Cos I can

Easiest solution:

Use $\displaystyle \cos \theta = \frac{e^{i\theta} + e^{-i\theta}}{2}$ and $\displaystyle \sum_{n=1}^{\infty} \frac{x^{n}}{n}$ . You'll get

$-\frac{1}{2} \ln((1-e^{i})(1-e^{-i})) = \boxed{-\frac{1}{2} \ln(2-2\cos 1)}$

We will make use of complex numbers and Taylor expansion of logarithmic function to solve this question.

We know that $\cos x = Re(e^{\iota x})$ Using above fact we can write that $\dfrac{\cos n}{n} = Re\Big(\dfrac{e^{\iota n}}{n}\Big)$ $\sum_{n=1}^{\infty}\dfrac{\cos n}{n} = \sum_{n=1}^{\infty} Re\Big( \dfrac{e^{\iota n}}{n} \Big)$ Using the fact that summation is distributive over Real and Imaginary parts of a complex number, the Real part function and summation can be interchanged. $\sum_{n=1}^{\infty}\dfrac{\cos n}{n} = Re\Big( \sum_{n=1}^{\infty} \dfrac{e^{\iota n}}{n}\Big)$ The above series so formed can be correlated with Taylor expansion of logarithmic function. $\ln(1-x) =- \sum_{n=1}^{\infty} \dfrac{x^{n}}{n}$ Put $x = e^{\iota}$ $-\ln(1-e^{\iota}) = \sum_{n=1}^{\infty} \dfrac{e^{\iota n}}{n}$ $\sum_{n=1}^{\infty} \dfrac{e^{\iota n}}{n} = -\ln(1-\cos1-\iota \sin1)$ $=-\ln(2\sin^{2} \dfrac{1}{2} - 2\iota \sin\dfrac{1}{2} \cos\dfrac{1}{2})$ $= -\ln(2\sin\dfrac{1}{2}) - \ln(\sin\dfrac{1}{2} - \iota \cos\dfrac{1}{2})$ $= -\ln(2\sin\dfrac{1}{2}) + \ln\Big(\dfrac{1}{\sin\dfrac{1}{2} - \iota \cos\dfrac{1}{2}}\Big)$ $= -\ln(2\sin\dfrac{1}{2}) + \ln\Big( \dfrac{\iota}{e^{\frac{\iota}{2}}}\Big)$ Now using the fact that $\iota = e^{\frac{\iota \pi}{2}}$ . $= -\ln(2\sin\dfrac{1}{2}) + \ln e^{\iota \dfrac{\pi-1}{2}}$ $\sum_{n=1}^{\infty} \dfrac{e^{\iota n}}{n} = -\ln(2\sin\dfrac{1}{2}) + \iota \dfrac{\pi - 1}{2}$ $Re\Big( \sum_{n=1}^{\infty} \dfrac{e^{\iota n}}{n}\Big) = -\ln(2\sin\dfrac{1}{2})$ $\sum_{n=1}^{\infty}\dfrac{\cos n}{n} = \dfrac{-1}{2}\ln(4\sin^{2}\dfrac{1}{2})$ $\sum_{n=1}^{\infty}\dfrac{\cos n}{n} = \dfrac{-1}{2}\ln(2-2\cos1)$

$Note that (Taylor series)$

$\sum_{n=1}^∞\frac{x^{n}}{n}=-log(1-x)$

$Let x=e^{i} we get$

$\sum_{n=1}^∞ \frac{x^{n}}{n}$ $=\sum_{n=1}^∞ \frac{e^{in}}{n}$ $=\sum_{n=1}^∞ \frac{(e^{i})^{n}}{n}$

$=-log(1-e^{i})$

$but \frac{e^{in}}{n}=\frac{cos(n)}{n}+i \frac{sin(n)}{n}$ $So \sum_{n=1}^∞ \frac{cos(n)}{n}$ $=- real \ part(\sum_{n=1}^∞ \frac{e^{in}}{n}$ $=-real \ part(log(1-e^{i}))$ )

$the \ log \ of \ a \ complex \ number \ Re^{in} \ is \ log(R)+in$

$then \ real \ part \ is \ log(R)$

$Now \ to \ find \ R$

$1-e^{i}=1-cos(1)-isin(1)$

$R=((1-cos(1))^{2}+(sin(1))^{2})^{\frac{1}{2}}$

$R=(2-2cos(1))^{\frac{1}{2}}$

$Finally$

$\sum_{n=1}^∞ \frac{cos(n)}{n}=-log(R)=-\frac{1}{2} log(2-2cos(1))$

$a=2,b=2,c=1$

$a+b+c=\boxed{5}$

I'll show a different method to solve this(bit lengthier)

We know that

$\displaystyle \int sin(nx) dx = -\dfrac{cos(nx)}{n} + C$

Since we have to work with $\frac{cos(n)}{n}$ we have to set lower(or upper) limit as 1. Upper(or lower) is upto us, for simplicity I'll take it as $\frac{\pi}{2}$ . So we have

$\displaystyle \int_{1}^{\frac{\pi}{2}} sin(nx) dx = \dfrac{cos(n)}{n}$

Now summing up for different value of 'n' upto a value say 'N'

$\displaystyle \int_{1}^{\frac{\pi}{2}} \left( \sum_{n = 1}^{N} sin(nx) \right) dx = \sum_{n = 1}^{N} \dfrac{cos(n)}{n}$

We know that

$\displaystyle \sum_{n = 1}^{N} sin(nx) = \dfrac{ \sin(Nx/2).\sin((N+1)x/2)}{\sin(x/2)} = \dfrac{\cos(x/2) - \cos( (2N+1)x/2)}{2sin(x/2)}$

Since we have to deal with infinite summation we will take limit

$\displaystyle \lim_{N \to \infty } \sum_{n = 1}^{N} \dfrac{cos(n)}{n} = \lim_{N \to \infty } \int_{1}^{\frac{\pi}{2}} \left( \dfrac{cos(x/2)}{2sin(x/2)} - \dfrac{cos((2N+1)x/2)}{2sin(x/2)} \right) dx$

The first integral is easy just substitute $sin(x/2) = t$ , for the second integral as N goes on increasing the integral will tends to 0.

Ok...too much work done here after solving the integral and reduce it in given closed form to get

$\displaystyle \sum_{n = 1}^{\infty } \dfrac{cos(n)}{n} = \frac{-1}{2} ln( 2 - 2cos1)$ .

Phew....