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March 5th, 2013, 03:45 AM   #1
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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.
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March 5th, 2013, 04:56 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 which grows almost like with an error bounded by 0 & 0.422, respectively. Hence, by some easy upper bounds, it trivially follows that is the only answer since all of the options above fails to follow the property.
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March 5th, 2013, 05:32 AM   #3
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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.

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 :

therefore its reciprocal gives you sum of H.P =
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March 5th, 2013, 05:37 AM   #4
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Re: Sequence and series

As far as I know, harmonic series has no closed form expression, has it?
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