Find function $f$ given $f(x+1) - f(x) = \frac{1}{x^2}$

Since $f(t)\to 0$ as $t\to \infty$, we see that $$\sum_{n=0}^\infty\,\frac{1}{(x+n)^2}=\sum_{n=0}^\infty\,\big(f(x+n+1)-f(x)\big)=-f(x)\,.$$ This shows that the desired function $f:\mathbb{R}_{>0}\to\mathbb{R}$ must satisfy $$f(x)=-\sum_{n=0}^\infty\,\frac{1}{(x+n)^2}\tag{*}$$ for all $x>0$. The rest in the hidden portion is the justification that $f$ given by (*) satisfies the continuity requirement, as well as the limit requirement.

Note that the function given by (*) is well defined since the infinite sum converges by the comparison test with the series $\sum\limits_{n=1}^\infty\,\dfrac{1}{n^2}=\dfrac{\pi^2}{6}$. The continuity of $f$ given in (*) follows from the observation that the sequence $\left\{f_k\right\}_{k\in\mathbb{Z}_{>0}}$ of functions $f_k:\mathbb{R}_{>0}\to\mathbb{R}$ defined by $$f_k(x):=-\sum_{n=0}^k\,\frac{1}{(x+n)^2}\text{ for all }k\in\mathbb{Z}_{\geq 0}\text{ and }x>0$$ uniformly converges to $f$. Clearly, we also have $$0>f(x)>-\frac{1}{x^2}-\sum_{n=1}^\infty\,\frac{1}{(x+n)(x+n-1)}=-\frac{1}{x^2}-\frac{1}{x}\,.$$ Ergo, $f$ indeed satisfies the limit requirement.

I apologise for the following but couldn't unsee :

Since $f(x+1) - f(x) = 1/x^2$ and $f$ is defined over $\mathbb R^+$, then $x>0$ and this means that $x+1 > x$ and also $1/x^2 >0$. Thus $f(x+1) - f(x) > 0 \Leftrightarrow f(x+1) > f(x)$ and since that holds for every $x \in \mathbb R^+$, then $f$ is increasing and also $f$ is bounded, since $\lim_{x \to + \infty} f(x) = 0$.

Now, consider an operator $T$ such that $Tf(x) = f(x+1)$ defined as $T : C^b(0,+\infty) \to C^b(0,+\infty)$ where $C^b(0,+\infty)$ is the space of the continuous bounded functions in $(0,+\infty)$. In this space the sup norm is well defined and this space is complete (basically since the uniform limit of continuous functions is continuous). This tells us that $C^b(0,+\infty)$ is a Banach space.

Now, let $f,g \in C^b(0,+\infty)$ and $\lambda \in \mathbb R$. Then :

$$T(\lambda f + g) = (\lambda f + g)(x+1) = (\lambda f)(x+1) + g(x+1)$$

$$=$$ $$ \lambda f(x+1) + g(x+1) = \lambda Tf + Tg$$

This tells us that the operator $T$ is linear.

Now, it also is

$$\|Tf(x)\|_\infty = \|f(x+1)\|_\infty \leq \|f(x)\|_\infty$$

and thus $T$ is a bounded linear operator (specifically with $\|T\| \leq 1$) as well as $T \in B(C^b(0,+\infty))$.

But, since $C^b(0,+\infty)$ is Banach and $T \in B(C^b(0,+\infty))$, the equation

$$f(x) = \frac{1}{x^2} + Tf(x) \Leftrightarrow f(x) - Tf(x) = \frac{1}{x^2}$$

has a unique solution in $C^b(0,+\infty)$.

For the resemblance with the answer posted above, notice that

$$f(x) - Tf(x) = -\frac{1}{x^2} \Leftrightarrow (\mathbf{1} - T)f(x) = -\frac{1}{x^2} \Rightarrow f(x) = (\mathbf{1}-T)^{-1}\bigg(-\frac{1}{x^2}\bigg) $$

where $\mathbf{1}$ is the identity operator. But :

$$(\mathbf{1}-T)^{-1} = \sum_{n=1}^\infty T^n, \quad |T| < 1$$

Note : We defined $\frac{1}{x^2} : \mathbb R^+ \to \mathbb R$.

Note 2: Per comment discussion, if any doubts/issues regarding bounds, the space $C^b([\varepsilon, + \infty))$ could be considered where $\varepsilon >0$ and by manipulating $\varepsilon$ accordingly we could still yield all results.