How to sum an infinite convergent series that has a term from the end

[tworitdash]

From my physical problem, I ended up having a sum that looks like the following.

$$S_N(\omega) = \sum_{q = 1}^{N-1} \left(1 - \frac{q}{N}\right) \exp{\left(-\frac{q^2\sigma^2}{2}\right)} \cos{\left(\left(\mu - \omega\right)q\right)}$$

I want to know what is the sum when $N \to \infty$. Here, $\omega$ is where this is computed and $\mu$ and $\sigma$ are constants. Can this be reduced to an expression (a function of variables $\omega$, $\mu$ and $\sigma$) ?

I proceeded with trying to show that it is indeed convergent. $$S_N(\omega) - S_{N - 1}(\omega) = (1 - \frac{N-1}{N}) \exp{\left(-\frac{(N-1)^2\sigma^2}{2}\right)} \cos{\left(\left(\mu - \omega\right)(N-1)\right)} + \sum_{q = 1}^{N-2} q(\frac{1}{N-1} - \frac{1}{N}) \exp{\left(-\frac{q^2\sigma^2}{2}\right)} \cos{\left(\left(\mu - \omega\right)q\right)}$$

This difference goes to $0$ when $N \to \infty$.

 

[Office_Shredder]

The definition of divergent (as you use in your thread title) is that as N goes to infinity, there is no value it converges to (which necessarily means no expression either).

And this looks pretty plausibly divergent to me inspecting the term $p=N-1$, so what do you want?

 

[tworitdash]

The definition of divergent (as you use in your thread title) is that as N goes to infinity, there is no value it converges to (which necessarily means no expression either).

And this looks pretty plausibly divergent to me inspecting the term $p=N-1$, so what do you want?

I have edited the question. I realized that I can have a better sum that is convergent by averaging with $N$.

 

[Mark44]

What does "has a term from the end" mean? Is there a word missing here?

 

[WWGD]

If you're trying to argue that the sequence is Cauchy, you need to work with $S_n -S_{n-k}$ (more accurately maybe, $S_m -S_j$ for generic m,j)and not just consecutive terms $S_n -S_{n-1}$. The Harmonic series are a standard (counter) example.

 

[mathman]

Assume $\sigma^2 \gt 0$, then $e^{\sigma^2/2}=x \gt 1$. Therefore series is bounded by $\sum_1^\infty x^{-q^2}$. Looks convergent.

 

[WWGD]

Just to clarify: what type of object is your $\omega$? A number, vector, etc? And, just to make sure, your rightmost cos expression is not multiplying part/of the exponent, right?

 

[tworitdash]

What does "has a term from the end" mean? Is there a word missing here?

I couldn't express myself clearly. The sum has the term $N$. That is the total number of terms in the series. I just wanted to convey that a term with $N$ is present in the series.

 

[tworitdash]

Assume $\sigma^2 \gt 0$, then $e^{\sigma^2/2}=x \gt 1$. Therefore series is bounded by $\sum_1^\infty x^{-q^2}$. Looks convergent.

Thank you for the response. I also think so. If I see the second term of the equation, that goes to $0$ when $N \to \infty$ because of $1/N$ . Therefore the series converges to the first term.

$$\lim_{N \to \infty} S(\omega) \approx \sum_{q = 1}^{N - 1} \exp(-q^2\sigma^2/2) \cos(q (\mu - \omega)) \leq \sum_{q = 1}^{N - 1} \exp(-q^2\sigma^2/2)$$

So, if I say that I want a tolerance of $\epsilon$, what should be my $N$.

 

[tworitdash]

Just to clarify: what type of object is your $\omega$? A number, vector, etc? And, just to make sure, your rightmost cos expression is not multiplying part/of the exponent, right?

The $\omega$ are observation points so they make a vector. And yes, the cosine term is outside of the exponential. It is a different term, not on the exponent.

 

[Mark44]

What does "has a term from the end" mean?

I couldn't express myself clearly. The sum has the term N.

That still doesn't make sense.

That is the total number of terms in the series. I just wanted to convey that a term with N is present in the series.

This makes more sense. Writing "has a term from the end" was just confusing. If all you meant that the series has N terms, that is clear when you write this:
$$\sum_{i = 1}^N \text{whatever}$$
The series you wrote has N - 1 terms.
$$\sum_{i = 1}^{N-1} \text{whatever}$$

 

[jgill]

Since each term involves the number of terms in your (partial) series I suggest you take a look at Tannery's theorem. Once the N is out of the way, what remains is clearly convergent. What it converges to, however, is another matter. Probably not a closed form.

 

[tworitdash]

That still doesn't make sense.

This makes more sense. Writing "has a term from the end" was just confusing. If all you meant that the series has N terms, that is clear when you write this:
$$\sum_{i = 1}^N \text{whatever}$$
The series you wrote has N - 1 terms.
$$\sum_{i = 1}^{N-1} \text{whatever}$$

Thank you. Now, I see the correct way of expressing this.

 

[tworitdash]

Since each term involves the number of terms in your (partial) series I suggest you take a look at Tannery's theorem. Once the N is out of the way, what remains is clearly convergent. What it converges to, however, is another matter. Probably not a closed form.

Can I still find a $N$ for which this sum converges with an error for example $\epsilon$ ?

I have tried it without the cosine term in the series. It is like the following. To find the $N$, we should show that

$$\sum_{q = N}^{\infty} \exp({\left( -\frac{\sigma^2 q^2}{2} \right)}) < \epsilon$$

Using the inequality $q(q-1) \geq N(N-1)$ for $q \geq N$, we have,

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

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

Using geometric series, we have,

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

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

$$N > \sqrt{ \frac{2}{\sigma^2} \left(-\log({\epsilon}) - \log({ 1 - \exp{(-\sigma^2/2)} }) \right) }$$

So,

$$N = \lceil { \sqrt{ \frac{2}{\sigma^2} (-\log({\epsilon}) - \log({ 1 - \exp{(-\sigma^2/2)} }) ) } } \rceil$$

However, I am not able to do it with the cosine term in it. I will try to show what I have tried so far Let's take $\mu - \omega = a$ and using series expansion of $\cos(aq)$ , we have,

$$\sum_{q = N}^{\infty} \exp({-( \sigma^2/2 q^2 )}) \sum_{p = 0}^{\infty} (-1)^{p} \frac{(aq)^{2 p}}{(2 p)!}$$

$$\sum_{p = 0}^{\infty} \frac{(-1)^{p}a^{2p}}{(2 p)!} \sum_{q = N}^{\infty} \exp({-( \sigma^2/2 q^2 )}) q^{2 p}$$

Using $v = q^2$, we have,

$$\sum_{p = 0}^{\infty} \frac{(-1)^{p}a^{2p}}{(2 p)!} \sum_{v = N^2}^{\infty} \exp({-( \sigma^2 v/2 )}) v^{p-1} v$$

$$\sum_{p = 0}^{\infty} \frac{(-1)^{p}a^{2p}}{(2 p)!} \int_{N^2}^{\infty} \exp({-( \sigma^2 v/2 )}) v^{p-1} dv$$

$$\sum_{p = 0}^{\infty} \frac{(-1)^{p}a^{2p}}{(2 p)!} ( \int_{0}^{\infty} \exp({-( \sigma^2 v/2 )}) v^{p-1} dv - \int_{0}^{N^2-1} \exp({-( \sigma^2 v/2 )}) v^{p-1} dv )$$ $$\sum_{p = 0}^{\infty} \frac{(-1)^{p}a^{2p}}{(2 p)!} ( 2^p (\sigma^2/2)^{-p} \Gamma(p) - \int_{0}^{N^2-1} \exp({-( \sigma^2v/2 )}) v^{p-1} dv )$$

I can't reduce further.

 

[pasmith]

If $\sum_{n=N}^\infty a_n < \epsilon$ and each $a_n > 0$ then $$\begin{split} \left|\sum_{n=N}^\infty a_n\cos(k(n-n_0))\right| &\leq \sum_{n=N}^\infty a_n \left|\cos(k(n-n_0)) \right| \\ &\leq \sum_{n=N}^\infty a_n \\ & < \epsilon. \end{split}$$ So your bound on $\sum_{n=N}^\infty a_n$ will work.

 

[tworitdash]

If $\sum_{n=N}^\infty a_n < \epsilon$ and each $a_n > 0$ then $$\begin{split} \left|\sum_{n=N}^\infty a_n\cos(k(n-n_0))\right| &\leq \sum_{n=N}^\infty a_n \left|\cos(k(n-n_0)) \right| \\ &\leq \sum_{n=N}^\infty a_n \\ & < \epsilon. \end{split}$$ So your bound on $\sum_{n=N}^\infty a_n$ will work.

Ah yes! I completely overlooked this. Thank you. I can indeed just use the term without the cosine.