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# Divergence of Infinite Series of Periodic Functions

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

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

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

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

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

$\displaystyle = \frac{\frac{1}{1-e^i} +\frac{1}{1-e^{-i}} }{2}$

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

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
1 year, 7 months ago

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This is because the geometric series will converge for $$|x| <1$$. The fact that you have assumed convergence and applied the formula is causing the anomaly. · 1 year, 7 months ago

Now look at this:

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

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

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

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

$\displaystyle = -\frac{1}{2} \ln (2-2\cos 1)$

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

I mean if we say that $$|e^i| = 1$$, this will contradict the convergence of $$\sum_{n=1}^{\infty} \frac{\cos n}{n}$$. · 1 year, 7 months ago

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

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

(I believe that) we get conditional convergence if we substitute $$|x| = 1, x \neq 1$$. In particular, $$x = e^i$$ is valid. Staff · 1 year, 7 months ago

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? · 1 year, 7 months ago