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# The sequence S1, S2, S3..., Sn ... is such that Sn= 1/n - 1/(n+1). If

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The sequence S1, S2, S3..., Sn ... is such that Sn= 1/n - 1/(n+1). If [#permalink]

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12 Jul 2011, 05:00
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The sequence S1, S2, S3..., Sn ... is such that Sn= 1/n - 1/(n+1). If k is a positive integer, is the sum of the first k terms of the sequence greater than 9/10?

(1) k > 10
(2) k < 19

OPEN DISCUSSION OF THIS QUESTION IS HERE: the-sequence-s1-s2-s3-sn-is-such-that-sn-1-n-167843.html
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Re: The sequence S1, S2, S3..., Sn ... is such that Sn= 1/n - 1/(n+1). If [#permalink]

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12 Jul 2011, 12:51
First see the sum of first 3 terms: (1 - 1/2) + (1/2 - 1/3 ) + (1/3 - 1/4) = 3/4
similarly first sum of first 2 terms: 2/3
therefore it will be of the form, sum of first n terms = n/n+1
since this is a proper fraction, adding one to Numerator and denominator increases the overall value.
consider stmt 1: k > 10; the sum of first 10 terms is = 10/11 which is (9+1)/(10+1) that means greater than 9/10.

but in stmt 2: sum of fist 9 (or fewer terms) is less than 9/10 and sum of terms more than 9 is greater than 9/10.

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Re: The sequence S1, S2, S3..., Sn ... is such that Sn= 1/n - 1/(n+1). If [#permalink]

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14 Jul 2011, 06:06
I could not do this in 2mins. picked a wrong strategy.. damn

Sn = 1/n - 1/(n+1)

Sum of first k terms = S1 + S2 ... Sk = 1/1 -1/2 + 1/2 - 1/3 ... + 1/(k-1) - 1/k + 1/k - 1(k+1) = 1 - 1/(k+1) = k/(k+1)

St#1 : k>10 , assume k=11, then Sum = 11/12, which is > 9/10, (also subsequent sums for k = 12, 13, 14 etc. will be greater that 9/10). Suff. there possible options AD

St #2: k<19, assume k=2, then Sum = 2/3, which is < 9/10 and we already saw that for k=11, sum > 9/10 so not suff. eliminate D => A.
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Re: The sequence S1, S2, S3..., Sn ... is such that Sn= 1/n - 1/(n+1). If [#permalink]

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25 Nov 2016, 13:53
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Re: The sequence S1, S2, S3..., Sn ... is such that Sn= 1/n - 1/(n+1). If [#permalink]

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25 Nov 2016, 23:19
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The sequence s1, s2, s3, ..., sn, ... is such that $$S_n=\frac{1}{n} - \frac{1}{n+1}$$ for all integers $$n\geq{1}$$. If k is a positive integer, is the sum of the first k terms of the sequence greater than 9/10?

Given: $$s_n=\frac{1}{n}-\frac{1}{n+1}$$ for $$n\geq{1}$$. So:
$$s_1=1-\frac{1}{2}$$;
$$s_2=\frac{1}{2}-\frac{1}{3}$$;
$$s_3=\frac{1}{3}-\frac{1}{4}$$;
...

If you sum the above 3 terms you'll get: $$s_1+s_2+s_3=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})=1-\frac{1}{4}$$ (everything but the first and the last numbers will cancel out). So the sum of first $$k$$ terms is fgiven by the formula $$sum_k=1-\frac{1}{k+1}$$.

Question: is $$sum_k=1-\frac{1}{k+1}>\frac{9}{10}$$? --> is $$\frac{k}{k+1}>\frac{9}{10}$$? --> is $$k>9$$?

(1) k > 10. Sufficient.
(2) k < 19. Not sufficient.

OPEN DISCUSSION OF THIS QUESTION IS HERE: the-sequence-s1-s2-s3-sn-is-such-that-sn-1-n-167843.html
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Re: The sequence S1, S2, S3..., Sn ... is such that Sn= 1/n - 1/(n+1). If   [#permalink] 25 Nov 2016, 23:19
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