Why is this sum equal to the Logarithmic Integral?

Solution 1:

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.

Solution 2:

First of all note that the term being added to the inner sum can be absorbed into the sum as follows:

$$ (-1)^{k-1} (n-1) + \sum_{j=1}^{k-1} (-1)^{j+k-1} n \frac{\log^j n}{j!} = (-1)^k \left( 1 - \sum_{j=0}^{k-1} (-1)^{j} n \frac{\log^j n}{j!} \right) $$ Now, write $1 = \sum_{j=0}^{\infty} (-1)^{j} n \frac{\log^j n}{j!}$. Thus the original sum becomes

$$ \mathcal{S} = \sum_{k=1}^\infty \frac{1}{k} \sum_{j=k}^{\infty} (-1)^{j} n \frac{\log^j n}{j!} $$ Now exchange the order of summation $\sum_{k=1}^\infty \sum_{j=k}^\infty \to \sum_{j=1}^\infty \sum_{k=1}^{j}$: $$ \mathcal{S} = \sum_{j=1}^\infty \sum_{k=1}^{j} \frac{1}{k} (-1)^{j} n \frac{\log^j n}{j!} = n \sum_{j=1}^\infty (-1)^{j} H_j \frac{\log^j n}{j!} $$

I am not sure at the spot how to manually convert this into logarithmic integral function, but Mathematica can solve this sum in terms of LogIntegral[n]:

In[179]:= 
n Sum[(-1)^j HarmonicNumber[j] Log[n]^j/j!, {j, 1, Infinity}] == 
    Log[Log[n]] + EulerGamma - LogIntegral[n] // FullSimplify[#, n > 1] &

Out[179]= True