For a positive integern, let

an=1+(1/2)+(1/3)+..........+(1/(2^n -1))

then

a) a200<=100

b) a200 >100

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- Oct 11th 2009, 11:15 AMIIT 2010analysis problem
__For a positive integer____n____, let__

__an____=1+(1/2)+(1/3)__+..........+(1/(2^n -1))

then

a) a200<=100

b) a200 >100

- Oct 12th 2009, 08:47 AMEnrique2
Here it is underlying the original proof of the divergence of the harmonic series.

Define

$\displaystyle s_1=1>1/2$

$\displaystyle s_2=1/2$

$\displaystyle s_3=1/3+1/4>2(1/4)=1/2$

$\displaystyle s_4 =1/5+1/6+1/7+1/8>4(1/8)=1/2$

$\displaystyle s_5=1/9+\cdots+1/16>8(1/16)=1/2$

$\displaystyle

s_n=(1/(2^{n-2}+1)+\cdots+1/2^{n-1}>(2^{n-1}-2^{n-2})1/2^{n-1}=1/2

$

$\displaystyle

a_n=s_1+\cdots+s_{n}>n(1/2)

$

Now it follows that the harmonic series diverges, and also that your correct answer is b)!