Exam 10: Infinite Sequences and Series

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Use the nth-Term Test for divergence to show that the series is divergent, or state that the test is inconclusive. - n=1nn+2\sum _ { n = 1 } ^ { \infty } \frac { n } { n + 2 }

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Determine if the sequence is monotonic and if it is bounded. - an=(n+7)!(7n+1)!a _ { n } = \frac { ( n + 7 ) ! } { ( 7 n + 1 ) ! }

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Find the smallest value of N that will make the inequality hold for all n > N. - an=tan1nena _ { n } = \frac { \tan ^ { - 1 } n } { e ^ { n } }

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Determine if the geometric series converges or diverges. If it converges, find its sum. - 1+25+(25)2+(25)3+(25)4+1 + \frac { 2 } { 5 } + \left( \frac { 2 } { 5 } \right) ^ { 2 } + \left( \frac { 2 } { 5 } \right) ^ { 3 } + \left( \frac { 2 } { 5 } \right) ^ { 4 } + \ldots

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Determine if the series converges or diverges. If the series converges, find its sum. - n=1(1n+11n+2)\sum _ { n = 1 } ^ { \infty } \left( \frac { 1 } { \sqrt { n + 1 } } - \frac { 1 } { \sqrt { n + 2 } } \right)

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Find a formula for the nth term of the sequence. - 0,0,2,2,0,0,2,20,0,2,2,0,0,2,2 (alternating 0's and 2's in pairs)

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Find the sum of the series. - n=087n\sum _ { n = 0 } ^ { \infty } \frac { 8 } { 7 ^ { n } }

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Determine if the series converges or diverges; if the series converges, find its sum. - n=0e4n\sum _ { n = 0 } ^ { \infty } e ^ { - 4 n }

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Find the limit of the sequence if it converges; otherwise indicate divergence. - an=6nnna _ { n } = \sqrt [ n ] { 6 ^ { n } \cdot n }

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Find the limit of the sequence if it converges; otherwise indicate divergence. - an=5+(1)n5a _ { n } = \frac { 5 + ( - 1 ) ^ { n } } { 5 }

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Determine if the series converges or diverges; if the series converges, find its sum. - n=14n+18n1\sum _ { n = 1 } ^ { \infty } \frac { 4 ^ { n + 1 } } { 8 ^ { n - 1 } }

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A recursion formula and the initial term(s) of a sequence are given. Write out the first five terms of the sequence. - a1=1,an+1=nann+5a _ { 1 } = 1 , a _ { n + 1 } = \frac { n a _ { n } } { n + 5 }

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Provide an appropriate response. -It can be shown that limnlnnnc=0\lim _ { \mathrm { n } \rightarrow\infty } \frac { \ln \mathrm { n } } { \mathrm { n } ^ { \mathrm { c } } } = 0 for c>0\mathrm { c } > 0 . Find the smallest value of N\mathrm { N } such that lnnnc<ε\left| \frac { \ln \mathrm { n } } { \mathrm { n } ^ { \mathrm { c } } } \right| < \varepsilon for all n>N\mathrm { n } > \mathrm { N } if ε=0.01\varepsilon = 0.01 and c=1.8c = 1.8 .

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Express the number as the ratio of two integers. - 1.8181811.818181 \ldots

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Find the limit of the sequence if it converges; otherwise indicate divergence. - an=(1+2n)na _ { n } = \left( 1 + \frac { 2 } { n } \right) ^ { n }

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A recursion formula and the initial term(s) of a sequence are given. Write out the first five terms of the sequence. - a1=1,an+1=6ana _ { 1 } = 1 , a _ { n + 1 } = 6 a _ { n }

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Find the limit of the sequence if it converges; otherwise indicate divergence. - an=(2nn+1)na _ { n } = \left( \frac { 2 n } { n + 1 } \right) ^ { n }

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Find the limit of the sequence if it converges; otherwise indicate divergence. - an=nn218na _ { n } = n - \sqrt { n ^ { 2 } - 18 n }

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Find the limit of the sequence if it converges; otherwise indicate divergence. - an=n!5n3na _ { n } = \frac { n ! } { 5 ^ { n } \cdot 3 ^ { n } }

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Find the limit of the sequence if it converges; otherwise indicate divergence. - an=(lnn)2na _ { n } = \frac { ( \ln n ) ^ { 2 } } { \sqrt { n } }

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