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Find the interval of convergence of
\[
\sum_{n=2}^{\infty} \frac{x^{3 n+2}}{\ln (n)}
\]
(Use symbolic notation and fractions where needed. Give your answers as intervals in the form $(*, *)$. Use symbol $\infty$ for infinity, $\cup$ for combining intervals, and appropriate type of parenthesis " (", ") ", " [" or "] " depending on whether the interval is open or closed. Enter DNE if interval is empty.)
\[
x \in
\]

Step 1 ：Let's denote the series as \(S\), so \(S = \sum_{n=2}^{\infty} \frac{x^{3n+2}}{\ln(n)}\).

Step 2 ：We can rewrite the series as \(S = \sum_{n=2}^{\infty} \frac{x^{3n+2}}{\ln(n)} = x^2 \sum_{n=2}^{\infty} \frac{x^{3n}}{\ln(n)}\).

Step 3 ：We can see that the series is a power series in the form of \(\sum_{n=2}^{\infty} a_n x^n\), where \(a_n = \frac{x^{3n}}{\ln(n)}\).

Step 4 ：We can use the Ratio Test to find the interval of convergence. The Ratio Test states that if \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1\), then the series converges.

Step 5 ：Applying the Ratio Test, we get \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{x^{3(n+1)}/\ln(n+1)}{x^{3n}/\ln(n)} \right| = \lim_{n \to \infty} \left| \frac{x^3 \ln(n)}{\ln(n+1)} \right|\).

Step 6 ：We want this limit to be less than 1, so we solve the inequality \(\left| \frac{x^3 \ln(n)}{\ln(n+1)} \right| < 1\).

Step 7 ：As \(n\) approaches infinity, \(\ln(n)\) and \(\ln(n+1)\) will be very close, so we can approximate the inequality as \(|x^3| < 1\).

Step 8 ：Solving this inequality gives us \(-1 < x < 1\).

Step 9 ：However, we need to check the endpoints to see if they are included in the interval of convergence. We substitute \(x = -1\) and \(x = 1\) into the original series.

Step 10 ：When \(x = -1\), the series becomes \(\sum_{n=2}^{\infty} \frac{(-1)^{3n+2}}{\ln(n)}\), which does not converge. So \(-1\) is not included in the interval of convergence.

Step 11 ：When \(x = 1\), the series becomes \(\sum_{n=2}^{\infty} \frac{1}{\ln(n)}\), which also does not converge. So \(1\) is not included in the interval of convergence.

Step 12 ：So, the interval of convergence is \(\boxed{(-1, 1)}\).