Use the Direct Comparison Test with ${\sum }_{n=1}^{\infty }\frac{1}{5^n}$ to decide ${\sum }_{n=1}^{\infty }\frac{1}{5^n+1}$.

A student notes that $\frac{1}{n^2+1}<\frac{1}{n}$ for every $n\ge 1$, and that ${\sum }_{n=1}^{\infty }\frac{1}{n}$ diverges. What does that comparison ESTABLISH about ${\sum }_{n=1}^{\infty }\frac{1}{n^2+1}$?

Apply the Limit Comparison Test to ${\sum }_{n=1}^{\infty }\frac{2n^2+3}{8n^4-n}$ using $b_n=\frac{1}{n^2}$. What is the limit, and what does it establish?

A student notes that $\frac{1}{n^2+8}<\frac{1}{n}$ for every $n\ge 1$, and that ${\sum }_{n=1}^{\infty }\frac{1}{n}$ diverges. What does that comparison ESTABLISH about ${\sum }_{n=1}^{\infty }\frac{1}{n^2+8}$?

Use the Direct Comparison Test with ${\sum }_{n=1}^{\infty }\frac{1}{4^n}$ to decide ${\sum }_{n=1}^{\infty }\frac{1}{4^n+4}$.

Which series is the natural comparison for ${\sum }_{n=1}^{\infty }\frac{9n^2-2}{n^5+4n^4}$?

For ${\sum }_{n=2}^{\infty }\frac{\ln n}{n^4}$ compared with $b_n=\frac{1}{n^3}$, the limit ${\lim }_{n\to \infty }\frac{a_n}{b_n}={\lim }_{n\to \infty }\frac{\ln n}{n}=0$. What follows?

Apply the Limit Comparison Test to ${\sum }_{n=1}^{\infty }\frac{4n+5}{9n^4-n}$ using $b_n=\frac{1}{n^3}$. What is the limit, and what does it establish?

Use the Direct Comparison Test with ${\sum }_{n=1}^{\infty }\frac{1}{n}$ to decide ${\sum }_{n=1}^{\infty }\frac{n+6}{n^2}$, for $n$ large enough that every term is positive.

Which series is the natural comparison for ${\sum }_{n=1}^{\infty }\frac{9n^2-2}{n^4+4n^3}$?

Apply the Limit Comparison Test to ${\sum }_{n=1}^{\infty }\frac{4n^2+5}{6n^3-n}$ using $b_n=\frac{1}{n}$. What is the limit, and what does it establish?

Apply the Limit Comparison Test to ${\sum }_{n=1}^{\infty }\frac{3n^2+4}{2n^4-n}$ using $b_n=\frac{1}{n^2}$. What is the limit, and what does it establish?

Use the Direct Comparison Test with ${\sum }_{n=1}^{\infty }\frac{1}{4^n}$ to decide ${\sum }_{n=1}^{\infty }\frac{1}{4^n+7}$.

For ${\sum }_{n=2}^{\infty }\frac{\ln n}{n^3}$ compared with $b_n=\frac{1}{n^2}$, the limit ${\lim }_{n\to \infty }\frac{a_n}{b_n}={\lim }_{n\to \infty }\frac{\ln n}{n}=0$. What follows?

A student notes that $\frac{1}{n^2+2}<\frac{1}{n}$ for every $n\ge 1$, and that ${\sum }_{n=1}^{\infty }\frac{1}{n}$ diverges. What does that comparison ESTABLISH about ${\sum }_{n=1}^{\infty }\frac{1}{n^2+2}$?

Use the Direct Comparison Test with ${\sum }_{n=1}^{\infty }\frac{1}{6^n}$ to decide ${\sum }_{n=1}^{\infty }\frac{1}{6^n+8}$.

Which series is the natural comparison for ${\sum }_{n=1}^{\infty }\frac{4n+6}{n^2+8n}$?

Which series is the natural comparison for ${\sum }_{n=1}^{\infty }\frac{6n^3-5}{n^5+7n^4}$?

Which series is the natural comparison for ${\sum }_{n=1}^{\infty }\frac{3n-3}{n^2+5n}$?

Apply the Limit Comparison Test to ${\sum }_{n=1}^{\infty }\frac{2n+3}{4n^4-n}$ using $b_n=\frac{1}{n^3}$. What is the limit, and what does it establish?

Apply the Limit Comparison Test to ${\sum }_{n=1}^{\infty }\frac{8n+9}{5n^4-n}$ using $b_n=\frac{1}{n^3}$. What is the limit, and what does it establish?

Apply the Limit Comparison Test to ${\sum }_{n=1}^{\infty }\frac{5n+6}{2n^2-n}$ using $b_n=\frac{1}{n}$. What is the limit, and what does it establish?

Use the Direct Comparison Test with ${\sum }_{n=1}^{\infty }\frac{1}{4^n}$ to decide ${\sum }_{n=1}^{\infty }\frac{1}{4^n+5}$.

For ${\sum }_{n=2}^{\infty }\frac{\ln n}{n^4}$ compared with $b_n=\frac{1}{n^2}$, the limit ${\lim }_{n\to \infty }\frac{a_n}{b_n}={\lim }_{n\to \infty }\frac{\ln n}{n^2}=0$. What follows?