How do you test for convergence of $Sigma (-1)^n/sqrt(lnn)$ from $n=[3,oo)$ ?

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How do you test for convergence of $Sigma (-1)^n/sqrt(lnn)$ from $n=[3,oo)$ ?

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Answer 1 · 2017-02-03T07:20:47

The series converges by the alternating series test.

Explanation:

The alternating series test states that a series $sum(-1)^na_n$ converges if $|a_n|$ decreases monotonically and $lim_(n->oo)a_n = 0$ .

In this case, we have an alternating series with $a_n = 1/sqrt(ln(n))$ . As $0 < 1/sqrt(ln(n)) < 1/sqrt(ln(n+1))$ , we have that $|a_n|$ is monotonically decreasing, and $lim_(n->oo)a_n = lim_(n->oo)1/sqrt(ln(n)) = 1/oo = 0$ . Thus, by the alternating series test, the series converges.

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How do you test for convergence of $Sigma (-1)^n/sqrt(lnn)$ from $n=[3,oo)$ ?

1 Answer

Explanation:

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How do you test for convergence of Sigma (-1)^n/sqrt(lnn)Sigma (-1)^n/sqrt(lnn) from n=[3,oo)n=[3,oo)?

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How do you test for convergence of $Sigma (-1)^n/sqrt(lnn)$ from $n=[3,oo)$ ?