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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

https://www.wolframalpha.com/input/?i=%5Csum+%7Bi%3D1%7D%5E%7Bn%2B1%7Di%5E%7B-3%7D%3D%5Csum+_%7Bi%3D1%7D%5En%5Cleft%28i%2B1%5Cright%29%5E%7B-3%7D%2B1.

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

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    $\begingroup$ 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
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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

$$\sum_{i=1}^n(i+1)^{-3}\,,$$

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

$$\sum_{i=1}^n(i+1)^{-3}=\sum_{j=2}^{n+1}j^{-3}\,.$$

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

$$\sum_{j=2}^{n+1}j^{-3}=\sum_{i=2}^{n+1}i^{-3}\,.$$

Thus,

$$\sum_{i=1}^n(i+1)^{-3}=\sum_{i=2}^{n+1}i^{-3}\,.$$

Finally,

$$\sum_{i=1}^{n+1}i^{-3}=1^{-3}+\sum_{i=2}^{n+1}i^{-3}=1+\sum_{i=2}^{n+1}i^{-3}=1+\sum_{i=1}^n(i+1)^{-3}\,.$$

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

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