Pregunta de Inducción Matemática, Ayuda para Pruebas

Question

Demostrar mediante Inducción Matemática que para todos los números naturales ( $n>0$ ):

$$ \frac 1 {\sqrt{1}} + \frac 1 {\sqrt{2}} + \cdots + \frac 1 {\sqrt{n}} \ge \sqrt{n}. $$

 Proof by Induction:
         Let P(n) denote    1/ 1 + 1/ 2 +  … + 1/ n   n
 Base Case: n = 1, P(1) = 1/1  1
  The base cases holds true for this case since the inequality for P(1) holds true.

  Inductive Hypothesis: For every n = k > 0 for some integer k 
  P(k) = 1/ 1 + 1/ 2 +  … + 1/ k   k, p(k) holds true for any integer k

 Inductive Step:
          P(k + 1)) = 1/ 1 + 1/ 2 +  … + 1/ k  + 1/ (k + 1)   k + (k+1)
         k + (k+1) > (k+1)   (this is where I got stuck)

Answer 1

Suponiendo que $\sum_{i=1}^n \frac{1}{\sqrt{i}} \geq \sqrt{n}$ tenemos \begin{align*} \sum_{i=1}^{n+1} \frac{1}{\sqrt{i}} &= \sum_{i=1}^n \frac{1}{\sqrt{i}} + \frac{1}{\sqrt{n+1}} \\ &\geq \sqrt{n} + \frac{1}{\sqrt{n+1}} \\ &= \frac{\sqrt{n^2+n} + 1}{\sqrt{n+1}} \\ &\geq \frac{\sqrt{n^2} + 1}{\sqrt{n+1}} = \frac{n+1}{\sqrt{n+1}} = \sqrt{n+1}. \end{align*}

Answer 2

Sugerencia: Utilice $\frac{1}{\sqrt{k+1}}\geq\sqrt{k+1}-\sqrt{k}$ en el paso inductivo.

Answer 3

Decir que es cierto para $n=k$ En otras palabras $$1/\sqrt{1}+1/\sqrt{2}+\cdots+1/\sqrt{k}\geq\sqrt{k}$$ entonces para $n=k+1$ $$1/\sqrt{1}+1/\sqrt{2}+\cdots+1/\sqrt{k}+1/\sqrt{k+1}\geq \sqrt{k}+1/\sqrt{k+1}=(\sqrt{k^2+k}+1)/\sqrt{k+1}\geq(\sqrt{k^2}+1)/\sqrt{k+1}=(k+1)/\sqrt{k+1}=\sqrt{k+1}$$

Answer 4

Usted sabe que

$$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{k}} \geq \sqrt{k}$$$$

y quieren demostrarlo:

$$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{k+1}} \geq \sqrt{k+1}$$$$

Añadir $\sqrt{k+1}-\sqrt{k}$ a ambos lados de la primera desigualdad, se obtiene

$$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{k}} +\sqrt{k+1}-\sqrt{k}\geq \sqrt{k+1}$$

Pero:

$$\sqrt{k+1}-\sqrt{k}=\frac{(\sqrt{k+1}+\sqrt{k})(\sqrt{k+1}-\sqrt{k})}{(\sqrt{k+1}+\sqrt{k})}=\frac{k+1-k}{(\sqrt{k+1}+\sqrt{k})}=\frac{1}{(\sqrt{k+1}+\sqrt{k})}\leq \\ \leq \frac{1}{\sqrt{k+1}}$$

Así que:

$$\sqrt{k+1} \leq \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{k}} +\sqrt{k+1}-\sqrt{k} \leq \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{k+1}}$$