Let's consider the given summation:
[tex]$\sum_{k=3}^{\infty} \left( \frac{3}{4} \right)^{k-4} \left( \frac{3}{4} \right)^{3-k}$[/tex]
First, we can simplify the expression inside the summation. Notice that:
[tex]$\left( \frac{3}{4} \right)^{k-4} \left( \frac{3}{4} \right)^{3-k}$[/tex]
In this expression, we can combine the exponents of the common base [tex]\(\frac{3}{4}\)[/tex]:
[tex]$ \left( \frac{3}{4} \right)^{k-4} \cdot \left( \frac{3}{4} \right)^{3-k} = \left( \frac{3}{4} \right)^{(k-4) + (3-k)}$[/tex]
Next, simplify the exponent:
[tex]$ (k-4) + (3-k) = k - 4 + 3 - k = -1$[/tex]
Thus, the expression simplifies to:
[tex]$ \left( \frac{3}{4} \right)^{-1} = \frac{1}{\left( \frac{3}{4} \right)} = \frac{4}{3}$[/tex]
So the summation becomes:
[tex]$ \sum_{k=3}^{\infty} \frac{4}{3} $[/tex]
This is now a summation of a constant value, [tex]\(\frac{4}{3}\)[/tex], from [tex]\(k = 3\)[/tex] to [tex]\(\infty\)[/tex]. We express the summation of a constant as:
[tex]$
\sum_{k=3}^{\infty} \frac{4}{3}
$[/tex]
When you sum a constant over an infinite number of terms, the series does not converge to a finite value. Instead, it diverges. Since we are summing an infinite series of the constant [tex]\(\frac{4}{3}\)[/tex], the series diverges to infinity.
Hence, the summation:
[tex]$
\sum_{k=3}^{\infty} \left( \frac{3}{4} \right)^{k-4} \left( \frac{3}{4} \right)^{3-k}
$[/tex]
diverges, and its value is:
[tex]$
\infty
$[/tex]
