14. Sequences & Series

Explain why the integral test does not apply to the series.

${\sum }_{n=0}^{\infty }\frac{n}{n^2-1}$

Determine whether the given series is convergent.

${\sum }_{n=1}^{\infty }\frac{1}{{}^{{}^5}\sqrt{n^3}}$

Determine whether the given series is convergent.

${\sum }_{n=1}^{\infty }\frac{5}{3n-7}$

Determine whether the given series are convergent.

${\sum }_{n=1}^{\infty }\frac{n^e}{n^2}$

Determine whether the given series is convergent.

${\sum }_{n=1}^{\infty }{n}^{-1}+n^{-3}$

Use the Direct Comparison Test to determine whether the series converges.

${\sum }_{n=1}^{\infty }\frac{3}{\sqrt{n}-2}$

Use the Direct Comparison Test to determine whether the series converges.

${\sum }_{n=1}^{\infty }\frac{{\cos }^{2}n}{n^2}$ Hint: Compare to $b_n=\frac{1}{n^2}$

Use the Limit Comparison Test to determine whether the series converges.

${\sum }_{n=1}^{\infty }\frac{\sqrt{n}}{n^2+3}$

Use the Limit Comparison Test to determine whether the series converges.

${\sum }_{n=1}^{\infty }\frac{n}{(n-1)3^{n+1}}$

Use the Limit Comparison Test to determine if the following series converges.

${\sum }_{n=1}^{\infty }\frac{1}{2n^2-n\cos \left(n\pi \right)}$

Use the Alternating Series Test to determine if the following series is conditionally convergent, absolutely convergent, or divergent.

${\sum }_{n=1}^{\infty }\frac{{(-1)}^n(n+1)}{\ln n}$

Use the Alternating Series Test to determine if the following series is conditionally convergent, absolutely convergent, or divergent.

${\sum }_{k=1}^{\infty }\frac{\sin \frac{n\pi }{2}}{n^3}$

Use the Alternating Series Test to determine if the following series is conditionally convergent, absolutely convergent, or divergent.

${\sum }_{n=1}^{\infty }{(-1)}^{n-1}\left(\frac{3}{4n+5}\right)$

Use the Alternating Series Test to determine the convergence or divergence of the series.

${\sum }_{n=1}^{\infty }\frac{{(-1)}^{n+1}}{\sqrt{n}+2}$

Use the Alternating Series Test to approximate the sum of the series using the first five terms. 

${\sum }_{n=1}^{\infty }\frac{{(-1)}^{n+1}}{\sqrt{n}+2}$