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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.

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  • $\begingroup$ What happens to the left-hand side when you put each of the brackets over a single denominator? $\endgroup$
    – Paradox
    Mar 8, 2015 at 17:27
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    $\begingroup$ Hint: $(1 + \frac{1}{j}) = \frac{j+1}{j}$ for $j >0.$ $\endgroup$ Mar 8, 2015 at 17:27
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    $\begingroup$ 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. $\endgroup$
    – Timbuc
    Mar 8, 2015 at 17:48

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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.

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

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

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

$=n+1$

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.

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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!

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

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