Answer :
To determine whether the series [tex]\(\sum_{n=1}^{\infty} n^3\left(2^{-3 \ln (n)}\right)\)[/tex] converges or diverges, we can simplify the general term of the series and analyze its behavior.
First, let's simplify the expression inside the summation:
[tex]\[ n^3\left(2^{-3 \ln (n)}\right) \][/tex]
We know that for any positive [tex]\(a\)[/tex],
[tex]\[ a^{-\ln(n)} = \frac{1}{n^{\ln(a)}} \][/tex]
Applying this to our series where [tex]\(a = 2^3\)[/tex]:
[tex]\[ 2^{-3 \ln (n)} = \left(2^{\ln (n)} \right)^{-3} = \left( n^{\ln (2)}\right)^{-3} = \frac{1}{n^{3 \ln (2)}} \][/tex]
Thus, the general term of our series can be rewritten as:
[tex]\[ n^3 \cdot \frac{1}{n^{3 \ln (2)}} = \frac{n^3}{n^{3 \ln (2)}} \][/tex]
We can simplify the exponent in the denominator:
[tex]\[ \frac{n^3}{n^{3 \ln (2)}} = n^{3 - 3 \ln (2)} \][/tex]
Denoting [tex]\(\alpha = 3 - 3 \ln(2)\)[/tex], our series becomes:
[tex]\[ \sum_{n=1}^{\infty} n^{\alpha} \][/tex]
To determine the convergence of this series, we analyze the value of [tex]\(\alpha\)[/tex]. The p-test for series [tex]\(\sum_{n=1}^{\infty} \frac{1}{n^p}\)[/tex] tells us that the series converges if and only if [tex]\(p > 1\)[/tex].
In our case, [tex]\(\alpha = 3 - 3 \ln(2)\)[/tex]. Calculating this value:
[tex]\[ 3 - 3 \ln(2) \approx 3 - 3(0.693) \approx 3 - 2.079 \approx 0.921 \][/tex]
Since [tex]\(\alpha \approx 0.921\)[/tex], which is greater than -1 but less than 1, the series does not meet the necessary condition for convergence (i.e., [tex]\(\alpha \leq -1\)[/tex] for the series to converge).
Therefore, the series diverges.
The correct answer is:
The series diverges.
First, let's simplify the expression inside the summation:
[tex]\[ n^3\left(2^{-3 \ln (n)}\right) \][/tex]
We know that for any positive [tex]\(a\)[/tex],
[tex]\[ a^{-\ln(n)} = \frac{1}{n^{\ln(a)}} \][/tex]
Applying this to our series where [tex]\(a = 2^3\)[/tex]:
[tex]\[ 2^{-3 \ln (n)} = \left(2^{\ln (n)} \right)^{-3} = \left( n^{\ln (2)}\right)^{-3} = \frac{1}{n^{3 \ln (2)}} \][/tex]
Thus, the general term of our series can be rewritten as:
[tex]\[ n^3 \cdot \frac{1}{n^{3 \ln (2)}} = \frac{n^3}{n^{3 \ln (2)}} \][/tex]
We can simplify the exponent in the denominator:
[tex]\[ \frac{n^3}{n^{3 \ln (2)}} = n^{3 - 3 \ln (2)} \][/tex]
Denoting [tex]\(\alpha = 3 - 3 \ln(2)\)[/tex], our series becomes:
[tex]\[ \sum_{n=1}^{\infty} n^{\alpha} \][/tex]
To determine the convergence of this series, we analyze the value of [tex]\(\alpha\)[/tex]. The p-test for series [tex]\(\sum_{n=1}^{\infty} \frac{1}{n^p}\)[/tex] tells us that the series converges if and only if [tex]\(p > 1\)[/tex].
In our case, [tex]\(\alpha = 3 - 3 \ln(2)\)[/tex]. Calculating this value:
[tex]\[ 3 - 3 \ln(2) \approx 3 - 3(0.693) \approx 3 - 2.079 \approx 0.921 \][/tex]
Since [tex]\(\alpha \approx 0.921\)[/tex], which is greater than -1 but less than 1, the series does not meet the necessary condition for convergence (i.e., [tex]\(\alpha \leq -1\)[/tex] for the series to converge).
Therefore, the series diverges.
The correct answer is:
The series diverges.
