To determine the values of \( x \) for which the series converges, we can use the Ratio Test. The Ratio Test involves taking the limit of the absolute value of the ratio of consecutive terms. If this limit is less than 1, the series converges. For the given series, the general term is \( a_k = \frac{5(7x)^k}{4k} \). We will compute the limit of \( \left| \frac{a_{k+1}}{a_k} \right| \) as \( k \to \infty \) and solve for \( x \).
We are given the series:
\[
\sum_{k=1}^{\infty} \frac{5(7 x)^{k}}{4 k}
\]
This is a power series in terms of \(x\). We will use the Ratio Test to determine the values of \(x\) for which the series converges.
The Ratio Test states that for a series \(\sum a_k\), the series converges absolutely if:
\[
\lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| < 1
\]
For our series, the general term \(a_k\) is:
\[
a_k = \frac{5(7x)^k}{4k}
\]
The next term \(a_{k+1}\) is:
\[
a_{k+1} = \frac{5(7x)^{k+1}}{4(k+1)}
\]
Now, compute the ratio \(\left| \frac{a_{k+1}}{a_k} \right|\):
\[
\left| \frac{a_{k+1}}{a_k} \right| = \left| \frac{\frac{5(7x)^{k+1}}{4(k+1)}}{\frac{5(7x)^k}{4k}} \right| = \left| \frac{(7x)^{k+1}}{(7x)^k} \cdot \frac{k}{k+1} \right|
\]
Simplify the expression:
\[
= \left| 7x \cdot \frac{k}{k+1} \right|
\]
Now, evaluate the limit as \(k\) approaches infinity:
\[
\lim_{k \to \infty} \left| 7x \cdot \frac{k}{k+1} \right| = \left| 7x \right| \cdot \lim_{k \to \infty} \frac{k}{k+1}
\]
Since \(\lim_{k \to \infty} \frac{k}{k+1} = 1\), we have:
\[
\lim_{k \to \infty} \left| 7x \cdot \frac{k}{k+1} \right| = \left| 7x \right|
\]
For convergence, we require:
\[
\left| 7x \right| < 1
\]
Solve the inequality \(\left| 7x \right| < 1\):
\[
-1 < 7x < 1
\]
Divide the entire inequality by 7:
\[
-\frac{1}{7} < x < \frac{1}{7}
\]
The series converges for:
\[
\boxed{-\frac{1}{7} < x < \frac{1}{7}}
\]
Answered
