Certainly! Let's evaluate the infinite series:
[tex]\[ \sum_{n=1}^{\infty} \frac{9^n}{n (-2)^{n+1}}. \][/tex]
1. Identify the General Term:
Let's denote the general term of the series by [tex]\( a_n \)[/tex]:
[tex]\[ a_n = \frac{9^n}{n (-2)^{n+1}}. \][/tex]
2. Simplify the General Term:
We notice that [tex]\( (-2)^{n+1} = (-2) \cdot (-2)^n \)[/tex]:
[tex]\[ a_n = \frac{9^n}{n \cdot (-2) \cdot (-2)^n} = \frac{9^n}{-2 \cdot n \cdot (-2)^n}. \][/tex]
3. Combine the Powers in the Denominator:
Combine [tex]\( (-2) \cdot (-2)^n \)[/tex]:
[tex]\[ (-2) \cdot (-2)^n = (-1)^{n+1} \cdot 2^n. \][/tex]
Therefore,
[tex]\[ a_n = \frac{9^n}{n \cdot (-1)^{n+1} \cdot 2^n}. \][/tex]
4. Write the Combined General Term:
Bring [tex]\( (-1)^{n+1} \)[/tex] to the numerator:
[tex]\[ a_n = \frac{9^n}{n \cdot 2^n} \cdot \frac{1}{(-1)^{n+1}} = \frac{9^n}{n \cdot 2^n} \cdot (-1)^{-(n+1)}. \][/tex]
5. Simplify the Exponent:
We know that [tex]\( (-1)^{-(n+1)} = (-1)^{-n-1} \)[/tex], so:
[tex]\[ a_n = (-1)^{-n-1} \cdot \frac{9^n}{n \cdot 2^n}. \][/tex]
6. Combine the Powers of the Numerator and Denominator:
Combine the powers of 9 / 2:
[tex]\[ a_n = \left(\frac{9}{2}\right)^n \cdot (-1)^{-n-1} \cdot \frac{1}{n}. \][/tex]
7. Express the General Term in a Simplified Form:
The simplified general term is:
[tex]\[ a_n = \frac{9^n}{n \cdot (-2)^{n+1}}. \][/tex]
Now, we can write the series:
[tex]\[ \sum_{n=1}^{\infty} \frac{9^n}{n (-2)^{n+1}} = \sum_{n=1}^{\infty} \frac{(9 / 2)^n}{n} \cdot (-1)^{-(n+1)}. \][/tex]
Final Summation Representation:
The sum of the given series can be expressed as:
[tex]\[ \boxed{\sum_{n=1}^{\infty} \frac{9^n}{n(-2)^{n+1}} = \sum_{n=1}^{\infty} \frac{(9 / 2)^n}{n} \cdot (-1)^{-(n+1)}}. \][/tex]
This is the most simplified form of the given series representation.
