MHB Proving Induction: (1+1/n+1) to 2-1/n+1

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The discussion focuses on proving the identity involving the product of terms (1 + 1/(n+j)) for j from 1 to n, equating it to 2 - 1/(n+1) using mathematical induction. The base case for n=1 is confirmed as true, establishing a foundation for the induction hypothesis. The process involves manipulating the product and re-indexing to demonstrate that the induction step holds for k+1. An alternative approach is suggested, simplifying the identity by expressing it in terms of products, which also confirms the equality. The initial confusion arose from the lack of clarity regarding the product notation in the original question.
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Show using induction that
(1 + 1 / n + 1).(1 + 1 / n + 2). ... . (1 + 1 / n + n) = 2 - 1 / n + 1, n >= 1.

I've tried everything with this question but the right hand side is not the same as the left hand side after substituting k+1 in the place of n, please help.
 
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I have moved this topic, as it is a better fit with discrete mathematics than number theory.

I am assuming (given the lack of bracketing symbols) that you are given to prove:

$$\prod_{j=1}^n\left(1+\frac{1}{n+j} \right)=2-\frac{1}{n+1}$$ where $$n\in\mathbb{N}$$.

The first thing we wish to do is demonstrate the base case $$P_1$$ is true:

$$\prod_{j=1}^1\left(1+\frac{1}{1+j} \right)=2-\frac{1}{1+1}$$

$$1+\frac{1}{1+1}=2-\frac{1}{1+1}$$

$$\frac{3}{2}=\frac{3}{2}$$

Thus, the base case is true.

Next, state the induction hypothesis $P_k$:

$$\prod_{j=1}^k\left(1+\frac{1}{k+j} \right)=2-\frac{1}{k+1}$$

Let's combine the two terms within the product:

$$\prod_{j=1}^k\left(\frac{k+j+1}{k+j} \right)=2-\frac{1}{k+1}$$

Let's pull out the first factor on the left.

$$\frac{k+2}{k+1}\prod_{j=2}^k\left(\frac{k+j+1}{k+j} \right)=2-\frac{1}{k+1}$$

This will allow us to re-index the product and replace $k$ with $k+1$:

$$\frac{k+2}{k+1}\prod_{j=1}^{k-1}\left(\frac{(k+1)+j+1}{(k+1)+j} \right)=\frac{2k+1}{k+1}$$

Next, multiply through by $$\frac{k+1}{k+2}$$:

$$\prod_{j=1}^{k-1}\left(\frac{(k+1)+j+1}{(k+1)+j} \right)=\frac{2k+1}{(k+1)+1}$$

Now, try as your induction step, multiplying by:

$$\prod_{j=k}^{k+1}\left(\frac{(k+1)+j+1}{(k+1)+j} \right)=\frac{2(k+1)+1}{2k+1}$$

What do you find?

Incidentally, there is an easier way to demonstrate the identity is true (if we hadn't been directed to use induction)...let's write the identity as:

$$\prod_{j=1}^n\left(\frac{n+j+1}{n+j} \right)=2-\frac{1}{n+1}$$

Now, we may choose to express this as:

$$\frac{\prod\limits_{j=1}^n\left(n+j+1 \right)}{\prod\limits_{j=1}^n\left(n+j \right)}=2-\frac{1}{n+1}$$

$$\frac{\prod\limits_{j=2}^{n+1}\left(n+j \right)}{(n+1)\prod\limits_{j=2}^{n}\left(n+j \right)}=2-\frac{1}{n+1}$$

$$\frac{(n+(n+1))\prod\limits_{j=2}^{n}\left(n+j \right)}{(n+1)\prod\limits_{j=2}^{n}\left(n+j \right)}=2-\frac{1}{n+1}$$

$$\frac{2n+1}{n+1}=2-\frac{1}{n+1}$$

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

$$2-\frac{1}{n+1}=2-\frac{1}{n+1}$$
 
The question did not include the product sign was why I couldn't figure it out, thank u so much for your help.
 
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