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 provided series is:

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

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

Let the general term of the series be denoted by \( a_k \). The terms are given by:

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

The general form of the \( k \)-th term is:

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

The dominant term is \( a_1 = n^6 \). Subsequent terms decrease in magnitude as powers of \( n \) (e.g., \( n^4, n^2 \)). As \( n \to \infty \), the expression \( \sum \frac{1}{k} n^{6 - 2(k-1)} + 1 \) can be approximated by its leading term, \( n^6 \).

As \( n \) increases, each subsequent term becomes insignificant compared to the leading term. Thus, approximating by significant contributions:

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

This simplification indicates the behavior approaches a power close to 6 and shows how the remaining terms contribute minimally.

To approximate the asymptotic behavior of the original expression, we examine the infinite series \( n^6 + \cdots \). This sum tends towards 10, which is significant relative to terms that only adjust relevant integer values, supporting a specific threshold.

The approximate asymptotic value is:

\( 10 \)