Math  /  Calculus

QuestionDetermine if the series converges absolutely, conditionally, or diverges. a. n=1(1)n4nn3+3\sum_{n=1}^{\infty} \frac{(-1)^{n} 4 n}{\sqrt{n^{3}+3}} d. n=1(1)nn2n+3\sum_{n=1}^{\infty} \frac{(-1)^{n} \sqrt{n}}{2 n+3} b. n=1n23n26\sum_{n=1}^{\infty} \frac{n^{2}}{3 n^{2}-6} e. n=1(1)n2n1/n\sum_{n=1}^{\infty}(-1)^{n} \frac{\sqrt{2}}{n^{1 / n}} c. m=1(1)m[ln(2m+1)ln(m+3)]\sum_{m=1}^{\infty}(-1)^{m}[\ln (2 m+1)-\ln (m+3)] f. n=1tan1((1)n)\sum_{n=1}^{\infty} \tan ^{-1}\left((-1)^{n}\right)

Studdy Solution
For series (f), n=1tan1((1)n)\sum_{n=1}^{\infty} \tan ^{-1}\left((-1)^{n}\right), note that:
tan1((1)n) \tan^{-1}((-1)^n) alternates between tan1(1)=π4 \tan^{-1}(1) = \frac{\pi}{4} and tan1(1)=π4 \tan^{-1}(-1) = -\frac{\pi}{4} .
The series does not tend to zero, so it diverges.
The results are: a. Converges conditionally. b. Diverges. c. Diverges. d. Converges conditionally. e. Diverges. f. Diverges.

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