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

AI Thread Summary
The discussion focuses on evaluating the infinite convergent series defined by S_N(ω) as N approaches infinity, specifically addressing the convergence of the series involving exponential and cosine terms. Participants clarify the meaning of "having a term from the end," leading to a better understanding that the series contains N terms. It is established that the series converges, particularly as the second term approaches zero when N increases, allowing the first term to dominate. The conversation also explores how to determine an appropriate N for a given tolerance ε, with suggestions for bounding the series. Ultimately, the conclusion is that the series can indeed be analyzed effectively without the cosine term for convergence purposes.
tworitdash
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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.
 
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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?
 
Office_Shredder said:
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.
 
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What does "has a term from the end" mean? Is there a word missing here?
 
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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.
 
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.
 
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?
 
Mark44 said:
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.
 
mathman said:
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.
 
  • #10
WWGD said:
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.
 
  • #11
Mark44 said:
What does "has a term from the end" mean?
tworitdash said:
I couldn't express myself clearly. The sum has the term N.
That still doesn't make sense.
tworitdash said:
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.
$$\sum_{i = 1}^{N-1} \text{whatever}$$
 
  • #12
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.
 
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  • #13
Mark44 said:
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.
 
  • #14
jgill said:
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.
 
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  • #15
If \sum_{n=N}^\infty a_n &lt; \epsilon and each a_n &gt; 0 then \begin{split}<br /> \left|\sum_{n=N}^\infty a_n\cos(k(n-n_0))\right| &amp;\leq \sum_{n=N}^\infty a_n \left|\cos(k(n-n_0)) \right| \\<br /> &amp;\leq \sum_{n=N}^\infty a_n \\<br /> &amp; &lt; \epsilon.<br /> \end{split} So your bound on \sum_{n=N}^\infty a_n will work.
 
  • #16
pasmith said:
If \sum_{n=N}^\infty a_n &lt; \epsilon and each a_n &gt; 0 then \begin{split}<br /> \left|\sum_{n=N}^\infty a_n\cos(k(n-n_0))\right| &amp;\leq \sum_{n=N}^\infty a_n \left|\cos(k(n-n_0)) \right| \\<br /> &amp;\leq \sum_{n=N}^\infty a_n \\<br /> &amp; &lt; \epsilon.<br /> \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.
 
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