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Evaluating a limit

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. :confused:

Any help is appreciated. Thanks!
 

chisigma

Well-known member
Feb 13, 2012
1,704
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. :confused:

Any help is appreciated. Thanks!
With the substitution $\displaystyle \frac{1}{\sqrt{x}} = u$ is...

$\displaystyle \int_{0}^{1} \{ \frac{1}{\sqrt{x}}\}\ d x = 2\ \int_{1}^{\infty} \frac{\{u\}}{u^{3}}\ d u = 2\ \sum_{k=1}^{\infty} \int_{k}^{k + 1} \frac{u - k}{u^{3}}\ d u = $

$\displaystyle = 2\ \sum_{k=1}^{\infty} \int_{0}^{1} \frac{u}{(u + k)^{3}}\ d u = \sum_{k=1}^{\infty} \frac{1}{k\ (k + 1)^{2}} = 2 - \frac{\pi^{2}}{6}$


Kind regards


$\chi$ $\sigma$
 

Pranav

Well-known member
Nov 4, 2013
428
With the substitution $\displaystyle \frac{1}{\sqrt{x}} = u$ is...

$\displaystyle \int_{0}^{1} \{ \frac{1}{\sqrt{x}}\}\ d x = 2\ \int_{1}^{\infty} \frac{\{u\}}{u^{3}}\ d u = 2\ \sum_{k=1}^{\infty} \int_{k}^{k + 1} \frac{u - k}{u^{3}}\ d u = $

$\displaystyle = 2\ \sum_{k=1}^{\infty} \int_{0}^{1} \frac{u}{(u + k)^{3}}\ d u = \sum_{k=1}^{\infty} \frac{1}{k\ (k + 1)^{2}} = 2 - \frac{\pi^{2}}{6}$


Kind regards


$\chi$ $\sigma$
This is great, thanks a lot chisigma! :) (Bow)