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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.

AvZ
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Rachel
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  • What happens to the left-hand side when you put each of the brackets over a single denominator? – Paradox Mar 08 '15 at 17:27
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    Hint: $(1 + \frac{1}{j}) = \frac{j+1}{j}$ for $j >0.$ – Geoff Robinson Mar 08 '15 at 17:27
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    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. – Timbuc Mar 08 '15 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.

Workaholic
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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.

Tim Raczkowski
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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.

Hasan Saad
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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!

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

mvw
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