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if 1/(n+1)<1/31+1/32+1/33<1/n, n=? 9 10 11 12 13 how

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Joined: 11 May 2008
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Kudos [?]: 216 [0], given: 0

if 1/(n+1)<1/31+1/32+1/33<1/n, n=? 9 10 11 12 13 how [#permalink]

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New post 11 Aug 2008, 02:13
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if 1/(n+1)<1/31+1/32+1/33<1/n, n=?
9
10
11
12
13
how does one solve such problems.. am not very good at convergent/divergent problems..
some body help me to get the concepts right of such problems....plss

Kudos [?]: 216 [0], given: 0

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Posts: 432

Kudos [?]: 164 [0], given: 1

Re: SERIES... [#permalink]

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New post 11 Aug 2008, 02:32
arjtryarjtry wrote:
if 1/(n+1)<1/31+1/32+1/33<1/n, n=?
9
10
11
12
13
how does one solve such problems.. am not very good at convergent/divergent problems..
some body help me to get the concepts right of such problems....plss


1/31 + 1/32 + 1/33 ~= 0.032 + 0.031 + 0.03 ~= 0.094

Therefore n = 10 , since 0.094 lies between 1/11 and 1/10

Kudos [?]: 164 [0], given: 1

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Kudos [?]: 1910 [1], given: 6

Re: SERIES... [#permalink]

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New post 11 Aug 2008, 05:44
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Expert's post
arjtryarjtry wrote:
if 1/(n+1)<1/31+1/32+1/33<1/n, n=?
9
10
11
12
13
how does one solve such problems.. am not very good at convergent/divergent problems..
some body help me to get the concepts right of such problems....plss


This can be answered by estimating, and using a common denominator. Notice:

1/33 + 1/33 + 1/33 < 1/31 + 1/32 + 1/33 < 1/30 + 1/30 + 1/30
3/33 < 1/31 + 1/32 + 1/33 < 3/30
1/11 < 1/31 + 1/32 + 1/33 < 1/10

So n should be 10.
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Kudos [?]: 1910 [1], given: 6

Re: SERIES...   [#permalink] 11 Aug 2008, 05:44
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