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# 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 \(b_{2008} \leq 2008+1=2009\).