# On Lamda function

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 vous and I'm sure it will lead me back to where I started. Of course contour integration over a square is out of question since the residue of the function $\dfrac{\pi \cot \pi z}{(2z+1)^s}$ is very tedious to calculate.

Any help?

• Hint: Write the infinite series formula for the Riemann $\zeta$ function, and then split it into two subseries, depending on the parity of the denominator. See also Dirichlet $\eta$ function. – Lucian May 14 '15 at 23:52
• @Thank you Lucian. Your hint was very helpful. I'll answer my own question a little bit later. – Tolaso May 15 '15 at 8:03

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}