Question

Determine convergence or divergence of the alternating series. 1) sum_{n=1}^{infinity} frac{(-1)^{n}}{n^{3/4}} A) Converges B) Diverges 2) sum_{n=1}^{infinity} (-1)^{n} ln left[ frac{5n+2}{4n+1} ight] A) Converges B) Diverges 3) sum_{n=1}^{infinity} (-1)^{n} left( frac{4n^{2}+1}{2n^{2}+5} ight) A) Converges B) Diverges

          Determine convergence or divergence of the alternating series.
1) sum_{n=1}^{infinity} frac{(-1)^{n}}{n^{3/4}}
A) Converges B) Diverges
2) sum_{n=1}^{infinity} (-1)^{n} ln left[ frac{5n+2}{4n+1} 
ight]
A) Converges B) Diverges
3) sum_{n=1}^{infinity} (-1)^{n} left( frac{4n^{2}+1}{2n^{2}+5} 
ight)
A) Converges B) Diverges

$Determine convergence or divergence of the alternating series. 1) sumn=1^infinity frac(-1)^nn^3/4 A) Converges B) Diverges 2) sumn=1^infinity (-1)^n ln left[ frac5n+24n+1 ight] A) Converges B) Diverges 3) sumn=1^infinity (-1)^n left( frac4n^2+12n^2+5 ight) A) Converges B) Diverges$

Added by Albert J.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Collin Parker

Step 1

We have $a_n = \frac{n^3}{4^n}$, which is positive and decreasing, and $\lim_{n\to\infty} a_n = 0$. Show more…

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Determine convergence or divergence of the alternating series. 1) sum_{n=1}^{infinity} ((-1)^n)/(n^{3/4}) A) Converges B) Diverges 2) sum_{n=1}^{infinity} (-1)^n ln[(5n + 2)/(4n + 1)] A) Converges B) Diverges 3) sum_{n=1}^{infinity} (-1)^n [(4n^2 + 1)/(2n^2 + 5)] A) Converges B) Diverges