Suppose that an=
(
1
+
ln
⁢3
𝑛
)
3
𝑛
.

What is the limit of the sequence \{a_n\}_{n=1}^\infty ?

Type your answer in the box. It should be a whole number.

Question

Suppose that an=
(
1
+
ln
⁢3
𝑛
)
3
𝑛
.

What is the limit of the sequence \{a_n\}_{n=1}^\infty ?

Type your answer in the box. It should be a whole number.

BlackTom AI · Accepted Answer

The standard exponential limit (1 + x/n)^n → e^x as n → ∞ is the key fact.

1) Relevant knowledge:
- Limit: lim_{n→∞} (1 + x/n)^n = e^x for any fixed real x.
- Properties of exponentials and logarithms: e^{ln a} = a, and (A^n)^m = A^{nm}.

2) Step-by-step reasoning:
Start with a_n = (1 + (ln 3)/n)^{3n}. Observe that 3n = 3·n, so this can be rewritten as a power of (1 + (ln 3)/n)^n:
a_n = [(1 + (ln 3)/n)^n]^3.

3) Intermediate limit calculation:
Apply the known limit with x = ln 3:
lim_{n→∞} (1 + (ln 3)/n)^n = e^{ln 3} = 3.
Therefore lim_{n→∞} a_n = [lim_{n→∞} (1 + (ln 3)/n)^n]^3 = 3^3 = 27.

(Alternatively, one can use logarithms: ln a_n = 3n·ln(1 + (ln 3)/n). For small t, ln(1+t) ~ t, so ln a_n → 3n·(ln 3 / n) = 3 ln 3, giving a_n → e^{3 ln 3} = e^{ln 27} = 27.)

4) Why this is correct:
The factorization into [(1 + (ln 3)/n)^n]^3 isolates the canonical form (1+x/n)^n whose limit is e^x. Raising that limit to the third power yields the final value. The alternative logarithmic approach confirms the same result.

5) Common incorrect ideas:
- Mistaking the exponent so as to get e^{ln 3} = 3 (forgetting the extra factor 3) would give 3 instead of 27.
- Misapplying the limit (1 + x/n)^{kn} without factoring out n can cause errors; proper factoring yields the cube of the basic limit.

Conclusion: The limit of {a_n} is 27, which matches the provided answer.

Suppose that an= ( 1 + ln ⁢3 𝑛 ) 3 𝑛 . What is the limit of the sequence \{a_n\}_{n=1}^\infty ? Type your answer in the box. It should be a whole number.Short answer

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