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=0cosn\displaystyle S=\sum_{n=0}^{\infty} \cos n

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

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

=11ei\displaystyle = \Re \frac{1}{1-e^i}

=11ei+11ei2\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??

Note by Hasan Kassim
4 years, 2 months ago

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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 - 4 years, 2 months ago

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Now look at this:

n=1cosnn\displaystyle \sum_{n=1}^{\infty} \frac{\cos n}{n}

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

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

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

=12ln(22cos1)\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=1xnn \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=1cosnn\sum_{n=1}^{\infty} \frac{\cos n}{n} .

Hasan Kassim - 4 years, 2 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 ln2 \ln 2 .

(I believe that) we get conditional convergence if we substitute x=1,x1 |x| = 1, x \neq 1 . In particular, x=ei x = e^i is valid.

Calvin Lin Staff - 4 years, 2 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 - 4 years, 2 months ago

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