Answer :
To determine the interval of convergence, we need to analyze the given series
[tex]\[ \sum_{n=1}^{\infty} \frac{4^n (x+2)^n}{\sqrt{n}}.\][/tex]
We already have the radius of convergence \( R = \frac{1}{4} \).
Next, we need to find the interval of convergence.
### Step 1: Ratio Test
To find where the series converges, we use the ratio test. Consider the general term of the series:
[tex]\[ a_n = \frac{4^n (x+2)^n}{\sqrt{n}}. \][/tex]
The ratio test involves finding the limit:
[tex]\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|. \][/tex]
First, we express \( a_{n+1} \):
[tex]\[ a_{n+1} = \frac{4^{n+1} (x+2)^{n+1}}{\sqrt{n+1}}. \][/tex]
Now, calculate the ratio:
[tex]\[ \frac{a_{n+1}}{a_n} = \frac{4^{n+1} (x+2)^{n+1}}{\sqrt{n+1}} \cdot \frac{\sqrt{n}}{4^n (x+2)^n}. \][/tex]
Simplify this expression:
[tex]\[ \frac{4^{n+1}}{4^n} \cdot \frac{(x+2)^{n+1}}{(x+2)^n} \cdot \frac{\sqrt{n}}{\sqrt{n+1}} = 4 \cdot (x+2) \cdot \frac{\sqrt{n}}{\sqrt{n+1}}. \][/tex]
Take the limit as \( n \) approaches infinity:
[tex]\[ \lim_{n \to \infty} \left| 4 \cdot (x+2) \cdot \frac{\sqrt{n}}{\sqrt{n+1}} \right| = \lim_{n \to \infty} 4 |x+2| \cdot \frac{\sqrt{n}}{\sqrt{n+1}}. \][/tex]
Observe that \(\frac{\sqrt{n}}{\sqrt{n+1}} \) approaches 1 as \( n \) goes to infinity, so:
[tex]\[ \lim_{n \to \infty} 4 |x+2| \cdot \frac{\sqrt{n}}{\sqrt{n+1}} = 4 |x+2|. \][/tex]
For convergence, we require:
[tex]\[ 4 |x+2| < 1. \][/tex]
Solve for \( |x+2| \):
[tex]\[ |x+2| < \frac{1}{4}. \][/tex]
So, the interval is:
[tex]\[ -\frac{1}{4} < x+2 < \frac{1}{4}. \][/tex]
Subtract 2 from all sides:
[tex]\[ -2 - \frac{1}{4} < x < -2 + \frac{1}{4}, \][/tex]
[tex]\[ -\frac{9}{4} < x < -\frac{7}{4}. \][/tex]
### Step 2: Check the Endpoints
Checking \( x = -\frac{9}{4} \):
For \( x = -\frac{9}{4} \),
[tex]\[ |x+2| = \left| -\frac{9}{4} + 2 \right| = \left| -\frac{9}{4} + \frac{8}{4} \right| = \left| -\frac{1}{4} \right| = \frac{1}{4}. \][/tex]
Thus, substituting \( x = -\frac{9}{4} \) in the series:
[tex]\[ \sum_{n=1}^{\infty} \frac{4^n \left( -\frac{1}{4} \right)^n}{\sqrt{n}} = \sum_{n=1}^{\infty} \frac{(4 \cdot -\frac{1}{4})^n}{\sqrt{n}} = \sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}. \][/tex]
The series \( \sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}} \) is the alternating harmonic series with terms that go to zero, so by the alternating series test, it converges.
Checking \( x = -\frac{7}{4} \):
For \( x = -\frac{7}{4} \),
[tex]\[ |x+2| = \left| -\frac{7}{4} + 2 \right| = \left| -\frac{7}{4} + \frac{8}{4} \right| = \left| \frac{1}{4} \right| = \frac{1}{4}. \][/tex]
Thus, substituting \( x = -\frac{7}{4} \) in the series:
[tex]\[ \sum_{n=1}^{\infty} \frac{4^n \left( \frac{1}{4} \right)^n}{\sqrt{n}} = \sum_{n=1}^{\infty} \frac{(4 \cdot \frac{1}{4})^n}{\sqrt{n}} = \sum_{n=1}^{\infty} \frac{1^n}{\sqrt{n}} = \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}. \][/tex]
The series \( \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \) is the p-series with \(p=\frac{1}{2}\), which diverges because \( p \leq 1 \).
### Conclusion
The interval of convergence is:
[tex]\[ \left[ -\frac{9}{4}, -\frac{7}{4} \right). \][/tex]
Thus:
[tex]\[ I = \left[ -\frac{9}{4}, -\frac{7}{4} \right). \][/tex]
[tex]\[ \sum_{n=1}^{\infty} \frac{4^n (x+2)^n}{\sqrt{n}}.\][/tex]
We already have the radius of convergence \( R = \frac{1}{4} \).
Next, we need to find the interval of convergence.
### Step 1: Ratio Test
To find where the series converges, we use the ratio test. Consider the general term of the series:
[tex]\[ a_n = \frac{4^n (x+2)^n}{\sqrt{n}}. \][/tex]
The ratio test involves finding the limit:
[tex]\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|. \][/tex]
First, we express \( a_{n+1} \):
[tex]\[ a_{n+1} = \frac{4^{n+1} (x+2)^{n+1}}{\sqrt{n+1}}. \][/tex]
Now, calculate the ratio:
[tex]\[ \frac{a_{n+1}}{a_n} = \frac{4^{n+1} (x+2)^{n+1}}{\sqrt{n+1}} \cdot \frac{\sqrt{n}}{4^n (x+2)^n}. \][/tex]
Simplify this expression:
[tex]\[ \frac{4^{n+1}}{4^n} \cdot \frac{(x+2)^{n+1}}{(x+2)^n} \cdot \frac{\sqrt{n}}{\sqrt{n+1}} = 4 \cdot (x+2) \cdot \frac{\sqrt{n}}{\sqrt{n+1}}. \][/tex]
Take the limit as \( n \) approaches infinity:
[tex]\[ \lim_{n \to \infty} \left| 4 \cdot (x+2) \cdot \frac{\sqrt{n}}{\sqrt{n+1}} \right| = \lim_{n \to \infty} 4 |x+2| \cdot \frac{\sqrt{n}}{\sqrt{n+1}}. \][/tex]
Observe that \(\frac{\sqrt{n}}{\sqrt{n+1}} \) approaches 1 as \( n \) goes to infinity, so:
[tex]\[ \lim_{n \to \infty} 4 |x+2| \cdot \frac{\sqrt{n}}{\sqrt{n+1}} = 4 |x+2|. \][/tex]
For convergence, we require:
[tex]\[ 4 |x+2| < 1. \][/tex]
Solve for \( |x+2| \):
[tex]\[ |x+2| < \frac{1}{4}. \][/tex]
So, the interval is:
[tex]\[ -\frac{1}{4} < x+2 < \frac{1}{4}. \][/tex]
Subtract 2 from all sides:
[tex]\[ -2 - \frac{1}{4} < x < -2 + \frac{1}{4}, \][/tex]
[tex]\[ -\frac{9}{4} < x < -\frac{7}{4}. \][/tex]
### Step 2: Check the Endpoints
Checking \( x = -\frac{9}{4} \):
For \( x = -\frac{9}{4} \),
[tex]\[ |x+2| = \left| -\frac{9}{4} + 2 \right| = \left| -\frac{9}{4} + \frac{8}{4} \right| = \left| -\frac{1}{4} \right| = \frac{1}{4}. \][/tex]
Thus, substituting \( x = -\frac{9}{4} \) in the series:
[tex]\[ \sum_{n=1}^{\infty} \frac{4^n \left( -\frac{1}{4} \right)^n}{\sqrt{n}} = \sum_{n=1}^{\infty} \frac{(4 \cdot -\frac{1}{4})^n}{\sqrt{n}} = \sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}. \][/tex]
The series \( \sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}} \) is the alternating harmonic series with terms that go to zero, so by the alternating series test, it converges.
Checking \( x = -\frac{7}{4} \):
For \( x = -\frac{7}{4} \),
[tex]\[ |x+2| = \left| -\frac{7}{4} + 2 \right| = \left| -\frac{7}{4} + \frac{8}{4} \right| = \left| \frac{1}{4} \right| = \frac{1}{4}. \][/tex]
Thus, substituting \( x = -\frac{7}{4} \) in the series:
[tex]\[ \sum_{n=1}^{\infty} \frac{4^n \left( \frac{1}{4} \right)^n}{\sqrt{n}} = \sum_{n=1}^{\infty} \frac{(4 \cdot \frac{1}{4})^n}{\sqrt{n}} = \sum_{n=1}^{\infty} \frac{1^n}{\sqrt{n}} = \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}. \][/tex]
The series \( \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \) is the p-series with \(p=\frac{1}{2}\), which diverges because \( p \leq 1 \).
### Conclusion
The interval of convergence is:
[tex]\[ \left[ -\frac{9}{4}, -\frac{7}{4} \right). \][/tex]
Thus:
[tex]\[ I = \left[ -\frac{9}{4}, -\frac{7}{4} \right). \][/tex]
