<s>[INST] Cesàro sum of $\sum\limits_{n = 0}^\infty {\cos n}=\dfrac{1}{2}$. Please check my work Thanks to the formula 
http://functions.wolfram.com/ElementaryFunctions/Cos/23/01/0001/
Partial sums
$$s_m=\sum\limits_{k = 0}^m {\cos k = } \frac{{\sin \left( {\frac{1}{2}\left( {m + 1} \right)} \right)\cos \frac{m}{2}}}{{\sin \frac{1}{2}}}+1$$
some rearrangement
$$\frac{{\frac{1}{2}\left( {\sin \left( {\frac{m}{2} + \frac{1}{2} + \frac{m}{2}} \right) + \sin \left( {\frac{m}{2} + \frac{1}{2} - \frac{m}{2}} \right)} \right)}}{{\sin \frac{1}{2}}} = \frac{{\frac{1}{2}\left( {\sin \left( {m + \frac{1}{2}} \right) + \sin \frac{1}{2}} \right)}}{{\sin \frac{1}{2}}}$$
and we get
$$s_m=\sum\limits_{k = 0}^m {\cos k = } \frac{{\sin \left( {m + \frac{1}{2}} \right)}}{{2\sin \frac{1}{2}}} + \frac{1}{2}$$
Let's now define
$${a_n} = \sum\limits_{m = 0}^n {\left[ {\frac{{\sin \left( {m + \frac{1}{2}} \right)}}{{2\sin \frac{1}{2}}} + \frac{1}{2}} \right]}$$
which is
$${a_n} = \frac{1}{{2\sin \frac{1}{2}}}\sum\limits_{m = 0}^n {\sin \left( {m + \frac{1}{2}} \right)}  + \sum\limits_{m = 0}^n {\frac{1}{2}} $$
using the formula that I found here
http://functions.wolfram.com/ElementaryFunctions/Sin/23/01/0003/
$$\sum\limits_{m = 0}^n {\sin \left( {m + \frac{1}{2}} \right)}  = \frac{{{{\sin }^2}\frac{{n + 1}}{2}}}{{\sin \frac{1}{2}}} = \frac{{1 - \cos \left( {n + 1} \right)}}{{2\sin \frac{1}{2}}}$$
I got
$${a_n} = \frac{{1 - \cos \left( {n + 1} \right)}}{{{{\left( {2\sin \frac{1}{2}} \right)}^2}}} + \frac{{\left( {n + 1} \right)}}{2} = \frac{{1 - \cos \left( {n + 1} \right)}}{{2\left( {1 - \cos 1} \right)}} + \frac{{\left( {n + 1} \right)}}{2}$$
and finally
$${a_n} = \frac{{\cos \left( {n + 1} \right) + \left( {n + 1} \right)\cos 1 - n - 2}}{{2\left( {\cos 1 - 1} \right)}}$$
Cesàro sum is the following limit
$$
  \mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{n} = \mathop {\lim }\limits_{n \to \infty } \frac{{\cos \left( {n + 1} \right) + n\left( {\cos 1 - 1} \right) + \cos 1 - 2}}{{2\left( {\cos 1 - 1} \right)}} $$
split into the sum of two limits
$$\mathop {\lim }\limits_{n \to \infty } \frac{{\cos \left( {n + 1} \right) + \cos 1 - 2}}{{2n\left( {\cos 1 - 1} \right)}} + \mathop {\lim }\limits_{n \to \infty } \frac{{n\left( {\cos 1 - 1} \right)}}{{2n\left( {\cos 1 - 1} \right)}} $$
the first is $0$ because numerator is limited and denominator goes to infinity, the second is $\dfrac{1}{2}$
therefore Cesàro sum is
$$\sum\limits_{n = 0}^\infty  {\cos n = } \frac{1}{2}$$
Thank you in advance for your attention
Any comment would be greatly appreciated
 [/INST] I haven't check your work, but it coincides with other results. Particularly, it is what you get when you do Ramanujan's summation of the series:
Let $f(x)=\cos x$. Then the various expressions for Ramanujan's summation are:
$$\sum _{x\ge0}f(x)=\sum _{n=1}^{\infty }{\frac {f^{(n-1)}(0)}{n!}}B_{n}(1)$$
$$\sum _{x\ge0}f(x)=-\sum _{k=1}^{\infty }{\frac {\Delta ^{k-1}f(x)}{k!}}(-1)_{k}$$
$$\sum _{x\ge0}f(x)=\int _{0}^{x}f(t)dt-{\frac {1}{2}}f(x)+\sum _{k=1}^{\infty }{\frac {B_{2k}}{(2k)!}}f^{(2k-1)}(1)$$
$$\sum _{x\ge0}f(x)=\int _{0}^{x}f(t)dt+\sum _{k=1}^{\infty }{\frac {c_{k}\Delta ^{k-1}f(1)}{k!}}$$
(where $c_{k}=\int _{0}^{1}{\frac {\Gamma (x+1)}{\Gamma (x-k+1)}}dx$)
In all these cases you also get 1/2.
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