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The series $\sum_{n=1}^\infty \frac{\cos \ n}{n}$ convergences. Mathematica gives as limit $$\sum_{n=1}^\infty \frac{\cos \ n}{n}=-\frac{\ln\left(2-2 \cos(1)\right)}{2} $$ What are the proofs of this limit?

David Lingard
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Since $\sum_{n \ge 1} \frac{z^n}{n}=-\log{(1-z)}, |z| < 1$, and the LHS converges for all $|z|=1, z \ne 1$, it follows by basic Abelian theory that the sum is $-\log(1-z)$ there too.

In particular if $z=e^{i}$ we get $\sum_{n \ge 1} \frac{\cos n + i \sin n}{n}=-\log(1-e^{i})$. Taking real parts and noting that $|1-e^i|^2=2-2 \cos 1$ we are done!

(Edit later: if one is not familiar with Tauberian theory and what are called Abelian vs Tauberian results in that context, the fundamental easy result due to Abel is that if $\sum a_nz^n=f(z), |z| <1$ and for some $|z_0|=1, \sum a_nz_0^n$ converges to a finite limit, then $f(z) \to \sum a_nz_0^n$ when $z \to z_0$ radially say. The converse is not true (if $f$ converges at $z_0$, the series definitely does not need to converge there as we can easily see with the geometric series) and the study of the conditions under which it holds is called Tauberian theory

Conrad
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    What do you mean by "basic Abelian theory"? – YiFan Tey Sep 24 '19 at 23:03
  • if $\sum a_nz^n=f(z), |z| <1$ and then for some $|z_0|=1$, $\sum a_nz_0^n$ converges to a finite limit, then $f(z) \to \sum a_nz_0^n$ when $z \to z_0$ radially say (Abel's theorem); the converse is not true and the study of the conditions under which it holds is called Tauberian theory – Conrad Sep 24 '19 at 23:07
  • Thanks for the response, I learned something new. :) – YiFan Tey Sep 24 '19 at 23:13
  • @YiFan great comment so added the explanation in the answer – Conrad Sep 24 '19 at 23:17