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Evaluate the following limit. $$\lim_{n \to \infty} \cfrac{\sum_{k=1}^n\lfloor kx\rfloor}{n^2} $$

I'm not sure how to proceed from here. Normal methods don't seem to apply here, because of the presence of $\lfloor\cdot\rfloor$ (greatest integer function).

Although, I do get a feeling that Sandwich Theorem may come into play here. Not sure how. May be taking interval of $\lfloor x\rfloor$ as : $x - 1 <\lfloor x\rfloor< x $ .

Any one who can guide me through?

Matcha Latte
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Kushashwa Ravi Shrimali
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3 Answers3

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As $x - 1 < [x] < x + 1$, we have

$$\cfrac{ x + 2x + 3x + \cdots + nx - n}{n^2} < \cfrac{[x] + [2x] + [3x] + \dots + [nx] }{n^2} < \cfrac{ x + 2x + 3x + \cdots + nx + n}{n^2} $$

See what to do now?

Simon S
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Recall that $y-1 < \lfloor y \rfloor \leq y$. Hence, we have $$\sum_{k=1}^n (kx-1) < \sum_{k=1}^n \lfloor kx \rfloor \leq \sum_{k=1}^n kx $$ This gives us $$\dfrac{n(n+1)}{2}x -n < \sum_{k=1}^n \lfloor kx \rfloor \leq \dfrac{n(n+1)}2 x$$ Hence, we have $$\dfrac{\dfrac{n(n+1)}{2}x -n}{n^2} < \dfrac{\sum_{k=1}^n \lfloor kx \rfloor}{n^2} \leq \dfrac{\dfrac{n(n+1)}2 x}{n^2}$$ Taking the limit as $n \to \infty$, we see that $$\dfrac{x}2 \leq \lim_{n \to \infty}\dfrac{\sum_{k=1}^n \lfloor kx \rfloor}{n^2} \leq \dfrac{x}2$$ Hence, $$\lim_{n \to \infty}\dfrac{\sum_{k=1}^n \lfloor kx \rfloor}{n^2} = \dfrac{x}2$$

Adhvaitha
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you can use the Stolz–Cesàro theorem, that states that if b_n is strictly monotone and $$b_n \rightarrow \inf$$ if $$\frac{a_{n+1}-a_n}{b_{n+1}-b_n} \rightarrow L$$ then we can say $$\frac{a_n}{b_n} \rightarrow L$$ its easy to prove that $$\frac{[nx]}{n^2-(n-1)^2} \rightarrow \frac{x}{2}$$ using Sandwich Theorem and then using Stolz–Cesàro theorem we get $$\frac{[x]+[2x].....[nx]}{n^2} \rightarrow \frac{x}{2}$$

gurtonn
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