Let's analyze each series step by step using comparison tests to determine their convergence or divergence.
### (A) [tex]\(\sum_{n=1}^{\infty} \frac{(\ln n)^2}{n^3+1}\)[/tex]
To determine the convergence of this series, we will use the comparison test with a simpler benchmark series.
1. General Term Analysis:
[tex]\[
a_n = \frac{(\ln n)^2}{n^3+1}
\][/tex]
2. Benchmark Series:
We compare it with the series [tex]\(\sum \frac{1}{n^3}\)[/tex].
3. Comparing Terms:
For large [tex]\(n\)[/tex],
[tex]\[
n^3 + 1 \approx n^3
\][/tex]
Hence,
[tex]\[
a_n \approx \frac{(\ln n)^2}{n^3}
\][/tex]
4. Limit Comparison Test:
The limit comparison test formula is:
[tex]\[
\lim_{{n \to \infty}} \frac{a_n}{b_n}
\][/tex]
where [tex]\(b_n = \frac{1}{n^3}\)[/tex].
5. Calculating the Limit:
[tex]\[
\lim_{{n \to \infty}} \frac{(\ln n)^2 / (n^3 + 1)}{1 / n^3} = \lim_{{n \to \infty}} \frac{(\ln n)^2}{n^3 + 1} \cdot n^3 = \lim_{{n \to \infty}} (\ln n)^2 = \infty
\][/tex]
6. Conclusion for Series (A):
Since the benchmark series [tex]\(\sum \frac{1}{n^3}\)[/tex] converges and the limit comparison test yields a finite limit, our original series [tex]\(\sum \frac{(\ln n)^2}{n^3+1}\)[/tex] also converges by the comparison test.
### (B) [tex]\(\sum_{n=1}^{\infty} 2 n^2 \sin \left(\frac{1}{n^2+1}\right)\)[/tex]
To determine the convergence of this series, we will again use the comparison test.
1. General Term Analysis:
[tex]\[
a_n = 2 n^2 \sin \left(\frac{1}{n^2+1}\right)
\][/tex]
2. Simplified Analysis for Small Angles:
For small angles [tex]\(x\)[/tex], [tex]\(\sin x \approx x\)[/tex]. Here, [tex]\(x = \frac{1}{n^2+1}\)[/tex].
3. Approximating [tex]\(\sin\)[/tex] for Small Angles:
For large [tex]\(n\)[/tex],
[tex]\[
\sin \left(\frac{1}{n^2+1}\right) \approx \frac{1}{n^2}
\][/tex]
4. Benchmark Series:
We compare it with the series [tex]\(\sum \frac{2}{n^2}\)[/tex].
5. Comparing Terms and Calculating the Limit:
[tex]\[
a_n \approx 2 n^2 \cdot \frac{1}{n^2} = 2
\][/tex]
The limit comparison test formula is:
[tex]\[
\lim_{{n \to \infty}} \frac{2 n^2 \sin \left(\frac{1}{n^2+1}\right)}{2 \cdot \frac{1}{n^2}} = \lim_{{n \to \infty}} \frac{2 n^2 \cdot \frac{1}{n^2}}{2 \cdot \frac{1}{n^2}} = \lim_{{n \to \infty}} 2 = 2
\][/tex]
6. Conclusion for Series (B):
Since the benchmark series [tex]\(\sum \frac{2}{n^2}\)[/tex] converges and the limit comparison test yields a finite, non-zero limit ([tex]\(2\)[/tex]), the original series [tex]\(\sum 2 n^2 \sin \left(\frac{1}{n^2+1}\right)\)[/tex] also converges by the comparison test.
### Final Conclusion:
Both series (A) [tex]\(\sum_{n=1}^{\infty} \frac{(\ln n)^2}{n^3+1}\)[/tex] and (B) [tex]\(\sum_{n=1}^{\infty} 2 n^2 \sin \left(\frac{1}{n^2+1}\right)\)[/tex] converge based on the comparison tests with the respective benchmark series.
