What is the value of the expression: $(1+\frac 12)(1+\frac 13)(1+\frac 14)...(1+\frac {1}{2004})(1+\frac {1}{2005})$? What is the value of the expression: $(1+\frac 12)(1+\frac 13)(1+\frac 14)...(1+\frac {1}{2004})(1+\frac {1}{2005})$? This question appeared on the UKMT senior maths challenge 2005, and I can't find an adequate method for a solution given its context i.e. without a calculator, and it should only take 3 to 4 minutes. I'm sure I'm missing something obvious.
Thanks!
 A: It can be seen directly that the finite products give a telescoping product as was demonstrated in the other answers. If this is unclear, another approach is to apply a logarithm to the product and work out the resulting series:
Let $L_{2005} = \left(1+\frac12\right)\left( 1+ \frac13 \right) \cdots \left( 1 + \frac1{2005}\right)$.
Then $$\ln(L_{2005}) = \sum_{n=2}^{2005} \ln\left( \frac{n+1}{n}\right)=\sum_{n=2}^{2005} \left( \ln(n+1) - \ln(n) \right).$$
This gives a telescoping sum. We can work out what the result should be by working out some smaller cases.
$$\ln(L_{3}) = \ln(3)-\ln(2) + \ln(4) - \ln(3) = \ln(4) - \ln(2)$$
$$\ln(L_4) = \ln(L_3) + (\ln(5) - \ln(4)) = \ln(5) - \ln(2)$$
Thus we see that this satisfies the general form:
$$\ln(L_N) = \ln(N+1) - \ln(2).$$
Thus $\ln(L_{2005}) = \ln(2006) - \ln(2) = \ln(1003)$ and $L_{2005} = 1003$.
Though, it was unnecessary to apply the logarithm to arrive at this result, often logarithms are used on an infinite (or large) product in order to apply our intuitions about the convergence or sums of series.
A: Using Mark Bennet's hint, we have: $$1 + \frac{1}{n} = \frac{n+1}{n}$$
So that your product becomes $$\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} \cdots \frac{2005}{2004} \cdot \frac{2006}{2005} = \frac{2006}{2} = 1003$$
A: Hint: $$1+\frac 1n=\frac {n+1}n$$
A: Hint. One may write
$$
\prod_2^{2005} \left(1+\frac 1k\right)=\prod_2^{2005} \frac {k+1}k
$$ and observe that factors telescope.
