Determine whether the sequence converges or diverges. If it is convergent, find its limit. If appropriate, enter "infinity," "-infinity," or "DNE" for the limit.

$a_n = \frac{n^4}{e^{-2n}}$

$\lim_{n \to \infty} a_n = \, \underline{\hspace{2cm}}$

Therefore, the sequence [converges/diverges].

Question :

Determine whether the sequence converges or diverges. if it is convergent, find its limit. if appropriate, enter "infinity," "-infinity," or "dne" for the limit.

$a_n = \frac{n^4}{e^{-2n}}$

$\lim_{n \to \infty} a_n = \, \underline{\hspace{2cm}}$

therefore, the sequence [converges/diverges].

Determine whether the sequence converges or diverges. if it is convergent, find | Doubtlet.com

Solution:

Neetesh Kumar | November 17, 2024

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This is the solution to DHW Calculus
Assignment: 8 Question Number 15
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Step-by-step solution:

Step 1: Simplify the general term

The given sequence is:

$a_n = \frac{n^4}{e^{-2n}}$ .

Using the property of exponents, rewrite the denominator:

$e^{-2n} = \frac{1}{e^{2n}}$ .

Substitute this into the expression:

$a_n = n^4 \cdot e^{2n}$ .

Thus, the sequence becomes:

$a_n = n^4 e^{2n}$ .

Step 2: Analyze the behavior as $n \to \infty$

As $n \to \infty$ , the term $n^4$ grows polynomially, while $e^{2n}$ grows exponentially. Since exponential growth dominates polynomial growth, $a_n$ grows without bound as $n \to \infty$ .

Conclusion:

The sequence diverges to infinity.

Final Answer:

$\displaystyle\lim_{n \to \infty} a_n = \boxed{\infty}$

Therefore, the sequence diverges.