= Solution
For odd $n$, the denominator is $(n+1)^2$; for even $n$, it is $(n-1)^2$. There is no zero denominator in the printed index range. For all $n\ge2$,
$$
0<\frac1{n^2+(-1)^{n+1}2n+1}\le\frac1{(n-1)^2}\le\frac4{n^2}.
$$
The <comparison test for series> with the convergent <p-series> gives \b[absolute convergence]. In fact the consecutive pair $n=2k-1,2k$ contributes $1/(2k)^2+1/(2k-1)^2$, so the sum is the same as $\sum_{m=1}^\infty m^{-2}$. Positivity justifies this paired reordering; an alternating-sign appearance in the denominator does not make this an alternating series.
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