Test whether the series is converging (uniformly) on the interval $\text{[}0,\infty \text{)}$.

$$\sum_{n=1}^\infty (-1)^\frac{n(n+1)}{2}\frac{1}{x+n-\ln(x^2+n^2)}$$

I'm guessing we could maybe do something with the comparison (Weierstrass test). However I'm having trouble finding the right thing to compare it with. Another guess is maybe Leibniz's alternating test since it seems to converge to $0$ as $n\rightarrow \infty$. The only problem is monoticity. Any ideas?


We can apply the Dirichlet test to prove uniform convergence for $x \in [0,\infty)$.

The term $(-1)^{n(n+1)/2}$ changes from $-1$ to $+1$ in the pattern $-,-,+,+,-,-, \ldots$ and the partial sums are uniformly bounded, $\left|\sum_{n=1}^m (-1)^{n(n+1)/2}\right| \leqslant 2.$

It remains to show that $(x + n - \ln(x^2 + n^2))^{-1}$ converges to $0$ not only montonically, but uniformly as well.

To prove monotonicity, examine the derivative of $f(y) = x+y - \ln(x^2 + y^2).$

Uniform convergence follows from the following inequality (when $n > 1):$

$$\tag{*}\frac{1}{x+n - \ln(x^2 + n^2)} \leqslant \frac{1}{n - \ln(n^2)}.$$

To prove (*), note that $e^x > 1 + x^2/n^2$ implies that

$$0 < x - \ln (1 + x^2/n^2) = x + n - \ln(x^2 +n^2) - (n - \ln(n^2).$$


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