Value of $\sum_{-\infty}^\infty e^{-i nt}=\text{?}$ Does anyone know if $\displaystyle\sum_{n=-\infty}^\infty e^{-i nt}$ converges to a known value? I don't have much to go with this so any assistance is appreciated. Thanks  
 A: It cannot converge, the things being summed always have magnitude $1$
A: Assuming $t$ is real, the absolute value of each term is $1$, so the series diverges.
A: Suppose you are working on the torus $\mathbb{T}$, then in some sense
\begin{align}
\sum^N_{n=-N} e^{int} = \frac{\sin ((N+1/2) t)}{\sin t/2}\rightarrow \delta(t)
\end{align}
as $N\rightarrow \infty$, where $\delta(t)$ is the Dirac delta function. 
Remark: However, I don't believe this is a good approximation of the delta function since the $L^1$ norm of the partial sum grows like $\log N$. 
Comment: I want to comment on @LutzL 's claim, but my comment is a little long for the comment section. 
The main point of this comment is to illustrate a subtle difference between smooth periodic function defined on $\mathbb{R}$ and smooth function defined on $\mathbb{T}$. To be honest, the difference is only a matter of language.   
For any sufficiently smooth function defined on $\mathbb{T}$ we can write down the Fourier series decomposition of $f(t)$
\begin{align}
f(t) =&\ C\sum^\infty_{n=-\infty}\hat f(n)e^{int} = C'\sum^\infty_{n=-\infty} \int^{\pi}_{-\pi} f(s)e^{in(t-s)}\ ds\\
 =&\ C' \int^\pi_{-\pi} f(s) K(t-s)\ ds  = C' (K\ast f)(t)  =: (T_K f)(t)
\end{align}
where
\begin{align}
K(t) := \sum^\infty_{n=-\infty} e^{in t} = \sum^\infty_{n=-\infty}e^{in(t+2\pi )} = K(t+2\pi ) \ \ \ (\text{This is completely formal.})
\end{align}
This shows that the convolution operator is actually the identity operator when restricted to functions defined on $[-\pi, \pi]$. With some abuse of notations, we usually write $K(t) = \delta(t)$ since the kernel of the identity operator (viewed as an integral operator) is usually identify with the $\delta$ function. 
But, if we want to extend the convolution operator $K$ to functions defined on all of $\mathbb{R}$ and not just functions on $[-\pi, \pi]$ then we need to be careful with our language. First note that $\delta(t) \neq \delta(t+2\pi)$! So of course $K(t)\neq \delta(t)$ when we view $T_K$ as an operator acting on functions defined on all of $\mathbb{R}$. As commented by LutzL, the kernel of $T_K$ as an operator acting on functions defined on all of $\mathbb{R}$ is actually given by
\begin{align}
K_\mathbb{R}(t) = \sum^\infty_{k=-\infty}\delta(t-2\pi k). \ \ \ \ (1)
\end{align}
To see that $(1)$ holds, observe that 
\begin{align}
(T_K f)(t) :=&\ \int^\infty_{-\infty} K_\mathbb{R}(t-s) f(s)\ ds = \sum^\infty_{k=-\infty} \int^{\pi + 2\pi k}_{-\pi +2\pi k} K_\mathbb{T}(t-s)f(s)\ ds \\
 =&\ \sum^\infty_{k=-\infty}\int^{\pi}_{-\pi} K_\mathbb{T}(t-s+2\pi k)f(s-2\pi k)\ ds\\
=&\  \sum^\infty_{k=-\infty}\int^{\pi}_{-\pi} K_\mathbb{T}(t-s)f(s-2\pi k) = \sum^\infty_{k=-\infty} f(t-2\pi k)\\
=&\ \int^\infty_{-\infty} \sum^\infty_{k=-\infty} \delta(t-2\pi k-s) f(s)\ ds. 
\end{align}
Conclusion: In short, on $\mathbb{T}$ we have that $\sum^\infty_{n=-\infty} e^{int} = \delta(t)$ but on $\mathbb{R}$ we have that $\sum^\infty_{n=-\infty}\delta(t-2\pi k)$.   
A: If one assumes $t$ is real, then $e^{int}$ does not approach $0$ as $n\to\infty$ not as $n\to-\infty,$ so the series cannot converge in the sense that you learned in high school.
If $t=0$ or $t$ is otherwise an integer multiple of $2\pi,$ then $e^{int}=1$ regardless of the value of $n,$ so the series is $\displaystyle \sum_{n\,\in\,\mathbb Z} 1 = +\infty.$
There are other kinds of convergence, including Cesàro summation, and in that sense the sum is $0$ except when $t$ is a multiple of $2\pi.$
So already we have some suggestions of the delta function.
Now recall that there is this inner product: $\displaystyle \langle f,g\rangle = \frac 1 {2\pi} \int_0^{2\pi} f(t) \overline{g(t)} \, dt.$
The Fourier series of $f(t)$ is a series $\sum\limits_{n\,\in\,\mathbb Z} c_n e^{int}$ for which $\Big\langle (t\mapsto e^{ikt}), (t\mapsto f(t))\Big\rangle$ is the same as $\Big\langle (t\mapsto e^{ikt}), (t\mapsto \sum\limits_{n\,\in\,\mathbb Z} c_n e^{int})\Big\rangle.$ According to that, the Fourier series of $\delta(t)$ is $\sum\limits_{n\,\in\,\mathbb Z} e^{int},$ provided you assume a bunch of stuff about linearity works as well for infinite sums as for finite sums.
But $\delta(t)$ is not really a function in the sense of giving an output for every input $t.$ It is characterized by
$$
\int_0^{2\pi} f(t)\delta(t)\,dt = f(0)
$$
(for infinitely differentiable $2\pi$-periodic functions $f$). Does this alleged Fourier series for $\delta(t)$ satisfy that characterization? It does, again if one assumes certain things work as well with infinite series as with finite sums.
A: For a $2\pi$ periodic, square integrable, continuous at zero, function $f$ you have that its Fourier series $$f(x)=\sum_{k\in\Bbb Z}c_k(f)\,e^{ikx}$$ will converge towards its value at $x=0$, that is
$$
\sum_{n\in\Bbb Z}\int_{-\pi}^\pi e^{-int}f(t)\,dt = \sum_{n\in\Bbb Z} 2\pi c_k(f) = 2\pi f(0)
$$
Now take some test function $\phi$ of the space of tempered test functions, so that it is, among other things, continuous and fast falling as $|\phi(t)|<C(1+|t|)^{-2}$. Its periodic summation $f(t)=\sum_{k\in\Bbb Z}\phi(t+2k\pi)$ is then a $2\pi$-periodic and continuous function. 
Then in the sense of tempered distributions the series in question actually converges and has the value
\begin{align}
\left\langle \sum_{n\in\Bbb Z}e^{-int},\phi\right\rangle 
&=\sum_{n\in\Bbb Z}\left\langle e^{-int},\phi\right\rangle 
=\sum_{n\in\Bbb Z}\int_{\Bbb R}e^{-int}ϕ(t)\,dt&&\text{(by topology of distributions)}
\\
&=\sum_{n\in\Bbb Z}\sum_{k\in\Bbb Z}\int_{(2k-1)\pi}^{(2k+1)\pi}e^{-int}ϕ(t)\,dt
\\
&=\sum_{n\in\Bbb Z}\int_{-\pi}^\pi e^{-int}f(t)\,dt
\\
&= 2\pi f(0)=2\pi \sum_{k\in\Bbb Z}\phi(2k\pi)
\end{align}
or
$$
\sum_{n\in\Bbb Z}e^{-int} = 2\pi \sum_{k\in\Bbb Z}\delta(t-2k\pi)
$$
This is one version of the Poisson summation formula. It is also summarized in that 

"the Fourier transform of a Dirac comb is again a Dirac comb,"

as, depending on the distributions of the constants in the Fourier transform,
$$
\sum_{n\in\Bbb Z}\widehat{δ_n}(t)=C\sum_{n\in\Bbb Z}e^{-int} ,\qquad δ_n(x)=δ(x-n).
$$
