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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?

  • 0
    Henning: True. The reason I kept the terms in its original, unnecessarily messy form, is that what I'm really trying to show is the relationship I've written up at http://www.icecreambreakfast.com/primecount/logintegral.html . Basically, it's a way of using Linnik's identity to connect the prime counting function to the logarithmic integral. I can get from the prime counting function up to the sum I just posted, but this last jump is stumping me.2011-09-12

2 Answers 2

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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.

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    I see, @Nathan. I do have to agree that those are cute-looking multiple integrals... :)2011-09-13
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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 
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    Regarding that last sum: I suspect manipulations similar to what Srivatsan did [here](http://math.stackexchange.com/questions/63466/64044#64044) might be useful. In particular, I ended up with the integral $\int_0^1 \frac{\exp(-zt)-\exp(-z)}{t-1}\mathrm dt$ after trying it out myself.2011-09-13