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Let $H_k = \sum_{i=1}^k \frac{1}{i} $ be the $k$-th Harmonic number. We have that $$\sum_{k=1}^n H_k = (n+1) H_{n+1}-(n+1)$$ My question is:

If $$\sum_{i=1}^{p-1} C_i^s = (s+p) C_p^s -(s+p+1)$$ Then $$C_p^s=1+H_{s+p}-H_s$$

I don't know how to get to this result... Is it so straightforward from the initial result or do I need to play a bit with the summations?

1 Answers1

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Note that$$C_p^s=\sum_{i=1}^pC_i^s-\sum_{i=1}^{p-1}C_i^s=(s+p+1)C_{p+1}^s-(s+p)C_p^s-1,$$so$$C_p^s=C_{p+1}^s-\frac{1}{s+p+1}\implies C_{p+1}^s=C_p^s+H_{s+p+1}-H_{s+p}.$$The desired result follows by induction provided $C_0^s=1$.

J.G.
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  • I still don't get how you get the final result... I tried $$C_p^s=\sum_{i=1}^{p} C_{i}^s-C_{i-1}^s = \sum_{i=1}^p H_{s+i}-\sum_{i=1}^p H_{s+i-1}$$ but I can't get nohere. Could you please expand your answer? –  Jun 11 '20 at 21:03
  • @Papagaio_da_Fauna My first display-line equation used your formula for $\sum_{i=1}^{p-1}C_i^s$; my second rearranged the first's result. You should be subtracting $\sum_{i=1}^{p-1}C_i^s$, not $C_{i-1}^s$. – J.G. Jun 11 '20 at 21:07