3
$\begingroup$

Using numeric experiments, I have found the following identity $$ k! H_k+\sum _{j=1}^k \frac{\sum _{i=j}^k (-1)^i \binom{k}{i} \left((j-i)^k-(-i-n+1)^k\right)}{j+n-1}=0 $$ where $H_k = \sum_{i=1}^k 1/i$. This seems to hold for all $k \ge 1$ and $n \ge 1$.

It's fairly easy to check this for any fix $k$. But I am not able to prove it for general $k$. Induction also does not seem to work well in this case.


Update, you can check this with the following Mathematica code

Table[k!*HarmonicNumber[k] + 
       Sum[((-1)^(1 + i + k)*(-1 + i + n)^k*Binomial[k, i] + 
              Eulerian[k, i])*HarmonicNumber[-1 + i + n], {i, 0, 
     k}] == 
     0, {k, 1, 6}, {n, 1, 10}]
$\endgroup$
1
  • 1
    $\begingroup$ Those sums look a bit inclusion-exclusion principle -ey $\endgroup$ – Vladimir Lenin Dec 5 '20 at 22:17
1
$\begingroup$

This is not a whole solution but I think it can be salvaged. If we apply the well known identity $a^k - b^k = (a-b) \cdot \sum\limits_{i=1}^{k-1} a^{k-1-i} b^i $ for $(a,b) = ((j-i),(-i-n+1))$ then the conjecture above boils down to proving that:

\begin{eqnarray} rhs_\eta := \sum\limits_{j=1}^k \sum\limits_{i=j}^k \sum\limits_{\xi=\eta}^{k-1} (-1)^i \binom{k}{i} \binom{\xi}{\eta} (j-i)^{k-1-\xi} (1-i)^{\xi-\eta} = -k! H_k \cdot \delta_{\eta,0} \end{eqnarray} for $\eta=0,1,2,\cdots$. This comes from the fact that the sum in question is now a polynomial in the variable $n$ whose all coefficients except for the zeroth term must vanish.

Let us take a look at the coefficient at the zeroth power of $n$ (meaning $\eta=0$). We have:

\begin{eqnarray} rhs_0 &=& -k + \sum\limits_{j=2}^k \sum\limits_{i=j}^k (-1)^i \binom{k}{i} \left( \frac{-(1-i)^k + (-i+j)^k}{j-1}\right) \\ &=& -k + \sum\limits_{i=1}^k (-1)^{i-1-k} (i-1)^k \binom{k}{i} \left( H_{i-1} - \psi^{(0)}_{i+1} + \psi^{(0)}_{k+1} \right) \\ &=& -k + \sum\limits_{i=0}^{k-1} (-1)^{i+k} i^k \binom{k}{i+1} \left(-\frac{1}{i+1} + H_k \right) \\ &=& -k+ \sum\limits_{i=0}^{k-1} (-1)^{i+k} i^k \binom{k}{i+1} \left(-\frac{1}{i+1} \right) + (1-k!) H_k \\ &=& -k+ (k - H_k) + (1-k!) H_k \end{eqnarray}

In the first line from the top we carried out a sum over $\xi$ since it is a geometric series. In the second line from the top we split the double sum into two sums and in the first sum we did the sum over $j$ whereas in the second sum we substituted for $(i-j) \rightarrow i $ and then we did the sum over $j$ as well. In the third line we substituted $(i-1) \rightarrow i $ and then we used the the relationship between the polygamma function and harmonic numbers. In the last two lines we did the remaining single sums by doing the usual tricks $i^k = \left. d_t^k \exp(i t) \right|_{t=0} $ and then using the binomial expansion to resum.

In order to complete the proof one needs to show that $rhs_\eta =0 $ for all integer $\eta >0 $ . This should not be hard to do.

In[530]:= n =.;

Table[Sum[(-1)^(i + xi - eta) Binomial[k, i] Binomial[xi, eta] If[
       i == j && k - 1 == xi, 1, (j - i)^(k - xi - 1)] If[
       i == 1 && xi == eta, 1, (i - 1)^(xi - eta)], {j, 1, k}, {i, j, 
      k}, {xi, eta, k - 1}] (-n)^eta, {k, 1, 5}, {eta, 0, k - 1}] // 
  Expand // MatrixForm

Table[-k + 
  Sum[(-1)^i Binomial[k, i] ((-(1 - i)^k + (-i + j)^k)/(-1 + j)) , {j,
     2, k}, {i, j, k}], {k, 1, 5}]

Table[-k + 
  Sum[(-1)^(i - 1 + k) ((i - 1)^k)  Binomial[k, 
     i] (HarmonicNumber[i - 1] - PolyGamma[0, 1 + i] + 
      PolyGamma[0, 1 + k]), {i, 1, k}], {k, 1, 5}]

Table[-k + 
  Sum[(-1)^(i + k) ((i)^k)  Binomial[k, 
     i + 1] (-1/(i + 1) + HarmonicNumber[k]), {i, 0, k - 1}], {k, 1, 
  5}]
Table[-k + 
  Sum[(-1)^(i + k) ((i)^k)  Binomial[k, i + 1] (-1/(i + 1)), {i, 0, 
    k - 1}] + (1 - k!) HarmonicNumber[k], {k, 1, 5}]
Table[-k + (k - HarmonicNumber[k]) + (1 - k!) HarmonicNumber[k], {k, 
  1, 5}]

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.