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Determine the value of [tex]\( p \)[/tex] for the series below, using the fact that [tex]\( a^{\log_b(n)} = b^{\log_b(n) \log_b(a)} = n^{\log_b(a)} \)[/tex].

[tex]\[ \sum_{n=1}^{\infty} n^2 \left( 4^{-3 \ln(n)} \right) \][/tex]

Be sure to include parentheses around the arguments of any logarithmic functions in your answer.

Sagot :

GinnyAnswer
To determine the value of [tex]\( p \)[/tex] for the given series, we start with the series:

[tex]\[ \sum_{n=1}^{\infty} n^2 \left( 4^{-3 \ln(n)} \right) \][/tex]

We can use the exponentiation property [tex]\( a^{\log_b(x)} = x^{\log_b(a)} \)[/tex] to simplify the expression inside the sum.

First, let's deal with the term [tex]\( 4^{-3 \ln(n)} \)[/tex]. We can rewrite 4 as [tex]\( 2^2 \)[/tex], so we have:

[tex]\[ 4^{-3 \ln(n)} = (2^2)^{-3 \ln(n)} \][/tex]

This can be simplified further. Using the power rule for exponents:

[tex]\[ (2^2)^{-3 \ln(n)} = 2^{2 \cdot (-3 \ln(n))} = 2^{-6 \ln(n)} \][/tex]

Next, we take advantage of the property of exponents involving logarithms. Recall that [tex]\( 2^{-6 \ln(n)} = n^{\log_2(2^{-6})} \)[/tex]. Therefore:

[tex]\[ 2^{-6 \ln(n)} = n^{-6 \ln(2)} \][/tex]

Since [tex]\( \ln(2) \)[/tex] is a constant, this can be interpreted as:

[tex]\[ 2^{-6 \ln(n)} = n^{-6} \][/tex]

Hence, our original series now looks like:

[tex]\[ \sum_{n=1}^{\infty} n^2 \cdot n^{-6} \][/tex]

We can combine the exponents for [tex]\( n \)[/tex]:

[tex]\[ n^2 \cdot n^{-6} = n^{2 + (-6)} = n^{-4} \][/tex]

So our series is now:

[tex]\[ \sum_{n=1}^{\infty} n^{-4} \][/tex]

This matches the form of a [tex]\( p \)[/tex]-series [tex]\(\sum_{n=1}^{\infty} \frac{1}{n^p}\)[/tex], where [tex]\( p = 4 \)[/tex].

So, the value of [tex]\( p \)[/tex] is:

[tex]\[ p = 4 \][/tex]
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