Define the function $f:(0,\infty)\rightarrow \mathbb{R}$ by
$$f(x)=\sum\limits_{n=1}^\infty\frac{1}{2^n}\cdot \frac{1}{\left(\frac{1}{n}-x\right)^2}I_{\{x: x \ne \frac{1}{n}\}}(x)$$.
What can we say about this function as it approaches zero? I think it is pretty clear that $\limsup\limits_{x \rightarrow 0^+}f(x)=\infty$, this is because no matter how close we are to zero, we can choose an $n^*$ closer to zero, and then approach it. But what about $\liminf\limits_{x \rightarrow 0^+}f(x)$?, is it possible to see what this is? As we get closer to zero, the tail in the sum has elements of $(1/n-x)^2$ that becomes very small, but $2^n$ becomes very big, so they might cancel them out?
Let $\epsilon > 0$. Take $x$ to be half way between $\frac1{m+1}$ and $\frac1m$. Then if $m$ is large enough, $$ \left|x-\frac1n\right| \ge \frac1{3m^2} \quad \text{if $n \ge \sqrt m$},$$ and $$ \left|x-\frac1n\right| \ge \frac{1-\epsilon} n \quad \text{if $n < \sqrt m$}.$$ Then $$ \sum_{n \ge \sqrt m} 2^{-n} \left(x-\frac1n\right)^{-2} \le \frac{18 m^4}{2^{\sqrt m}} \to 0 \quad \text{as $m \to \infty$,} $$ and $$ \sum_{n < \sqrt m} 2^{-n} \left(x-\frac1n\right)^{-2} \le \frac1{(1-\epsilon)^2}\sum_{n \le \sqrt m} \frac{n^2}{2^{n}} . $$ So $$ \liminf_{x\to 0+} f(x) \le \sum_{n=1}^\infty \frac{n^2}{2^{n}} .$$ But clearly $$ f(x) \ge \sum_{n=1}^\infty \frac{n^2}{2^{n}} .$$ Hence $$ \liminf_{x\to 0+} f(x) = \sum_{n=1}^\infty \frac{n^2}{2^{n}} = 6.$$