Two analysis problems

Question

I am doing these two problems to review my Analysis classes.

1. (a) Show that for each $n \geq 1$, there exists an $x > 0$ such that \begin{equation} \frac{1}{\sqrt{nx+1}} + \frac{1}{\sqrt{nx+2}} + \cdots + \frac{1}{\sqrt{nx+n}} = \sqrt{n}. \end{equation} (b) For each $n$, call solution to above equation $x_n$. Find $\lim_{n\to \infty} x_n$.

My attempt to Q1: I think (a) is easy. Write the LHS as $s_n$. Then we have $$ \frac{n}{\sqrt{nx+n}} -\sqrt{n} \leq s_n - \sqrt{n} \leq \frac{n}{\sqrt{nx+1}} - \sqrt{n}.$$ By solving the inequalities, we could find that if $x_n \geq 1 - \frac{1}{n}$, then $s_n - \sqrt{n} \leq \frac{n}{\sqrt{nx+1}} - \sqrt{n} \leq 0$; and if $x_n \leq 0$, then $0 \leq \frac{n}{\sqrt{nx+n}} -\sqrt{n} \leq s_n - \sqrt{n}$. Hence, by intermediate value theorem, $s_n - \sqrt{n}$ has a solution on $(0, 1-\frac{1}{n})$. However, I am not sure how to solve (b).

---

2. Let $(X,d)$ be a compact metric space, and let $\rho$ be another metric on $X$. Suppose that $\rho(x,y) \leq d(x,y)$ for every $x, y \in X$. Prove that for each $\epsilon > 0$, we have $\delta > 0$ such that $\rho(x,y) < \delta$ implies $d(x,y) < \epsilon$.

My thoughts to Q2: I think the assumption somehow looks like a continuity condition: say, if let $f: (X, d) \to \mathbb{R}$ and $f(x) := \rho(x,0)$, then $f$ is uniformly continuous. However, the problem seems require me to prove that $g: (X, \rho) \to \mathbb{R}$; $g(x) := d(x,0)$ is continuous. This looks weird to me; I think I should use the compactness somewhere.

Also, I think the assumption also says that the topology generated by $\rho$ is better than that generated by $d$. Then a finite open cover in $d$ is also a finite open cover in $\rho$. I am not sure whether this is helpful or not.

Answer 1

Answer to Q 2: let $f:(X,d)\to (X\rho)$ be the identity map. If $d(x_n,x) \to 0$ then $\rho(x_n,x) \to 0$, so $f$ is continuous. Recall that any continuous bijection from a compact metric space onto any metric space is a homeomorphism. This implies that the identity map from $(X\rho)$ to $(X,d)$ is continuous. This map is now a continuous map on a compact metric space, so it is uniformly continuous. What b) asserts is exactly uniform continuity of this map.

Answer 2

Answer to Q1b: The equation rewrites as $$\frac{1}{n}\cdot\frac{1}{\sqrt{x+\frac{1}{n}}}+\frac{1}{n}\cdot\frac{1}{\sqrt{x+\frac{2}{n}}}+\dots+\frac{1}{n}\cdot\frac{1}{\sqrt{x+\frac{n}{n}}}=1.$$ $$\sum\limits_{i=1}^n \frac{1}{n}\cdot \frac{1}{\sqrt{x+\frac{i}{n}}} =1.$$ Thus, taking the limit as $n$ approaches infinity on both sides, we get $$\int\limits_{x}^{x+1} \frac{1}{\sqrt{t}} \,\text{dt}=1\Leftrightarrow 2\sqrt{x+1}-2\sqrt{x}=1\Leftrightarrow x+1-2\sqrt{x^2+x}+x=\frac{1}{4}$$ $$2x+\frac{3}{4}=2\sqrt{x^2+x}\Leftrightarrow 4x^2+3x+\frac{9}{16}=4x^2+4x\Leftrightarrow x=\frac{9}{16},$$ which should be the limit you are looking for.