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$$\left( 1 + \frac{1}{2} \right) \left( 1 + \frac{1}{3} \right) \left( 1 + \frac{1}{4} \right) ... \left( 1 + \frac{1}{99} \right) $$

Of course I could do it factor for factor but there has got to be a more efficient way. Does anyone know it?

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$1 + \frac{1}{n} = \frac{n+1}{n}$ –  Daniel Fischer Jun 8 at 20:59

4 Answers 4

Write it as $$ \frac{3}{2}\cdot \frac{4}{3} \cdot \frac{5}{4}\cdots \frac{100}{99}. $$

Now things cancel out.

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We have

$$\left( 1 + \frac{1}{2} \right) \left( 1 + \frac{1}{3} \right) \left( 1 + \frac{1}{4} \right) ... \left( 1 + \frac{1}{99} \right) =\frac32\times\frac43\cdots\frac{100}{99}=\frac{100}2=50$$ by telescoping.

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Just for fun, there is a neat way to estimate products like this.

Recall that, $e^x \geq 1+x $ if $x \geq 0 $. Exploiting this we write,

$$ \left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{3}\right)\cdots\left(1+\frac{1}{99} \right) \leq e^{1/2+1/3+\cdots 1/99} =e^{H_{99}-1} $$

$H_{99}$ is the 99'th harmonic number which can be approximated using an identity due to Euler, $ \lim_{n\rightarrow \infty} H_n - \ln(n) = \gamma \approx .5772 $. Using this we estimate $H_{99}=\ln(99)+.5772$

$$ \left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{3}\right)\cdots\left(1+\frac{1}{99} \right) \leq e^{\ln(99)+.5772-1} = 99e^{.5772-1}\approx 99(.5772)\approx 57 $$

Which is within 14% of the actual answer. This can be improved if we don't approximate the first couple of factors with an exponential since this replacement contributes the greatest source of error (e.g. leave the $(1+1/2)$ term alone). There are also some better estimates of the harmonic numbers (e.g. Donald Knuth's "Concrete Mahtematics" equation 6.66) but these don't yield significant improvements.

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If you don't like the way mathematicians immediately rewrite fractions as $3(+)\frac17$ to $\frac{22}7$ (which suggests replacing $1+\frac1n$ by $\frac{n+1}n$), you might like the following approach. Contemplate the sequence $$ 2,3,4,5,6,7,\ldots,100. $$ The second term is $\frac12$ times larger than the first, then the third is $\frac13$ times larger than the second (so that it is $(1+\frac12)(1+\frac13)$ times as large as the first term), then the fourth term is $\frac14$ times larger again, etc. Now how many times as large is the final term as the first one?

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Hey that is a pretty cool way of looking at the problem. –  Spencer Jun 9 at 17:18

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