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Thread: Why does the series converge?

  1. #1
    Member Pranas's Avatar
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    Why does the series converge?

    Hello.

    How to prove that the series converge?

    $\displaystyle \[S = \sum\limits_{n = 2}^{ + \infty } {{{( - 1)}^n} \cdot {c_n}} = \sum\limits_{n = 2}^{ + \infty } {\frac{{{{( - 1)}^n}}}{{\sqrt n + {{( - 1)}^n}}}} \]$

    That's obviously an alternating series because $\displaystyle \[{c_n} > 0\]$

    In addition $\displaystyle \[\mathop {\lim }\limits_{n \to \infty } {c_n} = 0\]$

    But $\displaystyle \[{c_{n + 1}}\]$ is NOT always less than $\displaystyle \[{c_n}\]$

    and apparently it prevents me from using Leibniz's test.

    Without subtractions $\displaystyle \[\sum\limits_{n = 2}^{ + \infty } {{c_n}} \]$ diverges so I've got no way to go

    Am I missing something obvious?
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  2. #2
    Super Member girdav's Avatar
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    You know that the series $\displaystyle \sum_{n=2}^{+\infty}\frac{(-1)^n}{\sqrt n}$ is convergent. Hence $\displaystyle S$ is convergent if and only if $\displaystyle \sum_{n=2}^{+\infty}(-1)^nc_n-\frac{(-1)^n}{\sqrt n}$ is convergent.
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