Does the Sum of (-1)^n·n/(n^2+1) Converge? Conditionally

Both conditions, checked properly

The sizes are $b_n = \frac{n}{n^{2}+1}$. Dividing top and bottom by $n$ gives $\frac{1}{n + 1/n}$, which clearly tends to 0, so the second condition holds.

For the first, treat the size as a function of a real variable: $f(x) = \frac{x}{x^{2}+1}$ has $f'(x) = \frac{1-x^{2}}{(x^{2}+1)^{2}}$, which is negative for $x > 1$. So the sizes decrease from $n = 1$ onward, and the alternating series test applies.

The derivative route is worth the extra line here. Comparing $b_{n+1}$ with $b_n$ directly on a quotient like this is messy, and the sign of a derivative settles it at once.

Why the convergence is only conditional

The absolute series is $\sum \frac{n}{n^{2}+1}$. Limit comparison against the harmonic series gives $\lim \frac{n/(n^{2}+1)}{1/n} = \lim \frac{n^{2}}{n^{2}+1} = 1$, a finite nonzero number, so the two series share a fate.

The harmonic series diverges, so the absolute series diverges as well. The original converges only because of the cancellation the signs provide, which is exactly what conditional convergence means.