Divergence of Infinite Series of Periodic Functions

We all know that sums of periodic function are divergent. Take a look at this series:

S=∑n=0∞cos⁡n\displaystyle S=\sum_{n=0}^{\infty} \cos n

It is divergent yes, but look at the following manipulations:

=ℜ∑n=0∞ein\displaystyle = \Re \sum_{n=0}^{\infty} e^{in}

=ℜ11−ei\displaystyle = \Re \frac{1}{1-e^i}

=11−ei+11−e−i2\displaystyle = \frac{\frac{1}{1-e^i} +\frac{1}{1-e^{-i}} }{2}

=12\displaystyle = \frac{1}{2}

Shouldn't there be a mathematical error in some of the steps?? Analytically, the series should diverge since the summand is periodic and bounded, but what about the calculations??

#Calculus #brillianthelp #InfiniteSeries #Periodicfunctions

Note by Hasan Kassim
5 years, 10 months ago

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Comments

This is because the geometric series will converge for ∣x∣<1 |x| <1. The fact that you have assumed convergence and applied the formula is causing the anomaly.

Sudeep Salgia - 5 years, 10 months ago

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Now look at this:

∑n=1∞cos⁡nn\displaystyle \sum_{n=1}^{\infty} \frac{\cos n}{n}

=ℜ∑n=1∞einn\displaystyle = \Re \sum_{n=1}^{\infty} \frac{e^{in}}{n}

=−ℜln⁡(1−ei)\displaystyle = -\Re \ln (1-e^{i})

=−ln⁡(1−ei)+ln⁡(1−e−i)2\displaystyle = -\frac{\ln (1-e^{i}) +\ln (1-e^{-i}) }{2}

=−12ln⁡(2−2cos⁡1)\displaystyle = -\frac{1}{2} \ln (2-2\cos 1)

Now this series Converges to this value(you can check by wolfram alpha) .we know that the sum ∑n=1∞xnn \sum_{n=1}^{\infty} \frac{x^n}{n} converges iff ∣x∣<1 |x|<1 . But I used the same x=ei x = e^i used in the periodic sum. How can that be justified??

I mean if we say that ∣ei∣=1 |e^i| = 1, this will contradict the convergence of ∑n=1∞cos⁡nn\sum_{n=1}^{\infty} \frac{\cos n}{n} .

Hasan Kassim - 5 years, 10 months ago

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It is not true that ∑xnn \sum \frac { x^n } { n } converges iff ∣x∣<1 |x| < 1 .
The proper version of the statement is that ∑xnn \sum \frac { x^n } { n } converges absolutely iff ∣x∣<1 |x| < 1 .

For example, we know that ∑(−1)nn \sum \frac{ (-1)^n} { n} converges conditionally to ln⁡2 \ln 2 .

(I believe that) we get conditional convergence if we substitute ∣x∣=1,x≠1 |x| = 1, x \neq 1 . In particular, x=ei x = e^i is valid.

Calvin Lin Staff - 5 years, 10 months ago

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@Calvin Lin – Okay I got it.Thanks for the insight :)

Do you know a regularization to this series? "Assigning values to divergent series"?

And in General, for what reasons we assign values to divergent series? and doesn't that make any contradiction?

Hasan Kassim - 5 years, 10 months ago
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