Answer :
To determine the value of [tex]\( p \)[/tex] for the series
[tex]\[ \sum_{n=1}^{\infty} n^2\left(3^{-3 \ln(n)}\right), \][/tex]
we will simplify the term [tex]\( n^2 \left( 3^{-3 \ln(n)} \right) \)[/tex].
First, focus on the term [tex]\( 3^{-3 \ln(n)} \)[/tex].
Using the property of exponents and logarithms, [tex]\( a^{b \ln(c)} = c^{\ln(c) \cdot b} \)[/tex], where [tex]\(\Delta\)[/tex] is the base of the logarithms:
[tex]\[ 3^{-3 \ln(n)} \][/tex]
can be rewritten as:
[tex]\[ 3^{-3 \ln(n)} = (e^{\ln(3)})^{-3 \ln(n)}. \][/tex]
Using the property [tex]\( e^{\ln(a)} = a \)[/tex], this becomes:
[tex]\[ 3^{-3 \ln(n)} = e^{(\ln(3) \cdot -3 \ln(n))}. \][/tex]
Simplify it to:
[tex]\[ e^{-\ln(3) \cdot 3 \ln(n)} = e^{-\ln(3^3 \cdot \ln(n))} = e^{-\ln(27) \cdot \ln(n)}. \][/tex]
Since [tex]\( e^{\ln(a^b)} = a^{b} \)[/tex], this can be further rewritten as:
[tex]\[ e^{\ln(n^{-\ln(27)})} = n^{-\ln(27)}. \][/tex]
Recall that [tex]\(\ln(27) = \ln(3^3) = 3\ln(3)\)[/tex]. So we have:
[tex]\[ 3^{-3 \ln(n)} = n^{-3 \ln(3)}. \][/tex]
Hence the term [tex]\( n^2 \left( 3^{-3 \ln(n)} \right) \)[/tex] simplifies to:
[tex]\[ n^2 \cdot n^{-3 \ln(3)} = n^{2 - 3 \ln(3)}. \][/tex]
Therefore, the value of [tex]\( p \)[/tex] in the exponent of [tex]\( n \)[/tex] is:
[tex]\[ p = 2 - 3 \ln(3). \][/tex]
So, the value of [tex]\( p \)[/tex] is
[tex]\[ 2 - 3 \ln(3). \][/tex]
[tex]\[ \sum_{n=1}^{\infty} n^2\left(3^{-3 \ln(n)}\right), \][/tex]
we will simplify the term [tex]\( n^2 \left( 3^{-3 \ln(n)} \right) \)[/tex].
First, focus on the term [tex]\( 3^{-3 \ln(n)} \)[/tex].
Using the property of exponents and logarithms, [tex]\( a^{b \ln(c)} = c^{\ln(c) \cdot b} \)[/tex], where [tex]\(\Delta\)[/tex] is the base of the logarithms:
[tex]\[ 3^{-3 \ln(n)} \][/tex]
can be rewritten as:
[tex]\[ 3^{-3 \ln(n)} = (e^{\ln(3)})^{-3 \ln(n)}. \][/tex]
Using the property [tex]\( e^{\ln(a)} = a \)[/tex], this becomes:
[tex]\[ 3^{-3 \ln(n)} = e^{(\ln(3) \cdot -3 \ln(n))}. \][/tex]
Simplify it to:
[tex]\[ e^{-\ln(3) \cdot 3 \ln(n)} = e^{-\ln(3^3 \cdot \ln(n))} = e^{-\ln(27) \cdot \ln(n)}. \][/tex]
Since [tex]\( e^{\ln(a^b)} = a^{b} \)[/tex], this can be further rewritten as:
[tex]\[ e^{\ln(n^{-\ln(27)})} = n^{-\ln(27)}. \][/tex]
Recall that [tex]\(\ln(27) = \ln(3^3) = 3\ln(3)\)[/tex]. So we have:
[tex]\[ 3^{-3 \ln(n)} = n^{-3 \ln(3)}. \][/tex]
Hence the term [tex]\( n^2 \left( 3^{-3 \ln(n)} \right) \)[/tex] simplifies to:
[tex]\[ n^2 \cdot n^{-3 \ln(3)} = n^{2 - 3 \ln(3)}. \][/tex]
Therefore, the value of [tex]\( p \)[/tex] in the exponent of [tex]\( n \)[/tex] is:
[tex]\[ p = 2 - 3 \ln(3). \][/tex]
So, the value of [tex]\( p \)[/tex] is
[tex]\[ 2 - 3 \ln(3). \][/tex]
