Matematik B

Matematik B

Put b n=a 1+a 2++a nb_n=a_1 + a_2 + \ldots + a_n. Since a n+1=b n+1b na_{n+1}=b_{n+1} - b_{n} the b ib_i:s will be defined recursively by

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

We want to prove that a 2009>0.9995a_{2009} \gt 0.9995. Translating to the b ib_i:s we get the equivalent condition

(2)b 2009b 2008=2008b 2008>0.9995b 2008<20080.9995. b_{2009}-b_{2008} = \frac{2008}{b_{2008}} \gt 0.9995 \Leftrightarrow b_{2008} \lt \frac{2008}{0.9995}.

Now 2009<20080.99952009 \lt \frac{2008}{0.9995} so if we can prove that b 20082009b_{2008} \leq 2009 we are done.

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

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

we are led to consider the (maximum of) the function f(x)=x+mxf(x) = x + \frac{m}{x} for 2xm+12 \leq x \leq m+1. It’s easy to verify that ff has a minimum for x=mx=\sqrt{m}, declines to the left of this value, and grows to the right. Hence ff has it’s maximum in one of the endpoints of the interval.

Since 2b mm+12 \leq b_m \leq m+1 ((b ib_i) is obviously an increasing sequence, since a i>0a_i \gt 0 for all ii, so 2b m2 \leq b_m) we get

(4)b m+1=b m+mb mtextrmmax(2+m2,m+1+mm+1)=m+1+mm+1<m+2 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 20082008+1=2009b_{2008} \leq 2008+1=2009.