(a) Prove that $\sqrt{n+1}-\sqrt n<\frac1{2\sqrt n}<\sqrt n-\sqrt{n-1}$ for all $n\in\mathbb N$. (b) Prove that the integer part of the sum $1+\frac1{\sqrt2}+\frac1{\sqrt3}+\ldots+\frac1{\sqrt{m^2}}$, where $m\in\mathbb N$, is either $2m-2$ or $2m-1$.
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Tags: algebra
jasperE3
07.04.2021 17:22
Part (a).
homosapien1234
10.04.2021 00:04
The mean value theorem states that sqrt(n + 1) - sqrt(n) = 1/(2sqrt(c)) for some c greater than n, but less than n+1. n<c<n+1, which implies 1/(n+1)<1/c<1/n, this won't change if you square root this and then multiply by 0.5. do the same for the right hand side and then part a is proven
To find an upper bound for this sum, let's use the inequality from part a. This sum will be less than (sorry for the bad formatting, I don't know how to user latex yet) 2*(sqrtk - sqrt(k-1)) for integer k from 1 to k. Simplifying gives us an upper bound of 2m. Now for the lower bound, notice that the derivative of 1/sqrtx is always negative, meaning 1/sqrt(x) is always decreasing as a continuous function, Which means that this sum would be greater than the integral of 1/sqrtx from 1 to m^2, which turns out to be 2m-2. Since this function is in between 2m-2 and 2m, its integer part could either be 2m (if it's less than 2m-1) or it could be 2m-1 if it's greater than or equal to 2m-1
mop
10.04.2021 00:29
This is for 3rd graders?
Tong.Qiu
10.04.2021 01:45
homosapien1234 wrote:
The mean value theorem states that $\sqrt{n + 1} - \sqrt{n} = \frac{1}{2\sqrt{c}}$ for some $c$ greater than $n$, but less than $n+1$. $n<c<n+1$, which implies $\frac{1}{n+1}<\frac{1}{c}<\frac{1}{n}$, this won't change if you square root this and then multiply by $0.5$. do the same for the right hand side and then part a is proven
To find an upper bound for this sum, let's use the inequality from part a. This sum will be less than $2\cdot(\sqrt{k} - \sqrt{k-1})$ for integer $k$ from $1$ to $k$. Simplifying gives us an upper bound of $2m$. Now for the lower bound, notice that the derivative of $\frac{1}{\sqrt{x}}$ is always negative, meaning $\frac{1}{\sqrt{x}}$ is always decreasing as a continuous function, Which means that this sum would be greater than the integral of $\frac{1}{\sqrt{x}}$ from $1$ to $m^2$, which turns out to be $2m-2$. Since this function is in between $2m-2$ and $2m$, its integer part could either be $2m$ (if it's less than $2m-1$) or it could be $2m-1$ if it's greater than or equal to $2m-1$
$\LaTeX$ed for you
homosapien1234
10.04.2021 21:36
Tong.Qiu wrote: homosapien1234 wrote:
The mean value theorem states that $\sqrt{n + 1} - \sqrt{n} = \frac{1}{2\sqrt{c}}$ for some $c$ greater than $n$, but less than $n+1$. $n<c<n+1$, which implies $\frac{1}{n+1}<\frac{1}{c}<\frac{1}{n}$, this won't change if you square root this and then multiply by $0.5$. do the same for the right hand side and then part a is proven
To find an upper bound for this sum, let's use the inequality from part a. This sum will be less than $2\cdot(\sqrt{k} - \sqrt{k-1})$ for integer $k$ from $1$ to $k$. Simplifying gives us an upper bound of $2m$. Now for the lower bound, notice that the derivative of $\frac{1}{\sqrt{x}}$ is always negative, meaning $\frac{1}{\sqrt{x}}$ is always decreasing as a continuous function, Which means that this sum would be greater than the integral of $\frac{1}{\sqrt{x}}$ from $1$ to $m^2$, which turns out to be $2m-2$. Since this function is in between $2m-2$ and $2m$, its integer part could either be $2m$ (if it's less than $2m-1$) or it could be $2m-1$ if it's greater than or equal to $2m-1$
$\LaTeX$ed for you Thank you