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Determine convergence or divergence and explain which method you used for......
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\[\sum_{n=2}^{\infty} 1/ n \ln n-n\]
the whole term in in denominator?
no.. sorry I don't know how to make a division sign on here...... but its 1 over nln-n
thats what i want to confirm
yeah. 1 is in the numerator.
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have u read the integral test for convergence of the series?
If \[\int\limits_{1}^{\infty} f(x)dx\] converges then \[\sum_{n=1}^{\infty} a _{n} \] converges
right and function must be montonically decreasing
\[\int\limits_{2}^{\infty} (1/x)/(\ln x-1) dx\]
?
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solve the integral, the given series converges if the integral converges
\[\lim t \rightarrow \infty [\ln(lnx-1)] x varies from 2 \to t\]
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