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.

[tex] 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)} [/tex]

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

I proceeded with trying to show that it is indeed convergent. [tex] 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)}[/tex]

This difference goes to [itex]0[/itex] when [itex]N \to \infty[/itex].

 

[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 [itex] N [/itex].

 

[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 [itex] N [/itex]. That is the total number of terms in the series. I just wanted to convey that a term with [itex] N [/itex] 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 [itex] 0 [/itex] when [itex] N \to \infty [/itex] because of [itex] 1/N [/itex] . Therefore the series converges to the first term.

[tex] \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) [/tex]

So, if I say that I want a tolerance of [itex] \epsilon [/itex], what should be my [itex] N [/itex].

 

[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 [itex] \omega [/itex] 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 [itex] N [/itex] for which this sum converges with an error for example [itex] \epsilon [/itex] ?

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

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

Using the inequality [itex] q(q-1) \geq N(N-1) [/itex] for [itex] q \geq N [/itex], we have,

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

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

Using geometric series, we have,

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

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

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

So,

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

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 [itex] \mu - \omega = a [/itex] and using series expansion of [itex] \cos(aq) [/itex] , we have,

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

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

Using [itex] v = q^2 [/itex], we have,

[tex] \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 [/tex]

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

[tex] \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 )[/tex] [tex] \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 )[/tex]

I can't reduce further.

 

[pasmith]

If [itex]\sum_{n=N}^\infty a_n < \epsilon[/itex] and each [itex]a_n > 0[/itex] then [tex]\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}[/tex] So your bound on [itex]\sum_{n=N}^\infty a_n[/itex] will work.

 

[tworitdash]

If [itex]\sum_{n=N}^\infty a_n < \epsilon[/itex] and each [itex]a_n > 0[/itex] then [tex]\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}[/tex] So your bound on [itex]\sum_{n=N}^\infty a_n[/itex] will work.

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