https://file.aiquickdraw.com/mathbot/file/1728745817121w0d4qnoh.jpg

The series in the image is:

\[
\sum_{k=0}^{\infty} \frac{2^k - 3^k}{5^k}
\]

We can separate this into two simpler sums:

\[
\sum_{k=0}^{\infty} \frac{2^k}{5^k} - \sum_{k=0}^{\infty} \frac{3^k}{5^k}
\]

Each sum is a geometric series. The general form for a geometric series is:

\[
\sum_{k=0}^{\infty} ar^k = \frac{a}{1 - r} \quad \text{for} \quad |r| < 1
\]

For the first sum, \(\sum_{k=0}^{\infty} \frac{2^k}{5^k}\), we have \(a = 1\) and \(r = \frac{2}{5}\). Applying the geometric series formula:

\[
\sum_{k=0}^{\infty} \frac{2^k}{5^k} = \frac{1}{1 - \frac{2}{5}} = \frac{1}{\frac{3}{5}} = \frac{5}{3}
\]

For the second sum, \(\sum_{k=0}^{\infty} \frac{3^k}{5^k}\), we have \(a = 1\) and \(r = \frac{3}{5}\). Similarly:

\[
\sum_{k=0}^{\infty} \frac{3^k}{5^k} = \frac{1}{1 - \frac{3}{5}} = \frac{1}{\frac{2}{5}} = \frac{5}{2}
\]

Now, subtract the two results:

\[
\frac{5}{3} - \frac{5}{2} = \frac{10}{6} - \frac{15}{6} = -\frac{5}{6}
\]

Therefore, the sum of the series is:

\[
-\frac{5}{6}
\]

Would you like more details or explanations about any step?

Here are five related questions you might find interesting:
1. What conditions must be met for a geometric series to converge?
2. Can the sum of a geometric series ever be greater than 1?
3. How can we apply this approach to more complex infinite series?
4. What happens if the series involves a factor \(n^k\) instead of a constant \(r^k\)?
5. How can geometric series be applied to solve real-world problems?

**Tip:** When dealing with geometric series, always check that the absolute value of the common ratio \(r\) is less than 1 for the series to converge.

Find the sum of the convergent series given by sum of (2^k - 3^k)/5^k from k=0 to infinity.

This problem involves solving the convergent series sum of (2^k - 3^k)/5^k from k=0 to infinity. Using geometric series formula and applying convergence conditions, the sum is calculated to be -5/6. Ideal for university-level calculus students.

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