Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

Does the series Sum (k=2, oo) 1/((ln(k))^ln(k)) converge diverge or?

I'd like some help to decide if the series Sum (k=2, oo) 1/((ln(k))^ln(k)) converge or diverge.

Thank you

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  • 1 decade ago
    Favorite Answer

    a[k] = 1/ln(k)^ln(k)

    a[k+1] = 1/ln(k(1+1/k))^ln(k+1)

    = 1/( ln(k) + ln(1+1/k))^ln(k+1)

    = 1/ ( ln(k)^ln(k+1) ( 1+ (ln(1+1/k)/ln(k) )^ln(k+1)

    a[k]/a[k+1] =ln(k)(1+(ln(1+1/k)/ln(k) ) ^ ln(k+1) > ln(k)

    after the first term (or after the first few terms if we had some other base) ln(k) > 1

    in particular after 100 terms

    a[k+1]/a[k] < 1/ln[k] < 1/ln(100) < 1/4

    therefore the series converges

    Edit -

    In view of MathMark's answer posted about the same time I had a look at the sum in excel.

    I think its interesting that after the first 95 terms there are

    135 terms < 1/10^3 totalling < 0.135

    300 terms < 1/10^4 totalling < 0.0300

    640 terms < 1/10^5 totalling < 0.00640

    1348 terms < 1/10^6 totalling < 0.001348

    2770 terms < 1/10^7 totalling < 0.0002770

    5580 terms < 1/10^8 totalling < 0.00005580

    11126 terms< 1/10^9 totalling < 0.000011126

    21863 terms <1/10^10totalling <0.0000021863

    This doesn't prove anything of course, but its enough to convince me I don't have to spend more time looking for an error in my logic

    For info it has all the appearance of converging to 5.716970544

  • 3 years ago

    The shrink of (-a million)^n (n+a million)/sqrt(n) as n-->inf DNE, so the endless sequence diverges to boot. many times, the endless sum of phrases can purely exist if the shrink of the phrases themselves is 0 (even with the certainty that the shrink of the sums of the phrases need not be 0 of direction).

  • 1 decade ago

    we can compare it to 1/(lnk)^(k)

    ln(k) < k

    so

    ln(k) ^ ln(k) < (ln k) ^ k

    so

    1/ (ln k)^ln k > 1/ (ln k) ^ k

    Since 1 / (ln k) ^ k diverges, so does this series.

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