number theory - Why is this sum equal to the Logarithmic Integral?

I am using this sum:

$$\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}\left((-1)^{k-1} (n-1) + \sum_{j=1}^{k-1}\frac{(-1)^{j+k-1}n (\log n)^j}{j!}\right)$$

Empirically, this is precisely equal to

$$\sum_{k=1}^\infty \frac{(\log n)^k}{k! k}$$

which is the most significant term in this expansion of the logarithmic integral

$$\operatorname{li}(n) = \log \log n + \gamma + \sum_{k=1}^\infty \frac{(\log n)^k}{k! k}$$

where $\gamma$ is the Euler-Mascheroni constant.

Can anyone show why my sum is equal to the sum from the logarithmic integral?

Answer

I'll start the same way as Sasha, except that I'll first replace $n$ with $\exp\,z$:

$$\frac1{k}\left(-1+\exp\,z\sum_{j=0}^{k-1}\frac{(-z)^j}{j!}\right)$$

From here, we recall that the partial sums of the exponential function possess the following integral representation (see here for a proof):

$$\exp(-u)\sum_{j=0}^{k-1}\frac{u^j}{j!}=\frac1{(k-1)!}\int_u^\infty t^{k-1} \exp(-t)\mathrm dt$$

so we make the replacement:

$$-\frac1{k}+\frac1{k!}\int_{-z}^\infty t^{k-1} \exp(-t)\mathrm dt$$

Let's complicate things a bit:

$$-\frac{(k-1)!}{k!}+\frac1{k!}\int_{-z}^\infty t^{k-1} \exp(-t)\mathrm dt$$

and replace $(k-1)!=\Gamma(k)$ with its integral representation:

$$\frac1{k!}\left(-\int_0^\infty t^{k-1} \exp(-t)\mathrm dt+\int_{-z}^\infty t^{k-1} \exp(-t)\mathrm dt\right)$$

which simplifies:

$$\frac1{k!}\int_{-z}^0 t^{k-1} \exp(-t)\mathrm dt$$

We now treat the sum

$$\sum_{k=1}^\infty \frac1{k!}\int_{-z}^0 t^{k-1} \exp(-t)\mathrm dt$$

and swap summation and integration (justification left to the reader):

$$\int_{-z}^0\left(\sum_{k=1}^\infty \frac{t^{k-1}}{k!}\right)\exp(-t)\mathrm dt$$

which becomes

$$\int_{-z}^0\left(\frac{\exp\,t-1}{t}\right)\exp(-t)\mathrm dt=\int_{-z}^0\frac{1-\exp(-t)}{t}\mathrm dt=-\int_z^0\frac{1-\exp\,t}{-t}\mathrm dt=\int_0^z\frac{\exp\,t-1}{t}\mathrm dt$$

and since

$$\int_0^z\frac{\exp\,t-1}{t}\mathrm dt=\sum_{j=1}^\infty \frac{z^j}{j! j}$$

the claim is proven.