# 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

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

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

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

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

(3)$b_{m+1} = b_{m} + \frac{m}{b_{m}}$

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

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

and the proposition is proved (by induction).

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