Prove that \[\sum_{k=-N}^{N}|x+k|-|k| \ge x^2\] as \(N\to \infty\) and find equality cases.

**Details and Assumptions**

We take the limit as \(N\) approaches \(\infty\) because \(\displaystyle\sum_{k=-\infty}^{\infty}|x+k|-|k|\) is not well-defined.

Prove that \[\sum_{k=-N}^{N}|x+k|-|k| \ge x^2\] as \(N\to \infty\) and find equality cases.

**Details and Assumptions**

We take the limit as \(N\) approaches \(\infty\) because \(\displaystyle\sum_{k=-\infty}^{\infty}|x+k|-|k|\) is not well-defined.

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TopNewestSet : \[f_n(x)= \sum_{k=-n}^n |x+k|-|k| =|x|+\sum_{k=1}^n |x+k|+|x-k| -2|k|\] Note that \(f_n(-x)=f_n(x)\) for any real \(x\) and any integer \(n>0\) then we only need to consider the case \(x\geq 0\), and if \(k\geq x>0\) then \(|x+k|+|x-k|-2|k|=0\), therefore for \(n>x\) : \[f_n(x)=x+\sum_{k=1}^{\lfloor x\rfloor}(x+k)+(x-k)-2 k= x+2x \lfloor x \rfloor -\lfloor x\rfloor (\lfloor x\rfloor +1).\]

Clearly now, \(f_n(x)= x-\lfloor x\rfloor +2x \lfloor x\rfloor-\lfloor x\rfloor^2 \), the rest should follow easily.

\[f_n(x)-x^2 =x- \lfloor x\rfloor + x(\lfloor x\rfloor-x)+\lfloor x\rfloor(x-\lfloor x\rfloor)=(x-\lfloor x\rfloor)(1+\lfloor x\rfloor -x)\geq 0\]

Obviously, the equality is when \(x\in \mathbb{Z}\).

Remark :you don't need \(n\to \infty\) you just need \(n\geq x\), and the latter series does make sense and it well-defined (though I'd use the term 'convergent' because well defined usually used for mono-valued functions). – Haroun Meghaichi · 2 years, 2 months agoLog in to reply

– Daniel Liu · 2 years, 2 months ago

How is the latter sequence defined? I seem to be getting different values for plugging in the same \(x\) value if I compute it differently.Log in to reply

– Haroun Meghaichi · 2 years, 2 months ago

Can you give me an example ?Log in to reply