# Struggling to prove a property of an infinite sum

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.

• For $n=4$ this states that $$\frac1{1^3}+\frac1{2^3}+\frac1{3^3}+\frac1{4^3}+\frac1{5^3}=\frac1{2^3}+\frac1{3^3}+\frac1{4^3}+\frac1{5^3}+1.$$ – 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

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

$$i+1$$ 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,

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

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