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The Alternating Series Test is a crucial tool in determining the convergence of certain types of series. It applies to series where each term alternates in sign, such as \( (-1)^k \frac{3^k}{k!} \).
To utilize this test, the series must satisfy two conditions:

• The terms \( b_k = \frac{3^k}{k!} \) must be decreasing. This means that each term should be smaller than the preceding one.
• The limit of the absolute value of the terms, \( \lim_ {k \to \infty} b_k = 0 \), should equal zero.

In our exercise, both conditions are met: the sequence is decreasing and approaches zero as \( k \to \infty \). Therefore, by the Alternating Series Test, the series converges. This test helps determine if a series, which contains alternating signs, converges based only on these criteria.

The Ratio Test allows us to determine the absolute convergence of a series. This is particularly useful when terms involve factorials, as we see here with \( \frac{3^k}{k!} \).
The Ratio Test involves calculating the limit:\[\lim_ {k \to \infty} \left|\frac{a_{k+1}}{a_k}\right| = L\]where \(a_k\) denotes our general term of the series. If \(L < 1\), the series is absolutely convergent; if \(L > 1\) or \(L = \infty\), the series diverges; if \(L = 1\), the test is inconclusive.
For the series in question, applying this test results in:\[\lim_ {k \to \infty} \frac{\frac{3^{k+1}}{(k+1)!}}{\frac{3^k}{k!}} = \lim_ {k \to \infty} \frac{k}{3} = 0 < 1\]This conclusion confirms that the series is absolutely convergent.

A convergent series is one where the sum of its infinite terms approaches a specific finite value. In the context of our series, convergence is crucial for determining if the sum of the infinite terms calculates to a real finite number.
There are different types of convergence:

• Absolute Convergence: If the series formed by taking the absolute values of its terms also converges, then it is absolutely convergent.
• Conditional Convergence: If the series converges, but the series of its absolute terms does not, it is conditionally convergent.

In our exercise, the series \( \sum_{k=3}^{\infty}(-1)^{k} \frac{3^{k}}{k !} \) is determined to be absolutely convergent. This is because the series of absolute values \( \sum_{k=3}^{\infty} \frac{3^{k}}{k !} \) converges, as proven by the Ratio Test.

When you encounter a series with a factorial, such as \( \frac{3^k}{k!} \), it often suggests rapid growth in the denominator that can help with convergence.
Factorials grow extremely fast compared to exponential terms. Therefore, when you have an expression like \( 3^k / k! \), the factorial in the denominator usually outpaces the exponential function in the numerator as \( k \) becomes large.
This rapid growth in the denominator is why the Ratio Test is particularly effective in such situations. In our test, it helps demonstrate that:\[\lim_{k \to \infty} \frac{k}{3} = 0\]Ultimately, the factorials dominate, causing the terms to shrink towards zero rapidly, which is key to proving the absolute convergence of the series.