I am trying to prove the following property

$\displaystyle \sum_{i=1}^{n+1}i^{-3}=\sum_{i=1}^{n}\left(i+1\right)^{-3}+1$

holds for any $n\in\mathbb{N^{+}}$. When expanding out the series, no pattern emerges (instead a long polynomial). I understand the the LHS may be written as $H^{(3)}_{n+1}$, but don't know how the RHS may be simplified. Note: I am referring to wolfram alpha's interpretation


I would be grateful for any advice that you may give me.

  • 4
    $\begingroup$ For $n=4$ this states that $$\frac1{1^3}+\frac1{2^3}+\frac1{3^3}+\frac1{4^3}+\frac1{5^3}=\frac1{2^3}+\frac1{3^3}+\frac1{4^3}+\frac1{5^3}+1.$$ $\endgroup$ – Angina Seng Aug 5 at 2:32

This is just a matter of manipulating the index of summation; it has almost nothing to do with the expression being summed.

As $i$ runs from $1$ to $n$ in


$i+1$ runs from $2$ to $n+1$. Let $j=i+1$: then


Now just rename the index of summation back to $i$:






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  • 1
    $\begingroup$ Thank you. It was much simpler than I thought :) $\endgroup$ – user809100 Aug 5 at 2:37
  • 2
    $\begingroup$ @UNOwen: You’re welcome. $\endgroup$ – Brian M. Scott Aug 5 at 2:37

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