Question

?i?u ki?n c?a tham s? \( \alpha \) ?? chu?i ?an d?u \( \sum_{n=1}^{+\infty}(-1)^{n} \frac{n^{\alpha+1}+2}{\sqrt[3]{n^{4}+1}} \) h?i t? là: Select one: A. \( \alpha>\frac{4}{3} \) B. \( \alpha<\frac{4}{3} \) C. \( \alpha>\frac{1}{3} \) D. \( \alpha<\frac{1}{3} \)

          ?i?u ki?n c?a tham s? \( \alpha \) ?? chu?i ?an d?u \( \sum_{n=1}^{+\infty}(-1)^{n} \frac{n^{\alpha+1}+2}{\sqrt[3]{n^{4}+1}} \) h?i t? là:
Select one:
A. \( \alpha>\frac{4}{3} \)
B. \( \alpha<\frac{4}{3} \)
C. \( \alpha>\frac{1}{3} \)
D. \( \alpha<\frac{1}{3} \)

?i?u ki?n c?a tham s? α ?? chu?i ?an d?u ∑n=1^+∞(-1)^n(n^α+1+2)/(√(n^4+1)) h?i t? là:
Select one:
A. α>(4)/(3)
B. α<(4)/(3)
C. α>(1)/(3)
D. α<(1)/(3)

Added by Domingo C.

Comprehensive Mathematics for JEE Advanced

Ravi Prakash; Ajay Kumar; Usha Gupta 2019 Edition

Chapter 3

Instant Answer

Solved by Expert Zack A

Step 1

The given series is: \[ \sum_{n=1}^{+\infty}(-1)^{n} \frac{n^{\alpha+1}+2}{\sqrt[3]{n^{4}+1}} \] This is an alternating series of the form \((-1)^n a_n\), where: \[ a_n = \frac{n^{\alpha+1}+2}{\sqrt[3]{n^{4}+1}} \] Show more…

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Điều kiện của tham số \( \alpha \) đề chuỗi đan dấu \( \sum_{n=1}^{+\infty}(-1)^{n} \frac{n^{\alpha+1}+2}{\sqrt[3]{n^{4}+1}} \) hội tụ là: Select one: A. \( \alpha>\frac{4}{3} \) B. \( \alpha<\frac{4}{3} \) C. \( \alpha>\frac{1}{3} \) D. \( \alpha<\frac{1}{3} \)