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    Convergence and Divergence : Limits of Sequences and Series

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    Each series telescopes. In each case - express the nth partial sum Sn in terms of n and determine whether the series converges or diverges.

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    Solution to question 1.

    a) Sn=(1/3-1/5)+(1/5-1/7)+...+[1/(2n+1)-1/(2n+3)]
    =1/3-1/(2n+3)
    So, lim Sn=1/3
    So the series converges.

    b) Since 1/[(2k-1)(2k+1)]=[1/(2k-1)-1/(2k+1)]/2, we have
    Sn=(1/2-1/6)+(1/6-1/10)+...+{1/[2(2n-1)]-1/[2(2n+1)]}
    =1/2-1/[2(2n+1)]
    So, lim Sn=1/2
    So the series converges.

    c) Since [sqrt(k+1)-sqrt(k)]/sqrt(k^2+k)=1/sqrt(k)-1/sqrt(k+1), we ...

    Solution Summary

    Convergence and Divergence are shown in terms of the Limits of Sequences and Series are examined. The series telescopes expressed are provided.

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