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The series given is:

\[
\sum_{k=1}^{\infty} \frac{k^3}{5^k}
\]

To determine if this series converges or diverges, we can use the **Ratio Test**. The Ratio Test states that for a series \(\sum a_k\), if 

\[
L = \lim_{k \to \infty} \left|\frac{a_{k+1}}{a_k}\right|
\]

then:

- If \(L < 1\), the series converges.
- If \(L > 1\), the series diverges.
- If \(L = 1\), the test is inconclusive.

Here, \(a_k = \frac{k^3}{5^k}\).

Let's compute the ratio \(\frac{a_{k+1}}{a_k}\):

\[
\frac{a_{k+1}}{a_k} = \frac{\frac{(k+1)^3}{5^{k+1}}}{\frac{k^3}{5^k}} = \frac{(k+1)^3}{5^{k+1}} \cdot \frac{5^k}{k^3} = \frac{(k+1)^3}{k^3} \cdot \frac{1}{5}
\]

Simplifying further:

\[
\frac{a_{k+1}}{a_k} = \frac{(k+1)^3}{k^3} \cdot \frac{1}{5} = \left(1 + \frac{1}{k}\right)^3 \cdot \frac{1}{5}
\]

Taking the limit as \(k\) approaches infinity:

\[
\lim_{k \to \infty} \frac{a_{k+1}}{a_k} = \lim_{k \to \infty} \left(1 + \frac{1}{k}\right)^3 \cdot \frac{1}{5} = 1^3 \cdot \frac{1}{5} = \frac{1}{5}
\]

Since \(\frac{1}{5} < 1\), the series converges by the Ratio Test.

Would you like more details on the Ratio Test, or do you have any other questions?

### Follow-up Questions:
1. How would the series change if the exponent of \(k\) was higher or lower?
2. What if the base of the exponent was different from 5, say 2 or 10?
3. How would the test change if we used the Root Test instead of the Ratio Test?
4. What does convergence imply about the sum of the series?
5. How can you find an approximation for the sum of this series?

### Tip:
Understanding the convergence of series is crucial when dealing with infinite sums, especially in the context of power series and Fourier series, where such analysis often determines the behavior of functions.

Learn how to determine the convergence of the series \(\sum_{k=1}^{\infty} \frac{k^3}{5^k}\) using the Ratio Test. Understand the application of limits in sequence analysis. Suitable for advanced high school students and beyond.

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