Consider the infinite series $$\sum _{n=1}^\infty n 2^{-n}.$$ Apply the Ratio Test: $$\lim _{n \rightarrow \infty } \frac {\answer {(n+1) 2^{-n-1}}}{\answer {n 2^{-n}}} = \answer {\frac {1}{2}}.$$ Apply the Root Test: $$\lim _{n \rightarrow \infty } \left | \answer {n 2^{-n}} \right |^{\frac {1}{n}} = \answer {\frac {1}{2}}.$$ Both tests indicate that the series convergesdiverges .

Oftentimes the Ratio Test is easier to apply than the Root Test when dealing with factorials. Use the Ratio Test to determine convergence or divergence of the series $$\sum _{n=1}^\infty \frac {(n!)^2}{(2n)!}.$$ In the spaces below, record the eponymous “ratio” in the first blanks, simplify it in the second blanks, and then record the limit. $$\lim _{n \rightarrow \infty } \frac {\answer {\frac {((n+1)!)^2}{(2n+2)!}}}{\answer {\frac {(n!)^2}{(2n)!}}} = \lim _{n \rightarrow \infty } \answer {\frac {(n+1)^2}{(2n+1)(2n+2)}} = \answer {\frac {1}{4}}.$$ The test indicates convergenceindicates divergenceis inconclusive .

Apply the Ratio Test to the series given below. $$\sum _{n=0}^\infty \frac {5^n-3n}{4^n}$$ $$\lim _{n \rightarrow \infty } \frac {\answer {\frac {5^{n+1}-3(n+1)}{4^{n+1}}}}{\answer {\frac {5^n-3n}{4^n}}} = \lim _{n \rightarrow \infty } \answer {\frac {1}{4} \frac {5^{n+1} - 3(n+1)}{5^n - 3n}} = \answer {\frac {5}{4}}.$$ The test indicates convergenceindicates divergenceis inconclusive .

Apply the Ratio Test to the series given below. $$\sum _{n=1}^\infty \frac {1}{n^2+1}$$ $$\lim _{n \rightarrow \infty } \frac {\answer {((n+1)^2+1)^{-1}}}{\answer {(n^2+1)^{-1}}} = \lim _{n \rightarrow \infty } \answer {\frac {n^2+1}{(n+1)^2+1}} = \answer {1}.$$ The test indicates convergenceindicates divergenceis inconclusive .

Apply the Ratio Test to the series given below. $$\sum _{n=1}^\infty \frac {3^n}{n^2 2^n}$$ $$\lim _{n \rightarrow \infty } \frac {\answer {3^{n+1}/((n+1)^2 2^{n+1})}}{\answer {3^n/(n^2 2^n)}} = \lim _{n \rightarrow \infty } \answer { \frac {3}{2} \frac {n^2}{(n+1)^2}} = \answer {\frac {3}{2}}.$$ The test indicates convergenceindicates divergenceis inconclusive .

Apply the Ratio Test to the series given below. $$\sum _{n=0}^\infty \frac {4^n}{4^n+1}$$ $$\lim _{n \rightarrow \infty } \frac {\answer {\frac {4^{n+1}}{4^{n+1}+1}}}{\answer {\frac {4^n}{4^n+1}}} = \lim _{n \rightarrow \infty } \answer {4 \frac {4^n+1}{4^{n+1}+1}} = \answer {1}.$$ The test indicates convergenceindicates divergenceis inconclusive .

Oftentimes the Root Test is easier to apply when terms have large exponents (growing faster than a constant times $n$). Use the Root Test to determine the convergence or divergence of the series $$\sum _{n=1}^\infty 4^n \left ( 1 - \frac {1}{n} \right )^{n^2}.$$ First say what should quantity should have its $n$-th root taken, then simplify, and lastly record the value of the limit. $$\lim _{n \rightarrow \infty } = \left | \answer {4^n \left ( 1 - \frac {1}{n} \right )^{n^2}} \right |^{\frac {1}{n}} = \lim _{n \rightarrow \infty } \answer {4 \left ( 1 - \frac {1}{n} \right )^n} = \answer { \frac {4}{e}}.$$ The test indicates convergenceindicates divergenceis inconclusive .

Apply the Root Test to the series given below. $$\sum _{n=1}^\infty 2^{- \ln n}$$ $$\lim _{n \rightarrow \infty } \left | \answer {2^{-\ln n}} \right |^{\frac {1}{n}} = 2^{- \lim _{n \rightarrow \infty } \answer {\frac {\ln n}{n}}} = \answer {1}.$$ The test indicates convergenceindicates divergenceis inconclusive .

Apply the Root Test to the series given below. $$\sum _{n=1}^\infty 2^{-n^2}$$ $$\lim _{n \rightarrow \infty } \left | \answer {2^{-n^2}} \right |^{\frac {1}{n}} = \lim _{n \rightarrow \infty } \answer {2^{-n}} = \answer {0}.$$ The test indicates convergenceindicates divergenceis inconclusive .

Apply the Root Test to the series given below. $$\sum _{n=1}^\infty e^{\ln n} \left ( 1 - \frac {1}{n} \right )^{n^2}$$ $$\lim _{n \rightarrow \infty } \left | \answer {e^{\ln n} \left ( 1 - \frac {1}{n} \right )^{n^2}} \right |^{\frac {1}{n}} = \lim _{n \rightarrow \infty } \answer {e^{(\ln n)/n} \left ( 1 - \frac {1}{n} \right )^{n}} = \answer {1/e}.$$ The test indicates convergenceindicates divergenceis inconclusive .

Apply the Root Test to the series given below. $$\sum _{n=1}^\infty \frac {e^{n^2}}{n^n}$$ $$\lim _{n \rightarrow \infty } \left | \frac {e^{n^2}}{n^n} \right |^{\frac {1}{n}} = \lim _{n \rightarrow \infty } \answer {e^n/n} = \answer {\infty }$$ The test indicates convergenceindicates divergenceis inconclusive .