To determine if the given series converges, let's analyze it.

The series is:

$\sum_{n=1}^{\infty} \frac{5^{n-1}}{3^{n+1}}$

Step 1: Rewrite the General Term

We can rewrite the general term in a simpler form to see if it resembles a geometric series. Notice that:

$\frac{5^{n-1}}{3^{n+1}} = \frac{5^{n-1}}{3 \cdot 3^n} = \frac{1}{3} \cdot \frac{5^{n-1}}{3^n}$

Now we can rewrite $\frac{5^{n-1}}{3^n}$ as $(\frac{5}{3})^{n-1}$. Thus, the term becomes:

$\frac{1}{3} \cdot \left(\frac{5}{3}\right)^{n-1}$

Step 2: Recognize the Geometric Series Form

The series now looks like:

$\sum_{n=1}^{\infty} \frac{1}{3} \left(\frac{5}{3}\right)^{n-1}$

This is a geometric series with:

• First term $a = \frac{1}{3}$
• Common ratio $r = \frac{5}{3}$

Step 3: Check for Convergence

For a geometric series $\sum_{n=0}^{\infty} ar^n$ to converge, the common ratio $|r|$ must be less than 1. In this case:

$\left|\frac{5}{3}\right| > 1$

Since $\frac{5}{3}$ is greater than 1, the series diverges.

Conclusion

The series $\sum_{n=1}^{\infty} \frac{5^{n-1}}{3^{n+1}}$ diverges because the common ratio $r = \frac{5}{3}$ is greater than 1.

Would you like further details on this process or have any questions?

Here are some related questions to explore further:

1. What conditions are necessary for a geometric series to converge?
2. How does the ratio test apply to series with similar terms?
3. What methods exist for determining convergence of non-geometric series?
4. How can we simplify complex fractions in series terms?
5. How does the convergence of a series change if we alter the base of exponents?

Tip: Always check if a series has a common ratio and whether it’s less than 1 when dealing with geometric series to determine convergence.