To find the radius of convergence [tex]\( R \)[/tex] of the series [tex]\(\sum_{n=1}^{\infty} \frac{7^n (x-1)^n}{n}\)[/tex], we can use the Ratio Test. This test involves calculating the limit of the ratio of successive terms of the series.
Given series:
[tex]\[ \sum_{n=1}^{\infty} \frac{7^n (x-1)^n}{n} \][/tex]
First, identify the general term [tex]\( A_n \)[/tex]:
[tex]\[ A_n = \frac{7^n (x-1)^n}{n} \][/tex]
Next, consider the term [tex]\( A_{n+1} \)[/tex]:
[tex]\[ A_{n+1} = \frac{7^{n+1} (x-1)^{n+1}}{n+1} \][/tex]
Now, compute the ratio [tex]\(\left| \frac{A_{n+1}}{A_n} \right| \)[/tex]:
[tex]\[ \left| \frac{A_{n+1}}{A_n} \right| = \left| \frac{\frac{7^{n+1} (x-1)^{n+1}}{n+1}}{\frac{7^n (x-1)^n}{n}} \right| \][/tex]
[tex]\[= \left| \frac{7^{n+1} (x-1)^{n+1}}{n+1} \cdot \frac{n}{7^n (x-1)^n} \right| \][/tex]
[tex]\[= \left| 7 \cdot (x-1) \cdot \frac{n}{n+1} \right| \][/tex]
Simplify the ratio to find [tex]\( L \)[/tex] as [tex]\( n \)[/tex] approaches infinity:
[tex]\[ L = \lim_{n \to \infty} \left| 7 \cdot (x-1) \cdot \frac{n}{n+1} \right| \][/tex]
As [tex]\( n \)[/tex] approaches infinity, [tex]\(\frac{n}{n+1}\)[/tex] approaches 1:
[tex]\[ L = \left| 7 \cdot (x-1) \cdot 1 \right| \][/tex]
[tex]\[ L = 7 \left| x-1 \right| \][/tex]
For the series to converge, [tex]\( L \)[/tex] must be less than 1:
[tex]\[ 7 \left| x-1 \right| < 1 \][/tex]
Solve for [tex]\(\left| x-1 \right|\)[/tex]:
[tex]\[ \left| x-1 \right| < \frac{1}{7} \][/tex]
Therefore, the radius of convergence [tex]\( R \)[/tex] is:
[tex]\[ R = \frac{1}{7} = 0.14285714285714285 \][/tex]
Rounded to three decimal places, the radius of convergence is:
[tex]\[ R = 0.143 \][/tex]
Thus, the radius of convergence is:
[tex]\[ R = 0.143 \][/tex]
