Determine if $\sum_{n = 1}^{\infty} \frac{1}{x^{2} + n^{2}}$ converges on $\mathbb{R}$. Proof verification

Question

Determine if $\sum_{n = 1}^{\infty} \frac{1}{x^{2} + n^{2}}$ converges on $\mathbb{R}$.

Attempt:

Examining the behavior of $\frac{1}{x^{2} + n^{2}}$, we see that:

$$\|\frac{1}{x^{2} + n^{2}}\|_{\infty} = \sup_{x \in \infty}\Bigg|\frac{1}{x^{2} + n^{2}}\Bigg| \leq \frac{1}{n^2}.$$

If we let $M_{n} = \frac{1}{n^2}$ it has been established that $\sum_{k = 1}^{\infty}\frac{1}{n^{2}}$ converges. So by application of the M- test, our series converges.

My problem: I always have trouble deciding how to bound things in this form: $$\Bigg|\frac{1}{x^{2} + n^{2}}\Bigg| \leq \frac{1}{n^2}$$ as an example. What am I allowed to remove and what can I not remove from any sort of comparison? If my $x$ is being assumed as fixed am I allowed to remove that to bound my function or is it the $n$ that I can remove if need be? I may have answered my own question by stating that $x$ is fixed so that means any $n$ term would have to remain... But in a similar question my professor did a comparison of $k(k + x^2) \geq k$ to manipulate denominators. So again it comes down to what am I allowed to remove?

Answer 1

For any $x\in\Bbb R$, we have $x^2 \geq 0$. Hence, $x^2+n^2 \geq n^2$ and thus (by taking the reciprocal*) $$\frac1{x^2+n^2} \le \frac1{n^2}.$$

* For two strictly positive real numbers $r_1, r_2$, we have $$r_1\geq r_2 \iff \frac1{r_1} \le \frac1{r_2}.$$

Answer 2

Firstly, the series is running on $n$ therefore everything that is not a function of n is a constant.

Secondly, to "bound" these functions keep the following in mind.

$$\forall \space x,y\in \mathbf{R^+}$$ If $$ \space x>y$$ Then $$\frac{1}{x}<\frac{1}{y}$$

Now since both $x^2$ and $n^2$ are positive quantities And $$ x^2 +n^2 \geq n^2$$ $$ \implies \frac{1}{x^2+n^2}\leq \frac{1}{n^2}$$ $$\implies \sum\limits_{n=1}^{\infty} \frac{1}{x^2+n^2} \leq \sum\limits_{n=1}^{\infty} \frac{1}{n^2}$$ By comparision test i.e. If the Larger series converges then the smaller one converges too.

The given series is convergent.