= Solution
Compare with the divergent <series> from part (i). The ratio of its new summand to $1/(n\log n)$ is
$$
\frac{1/[n^{1+1/n}\log n]}{1/(n\log n)}=n^{-1/n}=e^{-(\log n)/n}\longrightarrow1.
$$
For completeness, $\log n=\int_1^n x^{-1}\,dx\le\int_1^n x^{-1/2}\,dx=2(\sqrt n-1)$, so $(\log n)/n\to0$. The <limit comparison test> now gives \b[divergence]. Although the exponent $1+1/n$ exceeds one for each summand, it approaches one and does not produce a convergent <p-series> with a fixed exponent greater than one.
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