When calculating the limit of the following characteristic function
$$ \frac{1}{n+1}\left[ \frac{1-\exp\left( \left(n+2 \right)it \right)}{ 1-\exp(it) } \right]$$
and taking its limit when $n\rightarrow\infty$ I have found its value equals to 0, since the complex exponential is a function valued between -1 and 1.
How can I interpret the random variable that has this limit characteristic function?
I guess the formula comes from the summation of $e^{ijt}$, but this is not valid if $e^{it}=0$. Let us call for $t\notin 2\pi\mathbf Z$, $$\varphi_n(t):=\frac{1}{n+1} \frac{1- \exp\left( \left(n+2 \right)it \right)}{1-\exp(it)}.$$ The limit $\lim_{t\to 0}\varphi_n(t)$ exits an is equal to $\frac{n+2}{n+1}$; by periodicity $\lim_{t\to 2j\pi}\varphi_n(t)=\frac{n+2}{n+1}$.
Note that if $t\notin 2\pi\mathbf Z$, then $$\left|\varphi_n(t)\right|\leqslant \frac 2{(n+1)|1-\exp(it)|},$$ therefore, $$\lim_{n\to\infty}\varphi_n(t)=\begin{cases} 1&\mbox{ if } x \in 2\pi\mathbf Z;\\ 0 &\mbox{ otherwise.} \end{cases} $$
Since the pointwise limit is not continuous, this cannot be a characteristic function.