# 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?

• Where you wrote $-1^{k-1}$, I am guessing you meant $(-1)^{k-1}$ and I changed it to that. Sep 12, 2011 at 18:30
• Note that when you write \log x rather than log x, then the backslash not only prevents "log" from being italicized, but it also provides proper spacing, so you don't need the spacing you did manually. Sep 12, 2011 at 18:31
• All of the $(-1)^{k-1}$ cancel out and there's only a $(-1)^j$ left in on the right numerator. Sep 12, 2011 at 18:33
• Michael - Thanks. I'm still getting used to Latex and MathJax. Sep 12, 2011 at 19:39
• 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 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. Sep 12, 2011 at 19:40

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.

• Wow. Thanks. Great to see that proven. Sep 13, 2011 at 10:20
• @Nathan: You're welcome. I haven't peered at how you implemented the computation of the logarithmic integral in your web page, but just in case: you might be interested in Ramanujan's series for the logarithmic integral (formula 15 here); I've found it slightly more efficient than the straightforward series for medium-sized arguments. Sep 13, 2011 at 10:23
• The series I provided is even slower than Formula 14 from that page. Instead, its interest to me is from the following (this is my argument from my webpage - I'll try to go quick): If you have the strict divisor functions $d_1(n) = 1$ and $d_k(n) = \sum_{j | n, 1 < j < n}d_1(j)d_{k-1}(n/j)$, then Linnik's identity says $\sum_{k=1} (-1^{k+1})/k \cdot d_k(n) =$1/a if n = $p^a$ where p is a prime, 0 otherwise. Sum this from 2 to n, and you have $\sum_{j=2}^n 1 - 1/2\sum_{j=2}^n \sum_{k=2}^{n/j} 1+ 1/3\sum_{j=2}^n \sum_{k=2}^{n/j}\sum_{l=2}^{n/(jk)} 1 - ... = \Pi(n)$ Sep 13, 2011 at 11:33
• The term on the right is the prime power counting function. Approximate the sums on the left as $\int_{1}^n dx - 1/2\int_{1}^n \int_{1}^{n/x} dydx+ 1/3\int_{1}^n \int_{1}^{n/x}\int_{1}^{n/xy} dz dy dx - ...$, evaluate these integrals, and you immediately have the sums I asked about. I knew empirically that my integrals here were equal to $li(x) - \log \log x - \gamma$, but... Sep 13, 2011 at 11:33
• I see, @Nathan. I do have to agree that those are cute-looking multiple integrals... :) Sep 13, 2011 at 11:40

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

Out= True

• Is exchanging the order of infinite sums sound when the terms have different signs? It's not clear to me that the double sum converges absolutely. Sep 12, 2011 at 19:09
• The exchange can be carried out with the upper summation bound fixed at $m$, and later $m$ would be sent to infinity: $\sum_{k=1}^m \sum_{j=k}^m \to \sum_{j=1}^m \sum_{k=1}^{j}$ Sep 12, 2011 at 19:21
• But is it valid to go from $\sum_{k=1}^{\infty}\sum_{j=k}^{\infty}$ to $\lim_{m\to\infty}\sum_{k=1}^m \sum_{j=k}^{m}$ in the first place? Sep 12, 2011 at 19:25
• Regarding that last sum: I suspect manipulations similar to what Srivatsan did here 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. Sep 13, 2011 at 10:43