DTFT of the unit step function If i apply the DTFT on unit step function, then i get follow:
$$DTFT\{u[n]\}=\sum_{n=-\infty}^{\infty}u[n]e^{-j\omega n}=\sum_{n=0}^{\infty}e^{-j\omega n} = \frac{1}{1-e^{-j\omega}}$$.
Now i have the problem, if $|e^{-j\omega}|$ = 1, the sum diverges.To handle this case, i know that $e^{-j\omega}$ is $2\pi$ periodic, i get 
$$\frac{1}{1-e^{-j\omega}}+\underbrace{e^{-j0}}_1
\sum_{k=-\infty}^{\infty}\delta(\omega+2\pi k)$$.
In books i found that the DTFT of the unit step is 
$$\frac{1}{1-e^{-j\omega}}+\pi
\sum_{k=-\infty}^{\infty}\delta(\omega+2\pi k)$$.
Can me anyone explain why get the $\pi$ in the DTFT of the unit step?

Firstly, you need to be aware that
$$u[n]=\sum_{n=0}^{\infty} \delta[n]$$
And according to Accumulation property of DTFT which implies that:
$$y[n]=\sum_{m=-\infty}^{n}x[m]$$
$$\sum_{m=-\infty}^{n}x[m]{\iff}\frac{1}{1-e^{-j\omega}}X(e^{j\omega}) + \pi X(e^{j0})\sum_{k=-\infty}^{\infty}\delta(\omega - 2\pi k)\ \ \ \ *$$
Now assume that: 
$$x[n] =\sum_{m=-\infty}^{n} g[m]$$
$$g[n] =\delta[n]$$
$$G({e^{j\omega}}) =  1$$
Substitute in *:
$$\frac{1}{1-e^{-j\omega}}G(e^{j\omega}) + \pi G(e^{j0})\sum_{k=-\infty}^{\infty}\delta(\omega - 2\pi k)$$
Now refine the last formula:
HINT: $G(e^{j\omega}) = 1\\ G(e^{j0}) = 1$
$$\frac{1}{1-e^{-j\omega}} + \pi \sum_{k=-\infty}^{\infty}\delta(\omega - 2\pi k)$$
