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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
2 years, 1 month 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. Sudeep Salgia · 2 years, 1 month ago

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@Sudeep Salgia 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 · 2 years, 1 month ago

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@Hasan Kassim 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 · 2 years, 1 month 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 · 2 years, 1 month ago

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