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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 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. 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. 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 = 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?  Tags sequence, series Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post naman Algebra 3 March 6th, 2013 08:22 PM Maths4ever Algebra 4 October 4th, 2012 01:35 AM suomik1988 Algebra 5 December 8th, 2010 07:38 AM 54088 Algebra 1 February 28th, 2010 07:27 AM suomik1988 Calculus 5 December 31st, 1969 04:00 PM

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