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Pranav
Well-known member
- Nov 4, 2013
- 428
Problem:
Compute:
$$\lim_{n\rightarrow \infty} \frac{1}{n^2}\sum_{i=1}^{n^2} \left\{ \frac{n}{\sqrt{i}} \right\}$$
where $\{ x\}$ denotes the fractional part of $x$.
Attempt:
I substitute $N=n^2$ i.e
$$\lim_{N\rightarrow \infty} \frac{1}{N}\sum_{i=1}^{N} \left\{ \frac{1}{\sqrt{i/N}} \right\}=\int_0^1 \left\{ \frac{1}{\sqrt{x}}\right\}\,dx$$
but I don't see how to evaluate the definite integral.
Any help is appreciated. Thanks!
Compute:
$$\lim_{n\rightarrow \infty} \frac{1}{n^2}\sum_{i=1}^{n^2} \left\{ \frac{n}{\sqrt{i}} \right\}$$
where $\{ x\}$ denotes the fractional part of $x$.
Attempt:
I substitute $N=n^2$ i.e
$$\lim_{N\rightarrow \infty} \frac{1}{N}\sum_{i=1}^{N} \left\{ \frac{1}{\sqrt{i/N}} \right\}=\int_0^1 \left\{ \frac{1}{\sqrt{x}}\right\}\,dx$$
but I don't see how to evaluate the definite integral.
Any help is appreciated. Thanks!