Sum to n terms and also to infinity of the following series:cosθ+2cos2θ+⋯+ncosnθthe solution provided by the book is Sn=(n+1)cosnθ−ncos(n+1)θ−12(1−cosθ)Can anyone help me to explain how to get Sn.
Thanks in advance.
Answer
To find closed form for series of cos and sin functions, approach using complex numbers helps very much. Consider, z=eiθ, where, i=√−1. We need to construct a function whose real part or imaginary part(otherwise we will have i in the sum, which we don't have in the sum we are interested to find out) will give the required sum. Now, if we take f(z)=1+z+z2+z3+⋯+zn,
z⋅f′(z)=z+2z2+3z3+⋯+nzn=cosθ+2cos2θ+⋯+cosnθ+in∑k=1k⋅sin(kθ)
So, notice that ℜ{z⋅f′(z)} is the required sum. On the other hand, f(z)=zn+1−1z−1 gives, \require{cancel}z\cdot f'(z)=z\cdot \frac{(z-1)\cdot (n+1)z^n-z^{n+1}+1}{(z-1)^2}=z\cdot \frac{(n+\cancel{1})z^{n+1}-(n+1)z^n-\cancel{z^{n+1}}+1}{(z-1)^2}\\= z^{n+1}\cdot \frac{n(z-1)-1}{(z-1)^2}
Now, \sum_{k=1}^{n}k\cdot \cos{(k\theta)}=\Re{\{z\cdot f'(z)\}}=\Re{\{z^{n+1}\cdot \frac{n(z-1)-1}{(z-1)^2}\}}. After some calculation you will get the result.
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