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??

#Calculus #brillianthelp #InfiniteSeries #Periodicfunctions

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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.

Sudeep Salgia - 4 years, 9 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 $\sum_{n=1}^{\infty} \frac{\cos n}{n}$ .

Hasan Kassim - 4 years, 9 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.

Calvin Lin Staff - 4 years, 9 months ago

@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, 9 months ago

@Calvin Lin @Michael Mendrin @Brian Charlesworth @Aman Raimann @Krishna Sharma @Ishan Dasgupta Samarendra

Any Thoughts??

Hasan Kassim - 4 years, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$