Last visit was: 18 Jul 2024, 19:53 It is currently 18 Jul 2024, 19:53
Toolkit
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

# The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1)

SORT BY:
Tags:
Show Tags
Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 94407
Own Kudos [?]: 642112 [223]
Given Kudos: 85997
Math Expert
Joined: 02 Sep 2009
Posts: 94407
Own Kudos [?]: 642112 [138]
Given Kudos: 85997
Math Expert
Joined: 02 Sep 2009
Posts: 94407
Own Kudos [?]: 642112 [61]
Given Kudos: 85997
Math Expert
Joined: 02 Sep 2009
Posts: 94407
Own Kudos [?]: 642112 [38]
Given Kudos: 85997
Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1) [#permalink]
19
Kudos
19
Bookmarks
SOLUTION

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.

Intern
Joined: 20 Jul 2010
Posts: 43
Own Kudos [?]: 144 [23]
Given Kudos: 32
Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1) [#permalink]
12
Kudos
11
Bookmarks
(1) Take K as 11.
So, Sum = S1 + S2 + S3 + S4 + S5+ S6 + S7 + S8 + S9 + S10 + S11
Where,
S1 = 1 - (1/2)
S2 = (1/2) - (1/3)
S3 = (1/3) - (1/4) ...
S11 = (1/11) - (1/12)

Which implies, Sum = 1 - (1/12) /* The terms like +1/2, -1/2, +1/3, -1/3 will be added to zero. Only the first and last numbers remains */
==> Sum = 1 - 0.0XXXX > 9/10

If you take k as 12, the SUM = 1 - (1/13) which is again > 0.9 Hence SUFFICIENT

(2) K < 19
Consider K as 2. Then the sum is = 1 - (1/2) + (1/2) - (1/3) = 1 - (1/3) which is Less than 9/10
Consider K as 11, Then the sum is greater than 9/10 /* We already proved this in (1) above */
Hence (2) is In Sufficient
General Discussion
Math Expert
Joined: 02 Sep 2009
Posts: 94407
Own Kudos [?]: 642112 [1]
Given Kudos: 85997
Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1) [#permalink]
1
Bookmarks
Bunuel wrote:
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?

(1) k > 10

(2) k < 19

Data Sufficiency
Question: 107
Category: Arithmetic Sequences
Page: 160
Difficulty: 650

The Official Guide For GMAT® Quantitative Review, 2ND Edition

Similar question to practice: https://gmatclub.com/forum/the-numbers- ... 06213.html
Manager
Joined: 31 Mar 2019
Posts: 87
Own Kudos [?]: 57 [0]
Given Kudos: 105
Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1) [#permalink]
Bunuel wrote:
SOLUTION

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.

Hi Bunuel

I understood your point but I approached the problem in the following way :

Point 1: I saw the word “Sequence” — that’s why I have written the formula of Sum of n integers as :
Sum of first n integers in s sequence =
n/2( 1st term + Last term )

We have to prove that : Sum of 1st K terms in the sequence is > 9/10 i.e greater than 0.9

1) as per the first statement , K>10

So I took k’s value as 11 to check whether the sun is greater than 0.9 or not ,

Sum (11 integers ) = 11/2{ 1/2 + [(1/11)-1/12] }
Which gives something around 5.6
So , it’s not true

Again I took k =100
Sum (100 integers ) =100/2{1/2 + [1/100-1/101]}
The answer is something around 55
Now it’s true

That’s why A is not sufficient

Plz guide me regarding this approach for which I was pretty confident initially

Posted from my mobile device
Math Expert
Joined: 02 Sep 2009
Posts: 94407
Own Kudos [?]: 642112 [3]
Given Kudos: 85997
Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1) [#permalink]
1
Kudos
2
Bookmarks
LeenaSai wrote:
Bunuel wrote:
SOLUTION

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.

Hi Bunuel

I understood your point but I approached the problem in the following way :

Point 1: I saw the word “Sequence” — that’s why I have written the formula of Sum of n integers as :
Sum of first n integers in s sequence =
n/2( 1st term + Last term )

We have to prove that : Sum of 1st K terms in the sequence is > 9/10 i.e greater than 0.9

1) as per the first statement , K>10

So I took k’s value as 11 to check whether the sun is greater than 0.9 or not ,

Sum (11 integers ) = 11/2{ 1/2 - [(1/11)-1/12] }
Which gives something around 5.6
So , it’s not true

Again I took k =100
Sum (100 integers ) =100/2{1/2-[1/100-1/101]}
The answer is something around 55
Now it’s true

That’s why A is not sufficient

Plz guide me regarding this approach for which I was pretty confident initially

Posted from my mobile device

The formula you quote, is for evenly spaced sets (aka arithmetical progression), and cannot be applied to all sequences. The sequence at hand is 1/2, 1/6, 1/12, 1/20, ... As you can see this is not an arithmetic progression, so you cannot use the formula you used.

12. Sequences

For other subjects:
ALL YOU NEED FOR QUANT ! ! !

Hope it helps.
Tutor
Joined: 17 Jul 2019
Posts: 1304
Own Kudos [?]: 1745 [1]
Given Kudos: 66
GMAT 1: 780 Q51 V45
GMAT 2: 780 Q50 V47
GMAT 3: 770 Q50 V45
Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1) [#permalink]
1
Kudos
Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
Intern
Joined: 31 Mar 2021
Posts: 21
Own Kudos [?]: 4 [0]
Given Kudos: 290
Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1) [#permalink]
avigutman wrote:
Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1

Hi
I don’t understand when we put k in expression Sn we solve to get 1/k(k+1)

We have to take LCM and cancel the rest. Which will result in 1/k - 1/k+1 = 1/k(1+k)

Correct me and help me with correct reasoning here please.

Posted from my mobile device
Tutor
Joined: 17 Jul 2019
Posts: 1304
Own Kudos [?]: 1745 [1]
Given Kudos: 66
GMAT 1: 780 Q51 V45
GMAT 2: 780 Q50 V47
GMAT 3: 770 Q50 V45
Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1) [#permalink]
1
Kudos
iamvishnu wrote:
I don’t understand when we put k in expression Sn we solve to get 1/k(k+1)

We have to take LCM and cancel the rest. Which will result in 1/k - 1/k+1 = 1/k(1+k)

Correct me and help me with correct reasoning here please.

Yes, your algebra is correct iamvishnu. But that's not a way to solve this problem in 2 minutes, IMHO.
Plugging in n=1, then n=2, etc (keeping the expression as it was, without doing the algebra) shows a pattern in which consecutive terms cancel one another, as I show in the video.
Manager
Joined: 08 Aug 2022
Posts: 75
Own Kudos [?]: 34 [1]
Given Kudos: 182
Location: Morocco
Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1) [#permalink]
1
Kudos
Bunuel wrote:
SOLUTION

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.

Hi Bunuel

Could you please explain to me the move highlighted in red? How did you move from is k/k+1>9/10 to is k>9?
Math Expert
Joined: 02 Sep 2009
Posts: 94407
Own Kudos [?]: 642112 [1]
Given Kudos: 85997
Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1) [#permalink]
1
Kudos
HoudaSR wrote:
Bunuel wrote:
SOLUTION

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.

Hi Bunuel

Could you please explain to me the move highlighted in red? How did you move from is k/k+1>9/10 to is k>9?

$$1-\frac{1}{k+1}>\frac{9}{10}$$;

$$\frac{(k+1) - 1}{k+1}>\frac{9}{10}$$;

$$\frac{k}{k+1}>\frac{9}{10}$$;

$$10k > 9(k+1)$$;

$$k>9$$.

Hope it helps.
Manager
Joined: 08 Aug 2022
Posts: 75
Own Kudos [?]: 34 [0]
Given Kudos: 182
Location: Morocco
Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1) [#permalink]
Bunuel wrote:
HoudaSR wrote:
Bunuel wrote:
SOLUTION

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.

Hi Bunuel

Could you please explain to me the move highlighted in red? How did you move from is k/k+1>9/10 to is k>9?

$$1-\frac{1}{k+1}>\frac{9}{10}$$;

$$\frac{(k+1) - 1}{k+1}>\frac{9}{10}$$;

$$\frac{k}{k+1}>\frac{9}{10}$$;

$$10k > 9(k+1)$$;

$$k>9$$.

Hope it helps.

It is clear. Thank you Bunuel for your prompt response.
Non-Human User
Joined: 09 Sep 2013
Posts: 34013
Own Kudos [?]: 852 [0]
Given Kudos: 0
Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1) [#permalink]
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1) [#permalink]
Moderator:
Math Expert
94407 posts