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 March 5th, 2013, 02:45 AM #1 Newbie   Joined: Mar 2013 Posts: 12 Thanks: 0 Sequence and series Let Sn= 1 + 1/2 + 1/3 + ....+1/(2^n-1), n belong to N then________ (a) S100<100 (b) S100> 100 (c) S200 = 100 (d) S200 > 200 Answer can be (A) but plz help me in its method so that I can understand it thoroughly. Thanks in advance.
March 5th, 2013, 03:56 AM   #2
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Re: Sequence and series

Quote:
 Originally Posted by naman Let Sn= 1 + 1/2 + 1/3 + ....+1/(2^n-1), n belong to N then________ (a) S100<100 (b) S100> 100 (c) S200 = 100 (d) S200 > 200
This is the harmonic series $\mathcal{H}_{2^n -1}$ which grows almost like $\log(2^n - 1) + \gamma$ with an error bounded by 0 & 0.422, respectively. Hence, by some easy upper bounds, it trivially follows that $\mathcal{S}_{100} < 100$ is the only answer since all of the options above fails to follow the property.

 March 5th, 2013, 04:32 AM #3 Senior Member   Joined: Feb 2013 Posts: 114 Thanks: 0 Re: Sequence and series Can we also go like this : S_n ( of an A.P arithmetic progression ) as the given progression is H.P therefore there reciprocal is in A.P. $S_n= \frac{n}{2}\{2a+(n-1)d\}$ where a is the first term of the series ; d is the common difference; n is the number of terms ; therefore putting these value in above we get : $S_n= \frac{100}{2}\{2 * 1 +(100-10 ) * 1\} = 50(2+90) = 92 * 50 = 4600$ therefore its reciprocal gives you sum of H.P = $\frac{1}{4600} < 100$
 March 5th, 2013, 04:37 AM #4 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Sequence and series As far as I know, harmonic series has no closed form expression, has it?

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