac{1}{k}n^{6 - 2(k-1)} $$

Thus, the maximum dominant term is \( a_1 = n^6 \), the next dominant terms are in descending powers of \( n \) like \( n^4, n^2 \), etc. As \( n \to \infty \), the expression \( \sum \frac{1}{k} n^{6 - 2(k-1)} + 1 \) can be approximated by its leading term \( n^6 \).

Each subsequent term becomes negligible when compared to the leading term as \( n \) grows larger. Therefore, approximating only by significant contributions:

$$ n^6 \approx n^6 + \frac{1}{2}n^4 + \frac{1}{3}n^2 + \cdots $$

This simplified approximation illustrates the convergence toward a power near 6, revealing the behavior of the residual terms.

Considering the original question attempts to approximate the asymptotic behavior, we need to analyze the contribution of the infinite series obtained \( n^6 + \cdots \to 10 \) for its significance over terms only rounding relevant integer shifts, supporting a known threshold.

Therefore, the correct answer is:

\( 10 \) approximately.