How to prove that $\lambda(s)=(1-2^{-s})\zeta(s)$? The Lambda function is defined as:
$$\lambda(s)=\sum_{n=0}^{\infty} \frac{1}{(2n+1)^s},\; \mathfrak{Re}(s)>1$$
How to prove that $\lambda(s)=(1-2^{-s})\zeta(s)$? 
Basically, I was dealing with the functional equation of $\eta$ and while I was trying to prove its functional equation when $\mathfrak{Re}(s)>1$ I came across this series. I googled it and found that is a known function under the name "lambda dirichlet function" and it is reduced down to that formula.
Trying to prove the formula I came across difficulties because I cannot manipulate the sum and split it apart somehow, unless I write it as an alternating sum but that is a deja vu and I'm sure it will lead me back to where I started. Of course contour integration over a square is out of the question since the residue of the function $\dfrac{\pi \cot \pi z}{(2z+1)^s}$ is very tedious to calculate. 
Any help?
 A: I give the solution according to Lucian's comment.
We start things off by using the $\zeta$ Riemann's function, defined as:
$$\zeta(s)= \sum_{n=1}^{\infty}\frac{1}{n^s}, \; \mathfrak{Re}(s)>1$$
which converges absolutely since all terms are positive. That the series converges if $\mathfrak{Re}(s)>1$ is an immediate consequence of the integral test.
Now, splitting the series into odd and even terms we get that:
$$\zeta(s)= \sum_{n=1}^{\infty}\frac{1}{(2n)^s}+ \sum_{n=0}^{\infty}\frac{1}{(2n+1)^s} \Leftrightarrow \zeta(s)= \frac{1}{2^s}\zeta(s) + \sum_{n=1}^{\infty}\frac{1}{(2n+1)^s} \Leftrightarrow$$
$$\Leftrightarrow \sum_{n=1}^{\infty}\frac{1}{(2n+1)^s} =(1-2^{-s})\zeta(s) \tag{1}$$
what we wanted.
Going one step further one can now prove the functional equation of $\eta$ Dirichlet function. This function is defined as:
$$\eta(s)= \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^s} , \; \mathfrak{Re}(s)>1$$
and converges absolutely since $\displaystyle \sum_{n=1}^{\infty}\left | \frac{(-1)^{n-1}}{n^s} \right |= \sum_{n=1}^{\infty}\frac{1}{n^s}=\zeta(s)$ as shown above.
Now, absolute convergence allows us to rearrange the terms. Hence:
$$\begin{aligned}
\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^s} &=\left ( 1+\frac{1}{3^s}+ \frac{1}{5^s}+\cdots \right )- \left ( \frac{1}{2^s}+ \frac{1}{4^s}+ \frac{1}{6^s}+\cdots \right ) \\ 
 &= \sum_{n=0}^{\infty}\frac{1}{(2n+1)^s}- \sum_{n=1}^{\infty}\frac{1}{(2n)^s}\\ 
 &=\sum_{n=0}^{\infty} \frac{1}{(2n+1)^s}- \frac{1}{2^s} \sum_{n=1}^{\infty}\frac{1}{n^s}\\ 
 &\overset{(1)}{=}\left ( 1-2^{-s} \right )\zeta(s)- \frac{1}{2^s}\zeta(s)\\
 &=\left ( 1-2^{1-s} \right )\zeta(s)
\end{aligned}$$
