# floor function and infinite series...

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.....

Note by Sarvesh Dubey
3 years, 2 months ago

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Here'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

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

- 3 years, 2 months ago

So as you see in the above example, we basically make use of inequalities while solving problems related to Floor/Ceil Function. Here are some other inequalities that can help you -

$x-1 < \lfloor x \rfloor \leq x$

$\lfloor x \rfloor \leq x < \lfloor x \rfloor +1$

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.

- 3 years, 2 months ago