a doubt with the series $ \sum_{n=0}^{\infty}e^{-nx} $

Question

I know that the series is equal to

$$ \sum_{n=0}^{\infty}e^{-nx}= \frac{1}{1-e^{-x}}$$

However, if I expand each exponential term into a Taylor series I get

$$ \sum_{n=0}^{\infty}e^{-nx}= \sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\frac{(-nx)^{k}}{k!} $$

Now if I use $ \sum_{n=0}^{\infty}n^{k}=\zeta (-k) $ I would get that

$$ \sum_{n=0}^{\infty}e^{-nx} = \sum_{k=0}^{\infty} \frac{(-1)^{k} \zeta (-k)}{k!}x^{k}$$

Is this last expression correct in the sense of Taylor series?

Answer 1

The derivation as it stands is not valid. Even though Bilou06 was probably speaking about convergent series, the relation $\sum_n\sum_m a_{n,m}=\sum_m\sum_n a_{n,m}$ can also fail for regularized divergent series.

In this case the conclusion is false - an error term is needed. We have two formulas at hand:

$$\zeta(-n)=-\frac{B_{n+1}}{n+1}$$

$$\frac{t}{e^t-1}=\sum_{k=0}^\infty B_k\frac{x^k}{k!}.$$

Therefore,

$$\begin{array}{ll} \sum_{k=0}^\infty\frac{\zeta(-k)}{k!}(-x)^k & =\zeta(0)+\sum_{k=1}^\infty-\frac{B_{k+1}}{(k+1)}\frac{(-x)^k}{k!} \\ & =-\frac{1}{2}+\frac{1}{x}\sum_{k=1}^\infty B_{k+1}\frac{(-x)^{k+1}}{(k+1)!} \\ & =-\frac{1}{2}+\frac{1}{x}\sum_{k=2}^\infty\frac{B_k}{k!}(-x)^k \\ & =-\frac{1}{2}+\frac{1}{x}\left[\frac{-x}{e^{-x}-1}-\frac{B_1}{1!}(-x)-\frac{B_0}{0!}(-x)^0\right] \\ & =\frac{1}{1-e^{-x}}-\frac{1}{x}-1.\end{array}$$

Answer 2

6 years later, but the given answer is i feel incorrect/incomplete.

The relation with swapping sums was correct, and $\sum_n\sum_m a_{n,m}=\sum_m\sum_n a_{n,m}$ is valid for regularization, as long as the shape of the sums/whatever is regularized stays the same.

The question made the mistake assuming Taylor series have a the lower limit of 0. Just like most general series, this should be in reality minus infinity and positive infinity.

Let's take the example where we start at n=1, you can add yourself n=0. Changed e^(-x) into c.

$$\sum_{n=1}^{\infty}c^{n}= \frac{c}{1-c}$$ $$\sum_{n=1}^{\infty}c^{n}= \sum_{n=1}^{\infty}\sum_{k \in \mathbb{Z}}\frac{(n\ln(c))^{k}}{k!}=$$ $$\sum_{k \in \mathbb{Z}}\frac{\zeta(-k)\ln^{k}(c)}{k!}=$$ if k<-2 it's 0. if k is -1, you need to look at the limit, which is $\frac{-1}{ln(c)}$.

$$\sum_{k=0}^{\infty}\frac{\zeta(-k)\ln^{k}(c)}{k!}=\frac{c}{1-c}+\frac{1}{ln(c)}$$ As can been seen writen in the answer above.

$$\sum_{k \in \mathbb{Z}}\frac{\zeta(-k)\ln^{k}(c)}{k!}=\frac{c}{1-c}$$