Interesting Sum of Unity

$\text{Method 1: Polynomial Transformation}$

By the method of transformation, $P(1-\frac{1}{z})$ has roots $\frac{1}{1-\alpha}, \frac{1}{1-\alpha^2}, \frac{1}{1-\alpha^3}, \cdots, \frac{1}{1-\alpha^n}$ .

$P(1-\frac{1}{z})= \left(1-\frac{1}{z} \right)^{n+1} -1 = 0$ $\Rightarrow (z-1)^{n+1} -z^{n+1} =0$

Expanding using the binomial theorem, the first three terms are:

$-z^{n+1} + z^{n+1} -(n+1)z^n + \frac{(n+1)n}{2}z^{n-1} +\cdots$

So the equation with the desired roots is:

$-(n+1)z^n+\frac{(n+1)n}{2}z^{n-1} +\cdots =0$

Using Vieta's Formulae, we have:

$\frac{2}{1-\alpha}+\frac{2}{1-\alpha^2}+\frac{2}{1-\alpha^3}+\cdots+\frac{2}{1-\alpha^n} = -2\frac{\frac{(n+1)n}{2}}{-(n+1)} = n$

$\text{Method 2: Careful Summation}$

Consider the following result: $\frac{2}{1-\beta}+\frac{2}{1-\frac{1}{\beta}}=2$

We also know $\alpha^k\alpha^{n+1-k} = 1$

Summing up the terms pairwise from end to end, we have:

$\frac{2}{1-\alpha}+\frac{2}{1-\alpha^n}=2,\frac{2}{1-\alpha^2}+\frac{2}{1-\alpha^{n-1}}=2,\cdots$

If $n$ is even, then all the terms pair perfectly and we are done. The resultant sum is equal to $n$

If $n$ is odd, there is the root $z=-1$ , whose transformation is just $1$ and we are done. The resultant sum is also equal to $n$ .

$\text{Method 3: Calculus}$

Consider the natural logarithm of $P(z)$ . By the fundamental theorem of algebra, we have:

$\log{(z^{n+1}-1)}=\log{(z-1)}+\log{(z-\alpha)}+\cdots+\log{(z-\alpha^n)}$

Differentiate both sides with respect to $z$ , giving us:

$\frac{(n+1)z^n}{z^{n+1}-1}=\frac{1}{z-1}+\frac{1}{z-\alpha}+\frac{1}{z-\alpha^2}+\cdots+\frac{1}{z-\alpha^n}$

Subtract from both sides to place the common singularity of $z=1$ on the left hand side.

$\frac{(n+1)z^n}{z^{n+1}-1}-\frac{1}{z-1}=\frac{1}{z-\alpha}+\frac{1}{z-\alpha^2}+\cdots+\frac{1}{z-\alpha^n}$

$\frac{nz^n - (z^{n-1}+z^{n-2}+\cdots+z+1)}{z^{n+1}-1} =\frac{1}{z-\alpha}+ \frac{1}{z-\alpha^2}+\cdots+\frac{1}{z-\alpha^n}$

Take the limit as $z \to 1$ , allowing us to invoke L'Hôpital's rule. The limit can be directly applied to the sum without any problem.

$\frac{2}{1-\alpha}+\frac{2}{1-\alpha^2}+\frac{2}{1-\alpha^3}+\cdots+\frac{2}{1-\alpha^n} = 2\lim_{z \to 1} \frac{nz^n - (z^{n-1}+z^{n-2}+\cdots+z+1)}{z^{n+1}-1}$

$\frac{2}{1-\alpha}+\frac{2}{1-\alpha^2}+\frac{2}{1-\alpha^3}+\cdots+\frac{2}{1-\alpha^n} = 2\lim_{z \to 1} \frac{n^2 z^{n-1} - ((n-1)z^{n-2}+(n-2)z^{n-3}+\cdots+2z+1)}{(n+1)z^n}$

$\frac{2}{1-\alpha}+\frac{2}{1-\alpha^2}+\frac{2}{1-\alpha^3}+\cdots+\frac{2}{1-\alpha^n} = 2\frac{n^2 - ((n-1)+(n-2)+\cdots+2+1)}{n+1}$

$\frac{2}{1-\alpha}+\frac{2}{1-\alpha^2}+\frac{2}{1-\alpha^3}+\cdots+\frac{2}{1-\alpha^n} = 2\frac{n^2 - \frac{(n-1)n}{2}}{n+1}=\frac{n^2+n}{n+1}=n$

Let's do it like little Gauss, taking the sum twice and dividing by two: $\sum_{k=1}^{n}(\frac{1}{1-\alpha^k}+\frac{1}{1-\alpha^{n+1-k}})$ $=\sum_{k=1}^{n}\frac{1-\alpha^{n+1-k}+1-\alpha^k}{1-\alpha^k-\alpha^{n+1-k}+1}=\sum_{k=1}^{n}1=n$ . All we need is $\alpha^k \alpha^{n+1-k}=\alpha^{n+1}=1$ .

This is just a refinement of Jack's "Method 2", of course, without having to deal with the even/odd cases.

@Jack Lam shouldn't in method 2 it be if n is odd then the terms perfectly match

Did it the same way as @Jack Lam (Method 1).....But for JEE Style, simply put n=1 and solve!!!