Let's solve the problem systematically:

Given:

The power series:
$\sum_{k=0}^\infty \frac{4}{e^{2k+5}}x^k$

We will address the parts:

(a) Determine the radius of convergence $R$:

For a power series $\sum_{k=0}^\infty a_k x^k$, the radius of convergence can be determined using the Ratio Test. Here, $a_k = \frac{4}{e^{2k+5}}$.

The ratio of consecutive terms is: $\left| \frac{a_{k+1}}{a_k} \right| = \left| \frac{\frac{4}{e^{2(k+1)+5}}}{\frac{4}{e^{2k+5}}} \right| = \frac{e^{2k+5}}{e^{2(k+1)+5}}$

Simplify the exponent: $\frac{e^{2k+5}}{e^{2(k+1)+5}} = \frac{e^{2k+5}}{e^{2k+7}} = \frac{1}{e^2}$

This ratio is constant and independent of $k$. The series converges if: $|x| < \frac{1}{\left|\frac{a_{k+1}}{a_k}\right|} = \frac{1}{\frac{1}{e^2}} = e^2$

Thus, the radius of convergence is: $R = e^2$

(b) Does the series converge at $x = 1$?

The series converges if $|x| < R$. Since $R = e^2 > 1$, $x = 1$ lies within the interval of convergence. Now check the series' behavior at $x = 1$:

The series becomes: $\sum_{k=0}^\infty \frac{4}{e^{2k+5}}(1)^k = \sum_{k=0}^\infty \frac{4}{e^{2k+5}}$

This is a geometric series where the ratio between consecutive terms is: $r = \frac{\frac{4}{e^{2(k+1)+5}}}{\frac{4}{e^{2k+5}}} = \frac{1}{e^2}$

Since $|r| = \frac{1}{e^2} < 1$, the series converges for $x = 1$.

(c) Does the series converge at $x = -2e^2$?

At $x = -2e^2$, check whether $|x|$ lies within the radius of convergence: $|x| = |-2e^2| = 2e^2$

Since $2e^2 > R = e^2$, $x = -2e^2$ lies outside the interval of convergence. Thus, the series diverges at $x = -2e^2$.

Final Answers:

(a) Radius of convergence: $R = e^2$
(b) The series converges at $x = 1$.
(c) The series diverges at $x = -2e^2$.

Would you like additional details or explanations?

Related Questions:

1. What is the general formula for the interval of convergence?
2. How does the Ratio Test work for determining convergence?
3. Why does the geometric series test apply in part (b)?
4. How does the radius of convergence relate to the ratio of consecutive terms?
5. Can the series converge at the endpoints of its radius?

Tip:

For power series, always check the boundary points separately as their behavior can differ from the interior of the radius of convergence.