Answer :
To determine whether the series
[tex]\[ \sum_{n=0}^{\infty} \frac{3 + \cos(n)}{5^n} \][/tex]
converges or diverges, we can apply the Ratio Test. The Ratio Test is useful for determining the convergence or divergence of series by examining the limit of the ratio of successive terms.
The general term of the series is:
[tex]\[ a_n = \frac{3 + \cos(n)}{5^n} \][/tex]
Let's apply the Ratio Test by computing the limit:
[tex]\[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \][/tex]
First, let's find the expression for [tex]\( \frac{a_{n+1}}{a_n} \)[/tex]:
[tex]\[ a_{n+1} = \frac{3 + \cos(n+1)}{5^{n+1}} \][/tex]
So, the ratio of successive terms is:
[tex]\[ \frac{a_{n+1}}{a_n} = \frac{\frac{3 + \cos(n+1)}{5^{n+1}}}{\frac{3 + \cos(n)}{5^n}} = \frac{3 + \cos(n+1)}{5^{n+1}} \cdot \frac{5^n}{3 + \cos(n)} = \frac{3 + \cos(n+1)}{5 \cdot (3 + \cos(n))} \][/tex]
Now, we take the limit of this ratio as [tex]\( n \)[/tex] approaches infinity:
[tex]\[ L = \lim_{n \to \infty} \left| \frac{3 + \cos(n+1)}{5 \cdot (3 + \cos(n))} \right| \][/tex]
The cosine function, [tex]\(\cos(n)\)[/tex], oscillates between -1 and 1 for all integer values of [tex]\( n \)[/tex]. Therefore, [tex]\(3 + \cos(n)\)[/tex] will oscillate between 2 and 4. Similarly, [tex]\( 3 + \cos(n+1) \)[/tex] will also oscillate between 2 and 4.
Because cosine functions are bounded, even as [tex]\( n \)[/tex] goes to infinity, the numerator [tex]\(3 + \cos(n+1)\)[/tex] will stay within the range [2, 4], and the denominator [tex]\( 5 \cdot (3 + \cos(n)) \)[/tex] will lie within [10, 20].
Thus, the limit of the ratio can be expressed as an interval, which is the ratio of bound intervals:
[tex]\[ L \in \left[ \frac{2}{20}, \frac{4}{10} \right] = \left[ \frac{1}{10}, \frac{2}{5} \right] \][/tex]
From the interval, we see that the largest possible value of [tex]\( L \)[/tex] is [tex]\( \frac{2}{5} \)[/tex], which is less than 1.
According to the Ratio Test, if [tex]\( L < 1 \)[/tex], the series converges.
Therefore, given that the limit ratio is always less than 1,
[tex]\[ \sum_{n=0}^{\infty} \frac{3 + \cos(n)}{5^n} \][/tex]
converges.
[tex]\[ \sum_{n=0}^{\infty} \frac{3 + \cos(n)}{5^n} \][/tex]
converges or diverges, we can apply the Ratio Test. The Ratio Test is useful for determining the convergence or divergence of series by examining the limit of the ratio of successive terms.
The general term of the series is:
[tex]\[ a_n = \frac{3 + \cos(n)}{5^n} \][/tex]
Let's apply the Ratio Test by computing the limit:
[tex]\[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \][/tex]
First, let's find the expression for [tex]\( \frac{a_{n+1}}{a_n} \)[/tex]:
[tex]\[ a_{n+1} = \frac{3 + \cos(n+1)}{5^{n+1}} \][/tex]
So, the ratio of successive terms is:
[tex]\[ \frac{a_{n+1}}{a_n} = \frac{\frac{3 + \cos(n+1)}{5^{n+1}}}{\frac{3 + \cos(n)}{5^n}} = \frac{3 + \cos(n+1)}{5^{n+1}} \cdot \frac{5^n}{3 + \cos(n)} = \frac{3 + \cos(n+1)}{5 \cdot (3 + \cos(n))} \][/tex]
Now, we take the limit of this ratio as [tex]\( n \)[/tex] approaches infinity:
[tex]\[ L = \lim_{n \to \infty} \left| \frac{3 + \cos(n+1)}{5 \cdot (3 + \cos(n))} \right| \][/tex]
The cosine function, [tex]\(\cos(n)\)[/tex], oscillates between -1 and 1 for all integer values of [tex]\( n \)[/tex]. Therefore, [tex]\(3 + \cos(n)\)[/tex] will oscillate between 2 and 4. Similarly, [tex]\( 3 + \cos(n+1) \)[/tex] will also oscillate between 2 and 4.
Because cosine functions are bounded, even as [tex]\( n \)[/tex] goes to infinity, the numerator [tex]\(3 + \cos(n+1)\)[/tex] will stay within the range [2, 4], and the denominator [tex]\( 5 \cdot (3 + \cos(n)) \)[/tex] will lie within [10, 20].
Thus, the limit of the ratio can be expressed as an interval, which is the ratio of bound intervals:
[tex]\[ L \in \left[ \frac{2}{20}, \frac{4}{10} \right] = \left[ \frac{1}{10}, \frac{2}{5} \right] \][/tex]
From the interval, we see that the largest possible value of [tex]\( L \)[/tex] is [tex]\( \frac{2}{5} \)[/tex], which is less than 1.
According to the Ratio Test, if [tex]\( L < 1 \)[/tex], the series converges.
Therefore, given that the limit ratio is always less than 1,
[tex]\[ \sum_{n=0}^{\infty} \frac{3 + \cos(n)}{5^n} \][/tex]
converges.
