Summing the sequence $a(n) = \sin(n x) \exp(-nt)$

Question

Consider the sequence $a(n)$ defined by $a(n) = \sin(n x) \exp(-nt)$, where $n = 0, 1, 2, 3, 4, \ldots$. The parameter $x$ is a real number. Parameter $t$ is a positive real number. It is clear that the sequence $a(n)$ converges to $0$ as $n \rightarrow \infty$.

We define $S(k)$ as the partial sum of the sequence $a(n)$ from $n = 0$ to $k$. It is straightforward to show that the $S(k)$, in the limit of $k \rightarrow \infty $, converges to the following value $S$:

$$S = \frac{\sin(x)}{2 - 2\cos(x) + (e^t -e^{-t})^2}$$

We see that $S = 0$ for $x = 0$. Now consider $S$ in the limit of $t \rightarrow +0$. We see that in general, except for the case $x = 0$, the result becomes:

$$S = 0.5*cot(x/2) \qquad x \neq 0$$

It is worth noting that for $t > 0$ there is not really a hyperbolic divergence at $x = 0$. The correct limit of $S$ when both $x$ and $t$ are small is $S = x/(x^2 + 4t^2)$. So actually $S$ is continuous! The result $S = 0$ for $x = 0$ is re-confirmed. Furthermore $S$ has a maximum $0.25/t$ fot $x= 2t$ and a minimum $-0.25/t$ for $x = -2t$. Only in the strict limit of $t$ to $0$ the interfacial region vanishes.

Question 1: Suppose we had defined $a(n)$ with a different convergence factor. So instead of the exponential factor $exp(-nt)$ we had used e.g. $(1 + nt)exp(-nt)$ or $1/(exp(nt)-nt)$ or a Gaussian. Would this lead to the same result for $S$ in the limit $t \rightarrow +0$, or to a different one?

Question 2: Under which conditions is it mathematically allowed to extend the result for $S$ in the limit $t \rightarrow +0$ to the case $t = 0$, where the sequence $a(n)$ becomes $sin(n x)$ which is no longer convergent?

EDIT: I now understand that the term "convergence factor" is rarely used in mathematics, and that the preferred terminology is "tempered distribution". I have been informed by Strants that the summation method used above is known as "Abel summation".

Answer 1

EDIT: I have learned the definition of convergence factor I assumed below is not the definition M. Wind intended.

Question 1

For question 1, the answer is that (at least for some not unreasonable choices of $x$) we can define a convergence factor such that that $\lim_{t\to 0} S \not= \frac{1}{2}\cot\left(\frac{x}{2}\right)$. To see this, let us consider the following question:

Question (1'): Does there exist a function $g:\mathbb{N} \times \mathbb{R}^{\ge 0}$ such that $g(n,0) = 0$ for all $n$, $\sum \sin(nx)g(n,t)$ converges for all $t > 0$ and $$\lim_{t \to 0} \sum_{n=0}^\infty \sin(nx) g(n,t) \not= 0.$$

The answer to this question is related to the answer to your question 1: if such a $g$ exists, then we can take the new convergence factor $\exp(-nt) + g(n,t)$ and get a new limit $\lim_{t \to 0} S$; alternatively, if the answer question 1' is no, then for any convergence factor $f(n,t)$, we have that for $g(n,t) = \exp(-nt) - f(n,t)$, $$\lim_{t \to 0} \sum_{n=0}^\infty \sin(nx) g(n,t) = 0$$ so $$\lim_{t \to 0} \sum_{n=0}^\infty \sin(nx) \exp(-nt) = \lim_{t \to 0} \sum_{n=0}^\infty \sin(nx) f(n,t).$$

I claim such a $g$ exists. Specifically, define $g$ by

$$g(n,t) = \left\{\begin{array}{cc} 1 & t \not= 0 \mbox{ and }n \ge \frac{1}{t} \mbox{ is the least positive integer such that } \sin(nx) \in \left[\frac{1}{2} - t, \frac{1}{2} + t\right]\\ 0 & \mbox{else}\end{array}\right.$$

For $\frac{x}{2\pi}$ irrational, $g$ is well-defined, since $\left\{\sin(nx)| n \in \mathbb{N}\right\}$ is dense in $[-1,1]$. (See here)

Then, $$\lim_{t \to 0} \sum_{n=0}^\infty \sin(nx)g(n,t) = \frac{1}{2}.$$ Question 2

As a thought on question 2, a series is defined to be Abel summable if $$\lim_{x \to 1^-} \sum_{n=0}^\infty a_nx^n$$ exists and is finite. If we set $x = e^{-t}$, the condition $x \to 1^-$ becomes $t \to 0^+$, so we are left with $$\lim_{t \to 0^+}\sum_{n=0}^{\infty}a_ne^{-nt},$$ which, if we let $a_n = \sin(nx)$, is exactly what you have. So, you can say the series is abel summable to sum $\frac{1}{2}\cot\left(\frac{x}{2}\right)$. In fact, this result is mentioned in the last exercise of this document.

Answer 2

I performed numerical tests on several convergence factors $g(n, t)$. Note that the parameter $t$ only appears in the product with $n$; therefore it is convenient to introduce $y = nt$. For the convergence factor $g(y)$ I set the following 4 criteria : [1] $g(0) = 1$; [2] $g(\infty) = 0$ ; [3] g is monotonically decreasing; [4] g is smooth.

I selected a broad range of candidate functions $g(y)$ and performed the summation in double precision for both the sine and the cosine function (for which the exact result is $S = 1/2$) for different values of $t$. In every case convergence to the exact value was seen.

The functions $g$ that led the sums to convergence with the greatest speed and with the highest accuracy, even for fairly large values of $t$ (0.001 for $x < 0.3$ and 0.002 otherwise), were of the following type. For small values of $y$ it behaves like $g(y) = 1 - y^N$ , with $N = 8$. Two excellent choices were found to be $g(y) = 1/(1 + y^8 + 0.25y^{16})$ and $g(y) = exp(-y^8)$.

Summarizing, I can now state with confidence that the answer to Question 1 is:

For every well-behaved convergence factor $g$ the sum $S$ converges in the limit of $t$ to $+0$ to the exact result, that is obtained for the exponential convergence factor $g(n, t) = exp(-nt)$.