# Direct proof for $\left (1+ \frac11\right )\left(1+ \frac12\right )\left(1+ \frac13\right )\cdots\left(1+ \frac1n \right) = n+1$

I have a question where I need to use a direct proof to show that:

$$\left (1+ \frac11\right )\left(1+ \frac12\right )\left(1+ \frac13\right )\cdots\left(1+ \frac1n \right) = n+1$$

I am not allowed to use mathematical induction. I have no idea where to start to prove this, any advice would be appreciated.

• What happens to the left-hand side when you put each of the brackets over a single denominator? Mar 8, 2015 at 17:27
• Hint: $(1 + \frac{1}{j}) = \frac{j+1}{j}$ for $j >0.$ Mar 8, 2015 at 17:27
• BTW, all the answers below use induction, more or less hidden, but very present when writing down the product. I can't see how this can formally be proved without any form of induction. Mar 8, 2015 at 17:48

## 5 Answers

Hint: Each term of the form $1+\tfrac1n$ can be written as $\tfrac{n+1}n$, hence your product is equal to :

$$\frac{\color{brown}2}{1}\frac{\color{royalblue}3}{\color{brown}2}\frac{\color{green}4}{\color{royalblue}3}\cdots\frac{\color{#C00}n}{\color{darkorange}{n-1}}\frac{n+1}{\color{#C00}n}.$$

Look at the mass cancellation.

• I think bright colours would look better :)
– AvZ
Mar 8, 2015 at 17:38
• @AvZ I do agree, but I didn't have access to color-pickers when I wrote my answer. :-) Mar 11, 2015 at 19:41

Hint: use the fact that $1+\frac1n=\frac{n+1}n$. For example:

$$(1+\frac11)(1+\frac12)(1+\frac13)=\frac21\cdot\frac32\cdot\frac43=4$$ by cancellation.

$(1+\frac{1}{1})(1+\frac{1}{2})\ldots(1+\frac{1}{n})$

$=\frac{2}{1}\frac{3}{2}\frac{4}{3}\ldots\frac{n+1}{n}$

$=n+1$

I don't think that this is as rigorous as it should be, but as you can see, each numerator cancels the denominator after it, and thus we'd get the wanted result.

Write the product as: $$\left(\frac{2}{1}\right)\left(\frac{3}{2}\right)\left(\frac{4}{3}\right)\dots\left(\frac{n+1}{n}\right)$$ You can simplify it!

$$P = \frac{2}{1}\frac{3}{2}\frac{4}{3}\cdots\frac{n}{n-1}\frac{n+1}{n} = n+1$$