Question

For \( c \in \mathbb{R} \), let the sequence \( \{ u_n \} \) be defined by \[ u_n = \frac{(1 + \frac{c}{n})^{n^2}}{(3 - \frac{1}{n})^n} \] Then the values of \( c \) for which the series \[ \sum_{n=1}^{\infty} u_n \] converges are

• A. \( c<\log_e 3 \)
• B. \( \log_e 6<c<\log_e 9 \)
• C. \( \log_e 9<c<\log_e 12 \)
• D. \( \log_e 3<c<\log_e 6 \)

Hint:

For series convergence tests, the root and ratio tests are effective for sequences with exponential growth.

Answer 1

The Correct Option is A

Step 1: Analyze the general term u_n.
We have u_n = (1 + c/n)^{n^2} / (3 - 1/n)^n.
Take natural logarithm to simplify:
ln(u_n) = n^2 ln(1 + c/n) - n ln(3 - 1/n).

Step 2: Use logarithm approximations for large n.
For large n, ln(1 + c/n) ≈ c/n, ln(3 - 1/n) ≈ ln 3 - 1/(3n).
So, ln(u_n) ≈ n^2 * (c/n) - n * (ln 3 - 1/(3n)) = cn - n ln 3 + 1/3.
Ignoring the constant 1/3 for large n:
ln(u_n) ≈ n(c - ln 3).

Step 3: Determine convergence condition.
For the series Σu_n to converge, u_n → 0 as n → ∞.
If c - ln 3 > 0, ln(u_n) → ∞ → u_n → ∞ → series diverges.
If c - ln 3 = 0, ln(u_n) ≈ 0 → u_n ≈ constant → series diverges.
If c - ln 3 < 0, ln(u_n) → -∞ → u_n → 0 → possible convergence.

Step 4: Conclude the range of c.
Hence, the series converges for:
c < ln 3.

Final Answer: c < log_e 3