Determine the convergence or divergence of the series.sum_{n=3}^inftyfrac{1}{n(ln n)[ln(ln n)]^4}

Ayaana Buck 2021-01-10 Answered

Determine the convergence or divergence of the series.
n=31n(lnn)[ln(lnn)]4

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Expert Answer

comentezq
Answered 2021-01-11 Author has 106 answers
A series, n=mf(n) converges if the value of the integral, mf(n) is finite and diverges if it is infinite.
For the series, n=31n(lnn)[ln(lnn)]4, the integral is 31x(lnx)[ln(lnx)]4dx
Let ln(lnx)=y,1xlnxdx=dy
At x=3,y=ln(ln3)
at x=,y=
So, the integral becomes:
31x(lnx)[ln(lnx)]4dx=ln(ln3)1x(lnx)(y)4x(lnx)dy
=ln(ln3)1y4dy
=15y5|ln(ln3)
=15(11(ln(ln3))5)
=15(ln(ln3))5
Since, the value of the integral is finite. Thus, the sum of the series converges.
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