Pls its a request to all the young mathematics mind to publish a note on floor function and how to solve an infinite series problem.....

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TopNewestHere's a simple example to gear you up. Prove that

\[\lim_{n \rightarrow \infty} \frac{1}{n^2}\sum_{k=1}^n \lfloor kx \rfloor = \frac{x}{2}\]

Proof.By the definition of Greatest Integer Function, \(x-1 < \lfloor x \rfloor \leq x\)Thus, we have \[kx-1 < \lfloor kx \rfloor \leq kx\] \[\Rightarrow \sum_{k=1}^n kx-1 < \sum_{k=1}^n \lfloor kx \rfloor \leq \sum_{k=1}^n kx\] \[\Rightarrow \frac{n(n+1)}{2}x-n < \sum_{k=1}^n \lfloor kx \rfloor \leq \frac{n(n+1)}{2}x\] \[\Rightarrow \frac{n+1}{2n}x-\frac{1}{n} < \frac{1}{n^2}\sum_{k=1}^n \lfloor kx \rfloor \leq \frac{n+1}{2n}x\]

Taking the limit, we find that \[\lim_{n \rightarrow \infty} \frac{n+1}{2n}x-\frac{1}{n} < \lim_{n \rightarrow \infty} \frac{1}{n^2}\sum_{k=1}^n \lfloor kx \rfloor \leq \lim_{n \rightarrow \infty} \frac{n+1}{2n}x\] \[\Rightarrow \frac{x}{2} < \lim_{n \rightarrow \infty} \frac{1}{n^2}\sum_{k=1}^n \lfloor kx \rfloor \leq \frac{x}{2}\]

And therefore by Squeeze Theorem our result follows as

\[\lim_{n \rightarrow \infty} \frac{1}{n^2}\sum_{k=1}^n \lfloor kx \rfloor = \frac{x}{2} \quad _\square\]

Also, you can read more about the floor function here – Kishlaya Jaiswal · 2 years, 1 month ago

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\[x-1 < \lfloor x \rfloor \leq x\]

\[\lfloor x \rfloor \leq x < \lfloor x \rfloor +1\]

Also, notice that, any integer can be written as

\[x = \lfloor x \rfloor + \{x\}\]

Since, \(0 \leq \{x\} < 1\). That's where we get the inequality \(\lfloor x \rfloor \leq x < \lfloor x \rfloor +1\) and \(\lfloor x \rfloor \leq x\)

Anyhow, if your concern was regarding any of the well-known series such as Hermite's Identity, then please let me know so that I can add examples regarding the same. – Kishlaya Jaiswal · 2 years, 1 month ago

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