Convergence as a function of error for the following function

Question

This question is about finding the number of samples in a sequence required for the convergence of a series as a function of an error tolerance $\epsilon$. I want to show what I have tried so far.

The function is

$$ \sum_{q=1}^{N-1} \exp(-q^2 \sigma^2/2)(1 - q/N) $$ for $\sigma > 0$ and $N \geq 1$

It is confirmed that this function converges with a large N. I want to find a function for $N$ at which the error is $\epsilon$.

So, the sum of this expression from $N$ to $\infty$ is less than $\epsilon$. The sum of this from $N$ to $\infty$ can be represented as,

$$ \sum_{q=1}^{\infty} \exp(-q^2\sigma^2/2) - \sum_{q=1}^{N-1} \exp(-q^2\sigma^2/2) (1-q/N) < \epsilon $$

For the first term, I tried something like this. If $q \geq 1$, $q(q-1) \geq 0$.

So,

$$ \sum_{q=1}^{\infty} \exp(-q(q-1)\sigma^2/2) \exp(-q \sigma^2/2) \leq \sum_{q=1}^{\infty} \exp(-q \sigma^2/2) = \frac{\exp(-\sigma^2/2)}{(1 - \exp(-\sigma^2/2))} $$

For the second term,

$$ \sum_{q=1}^{N-1} \exp(-q^2\sigma^2/2) (1-q/N) $$

I used the same trick as above, but now I say that as $q \leq N-1, \quad q(q-1) \leq (N-1)(N-2) $

For the second term, as there is a negative in equation (2), we will test for greater than equal to conditional properties of that function.

$$ \sum_{q=1}^{N-1} \exp(-q^2\sigma^2/2) (1-q/N) =\sum_{q=1}^{N-1} \exp(-q(q-1)\sigma^2/2) \exp(-q \sigma^2/2) (1-q/N) $$

$$ \sum_{q=1}^{N-1} \exp(-q^2\sigma^2/2) (1-q/N) \geq \exp(-(N-1)(N-2)\sigma^2/2) \sum_{q=1}^{N-1} (1-q/N) \exp(-q \sigma^2/2) $$

I typed this sum on wolfram, and the final expression I have is this,

$$ \sum_{q=1}^{N-1} \exp(-q^2\sigma^2/2) (1-q/N) \geq \exp(-(N-1)(N-2)\sigma^2/2) \frac{\exp(-\sigma^2/2) (N-1 + \exp(-N \sigma^2/2) - N \exp(-\sigma^2/2))}{((1 - \exp(-\sigma^2/2))^2 N}$$

Rearranging a bit,

$$ \sum_{q=1}^{N-1} \exp(-q^2\sigma^2/2) (1-q/N) \geq \exp(-(N^2-3N+3)\sigma^2/2) \frac{(N-1 + \exp(-N \sigma^2/2) - N \exp(-\sigma^2/2))}{((1 - \exp(-\sigma^2/2))^2 N}$$

If I plug this into the original inequality, things become complicated.

$$ \sum_{q=1}^{\infty} \exp(-q^2\sigma^2/2) - \sum_{q=1}^{N-1} \exp(-q^2\sigma^2/2) (1-q/N) \leq \frac{\exp(-\sigma^2/2)}{(1 - \exp(-\sigma^2/2))} - \exp(-(N^2-3N+3)\sigma^2/2) \frac{(N-1 + \exp(-N \sigma^2/2) - N \exp(-\sigma^2/2))}{((1 - \exp(-\sigma^2/2))^2 N} < \epsilon $$

I don't know how to proceed from here.

Answer 1

$\newcommand\ep\epsilon\newcommand{\si}{\sigma} $"I want to find a function for $N$ at which the error is $\epsilon$."

This question is stated very poorly.

Indeed, let $n:=N$ (there is no reason to use $N$ where $n$ will do.) The $n$th error is \begin{equation*} \ep_n:=s-s_n=\ep_{1n}+\ep_{2n}, \end{equation*} where \begin{equation*} s:=\sum_{q=1}^\infty a_q,\quad s_n:=\sum_{q=1}^{n-1} a_q(1-q/n), \quad a_q:=e^{-q^2\si^2/2}, \end{equation*} \begin{equation*} \ep_{1n}:=\sum_{q=n}^\infty a_q,\quad \ep_{2n}:=\frac1n\,\sum_{q=1}^{n-1} a_q q. \end{equation*}

Clearly, $\ep_n$ takes only countably many values; so, the equality $\ep_n=\ep$ can hold only for countably many values of $\ep$. Also, a closed-form expression for $\ep_n$ is not available. So, solutions of the equation $\ep_n=\ep$ for $n$ are not available in closed form, even when such solutions exist.

However, for any real $\ep>0$, we can provide an explicit lower bound $n_{\si,\ep}$ on $n$ such that $\ep_n\le\ep$ for $n\ge n_{\si,\ep}$ -- and this is what appears to have actually been tried to do in most of the OP.

Indeed, note that \begin{equation*} r_q:=\frac{a_{q+1}}{a_q}=e^{-(q+1/2)\si^2} \end{equation*} is decreasing in $q$. So, \begin{equation*} \ep_{1n}\le\sum_{q=n}^\infty a_n r_n^{q-n}=\frac{a_n}{1-r_n}=\frac{e^{-n^2\si^2/2}}{1-e^{-(n+1/2)\si^2}} \le2e^{-n^2\si^2/2} \end{equation*} if \begin{equation*} n\ge\frac{\ln2}{\si^2}-\frac12. \end{equation*} Next, \begin{equation*} \ep_{2n}\le\frac1n\,\sum_{q=1}^\infty a_q q \le\frac{h(\si)}n, \end{equation*} where \begin{equation} h(\si):=\sum_{q=1}^\infty a_1 r_1^{q-1} q =\frac{e^{5 \si ^2/2}}{(e^{3 \si^2/2}-1)^2}. \end{equation} Thus, if \begin{equation} n\ge n_{\si,\ep}:=\max\Big(\frac{\ln2}{\si^2}-\frac12,\sqrt{\frac2{\si^2}\, \max\Big(0,\ln\frac3\ep\Big)}, \frac{h(\si)}{\ep/3}\Big), \end{equation} then $\ep_n\le\ep$. $\quad\Box$