64 lines
1.4 KiB
TeX
64 lines
1.4 KiB
TeX
\beginsection{14.1}
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Determine which of the following series converge.
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Justify your answers.
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\medskip
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(a) $\sum n^4/2^n$
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\medskip
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Try the ratio test.
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$$
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{a_{n+1}\over a_n}
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={(n+1)^4/2^{n+1}\over n^4/2^n}
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={(n+1)^4\over2^{n+1}}\times{2^n\over n^4}
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={(n+1)^4\over2n^4}\rightarrow{1\over2}<1
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$$
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So by the ratio test $\sum n^4/2^n$ converges.
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\medskip
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(b) $\sum2^n/n!$
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\medskip
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Try the ratio test.
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$$
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{a_{n+1}\over a_n}={2^{n+1}/(n+1)!\over 2^n/n!}
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={2^{n+1}\over(n+1)!}\times{n!\over2^n}={2\over n+1}<1
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$$
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So by the ratio test $\sum2^n/n!$ converges.
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\medskip
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(c) $\sum n^2/3^n$
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\medskip
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Try the ratio test.
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$$
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{a_{n+1}\over a_n}
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={(n+1)^2/3^{n+1}\over n^2/3^n}
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={(n+1)^2\over 3^{n+1}}\times{3^n\over n^2}
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={(n+1)^2\over3n^2}\rightarrow{1\over3}<1
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$$
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So by the ratio test $\sum n^2/3^n$ converges.
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\medskip
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(d) $\sum n!/(n^4+3)$
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\medskip
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Try the ratio test.
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$$
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{a_{n+1}\over a_n}
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={(n+1)!/((n+1)^4+3)\over n!/(n^4+3)}
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={(n+1)!\over((n+1)^4+3)}\times{n^4+3\over n!}
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\rightarrow n+1>1
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$$
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So by the ratio test $\sum n!/(n^4+3)$ diverges.
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\medskip
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(e) $\sum\cos^2 n/n^2$
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\medskip
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Use the comparison test.
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Since $\sum1/n^2$ converges and
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$|\cos^2 n/n^2|\le1/n^2$ for all $n$, $\sum\cos^2 n/n^2$ must also converge.
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\medskip
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(f) $\sum_{n=2}^\infty1/(\log n)$
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\medskip
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Use the comparison test.
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Since $\sum(1/n)$ diverges and $1/(\log n)>1/n$ for all $n>1$,
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$\sum_{n=2}^\infty1/(\log n)$ must also diverge. |