The problem is asking to find the minimum value of $N$ such that the remainder of the series, $R_N = S - S_N$, is less than or equal to 0.00001, where:

$S = \sum_{n=1}^{\infty} 5e^{-\frac{3n}{2}}$

This is an infinite series, and we are tasked with estimating how many terms, $N$, are needed for the partial sum $S_N$ to approximate the total sum $S$ with an error less than or equal to $0.00001$.

Step-by-step approach:

1. The general term of the series is $5e^{-\frac{3n}{2}}$.
2. The remainder $R_N$ is approximately the next term after the partial sum, i.e., $R_N \approx 5e^{-\frac{3(N+1)}{2}}$, because the series converges exponentially.
3. We need to solve the inequality $5e^{-\frac{3(N+1)}{2}} \leq 0.00001$.

Solving for $N$:

We need to isolate $N$ in the inequality:

$5e^{-\frac{3(N+1)}{2}} \leq 0.00001$

Divide both sides by 5:

$e^{-\frac{3(N+1)}{2}} \leq \frac{0.00001}{5} = 0.000002$

Now, take the natural logarithm (ln) of both sides:

$-\frac{3(N+1)}{2} \leq \ln(0.000002)$

Calculate the logarithm:

$\ln(0.000002) \approx -13.8155$

Thus, the inequality becomes:

$-\frac{3(N+1)}{2} \leq -13.8155$

Multiply both sides by -1 (reversing the inequality):

$\frac{3(N+1)}{2} \geq 13.8155$

Multiply both sides by 2:

$3(N+1) \geq 27.631$

Now divide by 3:

$N+1 \geq 9.2103$

Finally, subtract 1:

$N \geq 8.2103$

Conclusion:

Since we are instructed to round up to the nearest whole number, the minimum value of $N$ is 9.

Would you like a more detailed explanation or have any further questions? Here are some related topics you might find useful:

1. What is the concept of convergence in infinite series?
2. How is the remainder term estimated in a convergent series?
3. How does exponential decay affect series convergence?
4. What are some techniques for approximating sums of infinite series?
5. Can this method be used for other types of series, such as geometric series?

Tip: Always check the remainder term when working with infinite series to ensure accurate approximations.