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

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.

Answer: A.

(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

Similar question to practice: https://gmatclub.com/forum/the-numbers- ... 06213.html

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.

Check links below for more.

12. Sequences

• Theory
  Sequences Made Easy - All in One Topic!
  Math : Sequences & Progressions
  Time to Tackle Geometric Progressions!
  
  • Questions
    Special Numbers and Sequences Problems
    DS Questions
    PS Questions

For other subjects:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread

Hope it helps.

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

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.

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.

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