Proving a Proposition

Question

This is my first question. I hope it wouldn't be bad received.

I found this question in an old math book and I really got stuck

I have to demonstrate the following propositions : $$ \forall n \in \mathbb{N^*}$$

$1)$ $\qquad $ $ (1- \frac{1}{2^2})(1- \frac{1}{3^2})* ...*(1- \frac{1}{n^2}) = \frac{1}{2} ( \frac{n+1}{n})$

$2)$ $\qquad\displaystyle\sum_{k=1}^{k=n} \frac{1}{\sqrt{k}} \le 2\sqrt{n} $

I hope I asked this question correctly because it's my first attempt.

Answer 1

• Since $1-\dfrac{1}{n^2}=\dfrac{n^2-1}{n^2}=\dfrac{n-1}{n}\cdot\dfrac{n+1}{n}$ for all $n\neq 0$, we have $$\displaystyle\prod_{i=2}^n\left(1-\dfrac{1}{i^2}\right)=\displaystyle\prod_{i=2}^n\left(\dfrac{i-1}{i}\cdot\dfrac{i+1}{i}\right)$$

By an analogue of telescoping series to that of telescoping product (I will leave this as an exericise), we can write $$\displaystyle\prod_{i=2}^n\left(\dfrac{i-1}{i}\cdot\dfrac{i+1}{i}\right)=\dfrac{1}{2}\cdot \dfrac{n+1}{n}$$

• For each $n\in\mathbb{N}$, let $f_n:\{y\in\mathbb{R}:0\le y\le n\}\to\mathbb{R}$ be a function defined by $f_n(x):=\dfrac{1}{\sqrt{x}}$ for all $x\in\{y\in\mathbb{R}:0\le y\le n\}$.

To show that $\displaystyle\sum_{k=1}^{n}\dfrac{1}{\sqrt{k}}\le 2\sqrt{n}$, we can use the lower Riemann sum of $f_n$ and the integral $\displaystyle\int_0^n f_n(x)\mathrm{dx}$.

Let $$\mathcal{F}=\{[i-1,i]:i\in\mathbb{N},1\le i\le n\}$$ be a partition of the interval $[0,n]$.

The lower Riemann sum $S$ of $f_n$ is given by $$S=\displaystyle\sum_{k=1}^{n}\dfrac{1}{\sqrt{k}}$$

We have $$\displaystyle\int_0^n f_n(x)\mathrm{dx}=\displaystyle\int_0^n \dfrac{1}{\sqrt{x}}\mathrm{dx}=2\sqrt{n}$$

Using the fact that $S\le\displaystyle\int_0^n f_n(x)\mathrm{dx}$ we have $$\displaystyle\sum_{k=1}^{n}\dfrac{1}{\sqrt{k}}\le 2\sqrt{n}$$

Answer 2

$(1)$ Put $n=2$ to verify that

$(1-\frac{1}{2^2})=\frac{1}{2}$.$\frac{2+1}{2}$.

Let the result be true for $n=k$ i.e.

$(1-1/2^2).....(1-1/k^2)=\frac{1}{2}.\frac{k+1}{k}$..........$(A)$

It remains to prove that the result is also true for $n=k+1$.

Now, $(1-1/2^2).....(1-1/k^2)(1- 1/(k+1)^2)=\frac{1}{2}.\frac{k+1}{k}\left(1-\frac{1}{(k+1)^2}\right)$....using $(A)$.

$=\frac{1}{2}\frac{k+2}{k+1}$.

Proceed similarly for $(2)$.