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Sagot :
To determine the radius of convergence \( R \) and the interval of convergence \( I \) of the series
[tex]\[ \sum_{n=1}^{\infty} \frac{(8x+5)^n}{n^2}, \][/tex]
we can use the Ratio Test. The Ratio Test is applied to the terms \( a_n \) of the series.
First, let's denote the general term of the series:
[tex]\[ a_n = \frac{(8x+5)^n}{n^2}. \][/tex]
The Ratio Test requires us to calculate the limit:
[tex]\[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|. \][/tex]
First, we find \( a_{n+1} \):
[tex]\[ a_{n+1} = \frac{(8x+5)^{n+1}}{(n+1)^2}. \][/tex]
Next, we form the ratio \( \frac{a_{n+1}}{a_n} \):
[tex]\[ \frac{a_{n+1}}{a_n} = \frac{\frac{(8x+5)^{n+1}}{(n+1)^2}}{\frac{(8x+5)^n}{n^2}} = \frac{(8x+5)^{n+1}}{(n+1)^2} \cdot \frac{n^2}{(8x+5)^n}. \][/tex]
Simplifying the expression:
[tex]\[ \frac{a_{n+1}}{a_n} = \frac{(8x+5) \cdot (8x+5)^n}{(n+1)^2} \cdot \frac{n^2}{(8x+5)^n} = \frac{(8x+5) \cdot n^2}{(n+1)^2}. \][/tex]
Now, let's take the limit as \( n \) approaches infinity:
[tex]\[ L = \lim_{n \to \infty} \left| \frac{(8x+5) \cdot n^2}{(n+1)^2} \right|. \][/tex]
To simplify this limit, divide the numerator and the denominator inside the limit by \( n^2 \):
[tex]\[ L = \lim_{n \to \infty} \left| (8x+5) \cdot \frac{n^2}{(n+1)^2} \right| = |8x+5| \lim_{n \to \infty} \frac{n^2}{(n+1)^2}. \][/tex]
As \( n \) approaches infinity, \( \frac{n^2}{(n+1)^2} \) approaches 1:
[tex]\[ L = |8x+5| \cdot \lim_{n \to \infty} \frac{n^2}{(n+1)^2} = |8x+5| \cdot 1 = |8x+5|. \][/tex]
For the series to converge, according to the Ratio Test, we need \( L \) to be less than 1:
[tex]\[ |8x+5| < 1. \][/tex]
Now, solve this inequality for \( x \):
[tex]\[ -1 < 8x+5 < 1. \][/tex]
Subtract 5 from all parts of the inequality:
[tex]\[ -6 < 8x < -4. \][/tex]
Divide every part by 8:
[tex]\[ -\frac{6}{8} < x < -\frac{4}{8}, \][/tex]
which simplifies to:
[tex]\[ -\frac{3}{4} < x < -\frac{1}{2}. \][/tex]
Therefore, the radius of convergence \( R \) is the distance from the center of this interval to either endpoint. The center of the interval is at \( x = -\frac{5}{8} \) and each endpoint is \( 1/8 \) away from this center. Thus, the radius of convergence \( R \) is:
[tex]\[ R = \frac{1}{8}. \][/tex]
Thus,
[tex]\[ R = \frac{1}{8}. \][/tex]
The interval of convergence, \( I \), is:
[tex]\[ I = \left( -\frac{3}{4}, -\frac{1}{2} \right). \][/tex]
To summarize:
[tex]\[ R = \frac{1}{8}, \][/tex]
[tex]\[ I = \left( -\frac{3}{4}, -\frac{1}{2} \right). \][/tex]
[tex]\[ \sum_{n=1}^{\infty} \frac{(8x+5)^n}{n^2}, \][/tex]
we can use the Ratio Test. The Ratio Test is applied to the terms \( a_n \) of the series.
First, let's denote the general term of the series:
[tex]\[ a_n = \frac{(8x+5)^n}{n^2}. \][/tex]
The Ratio Test requires us to calculate the limit:
[tex]\[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|. \][/tex]
First, we find \( a_{n+1} \):
[tex]\[ a_{n+1} = \frac{(8x+5)^{n+1}}{(n+1)^2}. \][/tex]
Next, we form the ratio \( \frac{a_{n+1}}{a_n} \):
[tex]\[ \frac{a_{n+1}}{a_n} = \frac{\frac{(8x+5)^{n+1}}{(n+1)^2}}{\frac{(8x+5)^n}{n^2}} = \frac{(8x+5)^{n+1}}{(n+1)^2} \cdot \frac{n^2}{(8x+5)^n}. \][/tex]
Simplifying the expression:
[tex]\[ \frac{a_{n+1}}{a_n} = \frac{(8x+5) \cdot (8x+5)^n}{(n+1)^2} \cdot \frac{n^2}{(8x+5)^n} = \frac{(8x+5) \cdot n^2}{(n+1)^2}. \][/tex]
Now, let's take the limit as \( n \) approaches infinity:
[tex]\[ L = \lim_{n \to \infty} \left| \frac{(8x+5) \cdot n^2}{(n+1)^2} \right|. \][/tex]
To simplify this limit, divide the numerator and the denominator inside the limit by \( n^2 \):
[tex]\[ L = \lim_{n \to \infty} \left| (8x+5) \cdot \frac{n^2}{(n+1)^2} \right| = |8x+5| \lim_{n \to \infty} \frac{n^2}{(n+1)^2}. \][/tex]
As \( n \) approaches infinity, \( \frac{n^2}{(n+1)^2} \) approaches 1:
[tex]\[ L = |8x+5| \cdot \lim_{n \to \infty} \frac{n^2}{(n+1)^2} = |8x+5| \cdot 1 = |8x+5|. \][/tex]
For the series to converge, according to the Ratio Test, we need \( L \) to be less than 1:
[tex]\[ |8x+5| < 1. \][/tex]
Now, solve this inequality for \( x \):
[tex]\[ -1 < 8x+5 < 1. \][/tex]
Subtract 5 from all parts of the inequality:
[tex]\[ -6 < 8x < -4. \][/tex]
Divide every part by 8:
[tex]\[ -\frac{6}{8} < x < -\frac{4}{8}, \][/tex]
which simplifies to:
[tex]\[ -\frac{3}{4} < x < -\frac{1}{2}. \][/tex]
Therefore, the radius of convergence \( R \) is the distance from the center of this interval to either endpoint. The center of the interval is at \( x = -\frac{5}{8} \) and each endpoint is \( 1/8 \) away from this center. Thus, the radius of convergence \( R \) is:
[tex]\[ R = \frac{1}{8}. \][/tex]
Thus,
[tex]\[ R = \frac{1}{8}. \][/tex]
The interval of convergence, \( I \), is:
[tex]\[ I = \left( -\frac{3}{4}, -\frac{1}{2} \right). \][/tex]
To summarize:
[tex]\[ R = \frac{1}{8}, \][/tex]
[tex]\[ I = \left( -\frac{3}{4}, -\frac{1}{2} \right). \][/tex]
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