Summation of function subtraction with a limit as $\epsilon \rightarrow 0$

Question

I got stuck in a problem in the middle of my calculations of integrals and sums.

$$\lim_{\epsilon \rightarrow 0} \sum_{n=1}^\infty f(n-\epsilon)-f(n+\epsilon)=0$$

where $f$ is continuous on all of the values in the summation

I am not sure for the general case of any $f$, but I needed verification for $f(n) = -\frac{1}{n}$ will be gratefull for an explanation for the any function

First I tried to verify for no sum and here what I got;

$$\forall_{n \in \mathbb{R}} \;\lim_{\epsilon \rightarrow 0} \frac{1}{n+\epsilon} - \frac{1}{n-\epsilon} = \frac{1}{n} - \frac{1}{n} = 0$$

$$\lim_{\epsilon \rightarrow 0} \sum_{n=1}^\infty \frac{1}{n+\epsilon} - \frac{1}{n - \epsilon}= \lim_{\epsilon \rightarrow 0} \sum_{n=1}^\infty\frac{-2\epsilon}{(n+\epsilon)(n-\epsilon)} = 0$$

Is this correct? And what about any other function

Answer 1

For this $f(x)=1/x$ you have $$\left|\sum_{n=1}^\infty f(n-\epsilon)-f(n+\epsilon)\right|=2\epsilon\left|\sum_{n=1}^\infty \frac{1}{n^2-\epsilon^2}\right|$$ and one way to finish: for any $\epsilon<1/2$ the sum in the modulus bars is positive and bounded uniformly in $\epsilon$ by $\sum_{n=1}^\infty \frac1{n^2-1/2}$ which is convergent, so by squeeze rule the limit is $0$.

The result is not true in general without something about $f$ becoming smaller as $n\to\infty$. For example if $f(x)=x$ then $$ f(n-\epsilon) - f(n+\epsilon) = n-\epsilon - n - \epsilon = -2\epsilon$$ which means $\sum_{n=1}^\infty f(n-\epsilon)-f(n+\epsilon)=-\infty$ for all $\epsilon>0$.

It's e.g. worse if $f(x)=x^2$, $f(n-\epsilon)-f(n+\epsilon) = -4\epsilon n$, and its better if $f(x)=\sqrt x$, $f(n-\epsilon)-f(n+\epsilon)=\sqrt{n-\epsilon}-\sqrt{n+\epsilon}=\frac{-2\epsilon}{\sqrt{n-\epsilon}+\sqrt{n+\epsilon}}\sim \frac{-2\epsilon}{\sqrt n}$, but not good enough.