# Matematik B

Put $$b_n=a_1 + a_2 + \ldots + a_n$$. Since $$a_{n+1}=b_{n+1} - b_{n}$$ the $$b_i$$:s will be defined recursively by

$$b_1=2, , b_{n+1} = b_{n} + \frac{n}{b_n}.$$

We want to prove that $$a_{2009} \gt 0.9995$$. Translating to the $$b_i$$:s we get the equivalent condition

$$b_{2009}-b_{2008} = \frac{2008}{b_{2008}} \gt 0.9995 \Leftrightarrow b_{2008} \lt \frac{2008}{0.9995}.$$

Now $$2009 \lt \frac{2008}{0.9995}$$ so if we can prove that $$b_{2008} \leq 2009$$ we are done.

We propose that $$b_{n} \leq n+1$$ for $$n = 1,2,\ldots$$. This is clearly true for $$n=1$$. Suppose that $$b_{m} \leq m+1$$ for some $$m$$. Since

$$b_{m+1} = b_{m} + \frac{m}{b_{m}}$$

We are led to consider the (maximum of) the function $$f(x) = x + \frac{m}{x}$$ for $$2 \leq x \leq m+1$$. It's easy to verify that $$f$$ has a minimum for $$x=\sqrt{m}$$, declines to the left of this value, and grows to the right. Hence $$f$$ has it's maximum in one of the endpoints of the interval.

Since $$2 \leq b_m \leq m+1$$ $$b_i$$ is obviously an increasing sequence, since $$a_i \gt 0$$ for all $$i$$, so $$2 \leq b_m$$) we get

$$b_{m+1} = b_{m} + \frac{m}{b_{m}} \leq \textrm{max}(2+\frac{m}{2}, m+1 + \frac{m}{m+1}) = m+1 +\frac{m}{m+1} \lt m+2$$

and the proposition is proved (by induction).

In particular $$b_{2008} \leq 2008+1=2009$$.