If S = 1 + 1/2^2 + 1/3^2 + 1/4^2 + 1/5^2 + 1/6^2 + 1/7^2 + 1/8^2 + 1/9

Question

If  S = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \frac{1}{7^2} + \frac{1}{8^2} + \frac{1}{9^2} + \frac{1}{10^2} , which of the following is true?

A. S > 3
B. S = 3
C. 2 < S < 3
D. S = 2
E. S < 2

(PS06243)

BrentGMATPrepNow · Accepted Answer

S = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \frac{1}{7^2} + \frac{1}{8^2} + \frac{1}{9^2} + \frac{1}{10^2}  = 1 + 1/4 + 1/9 + 1/16 + 1/25 + 1/36 + . . . .

Now, let's convert a few of the fractions to decimal approximations...
S = 1 + 0.25 + 0.11 + 0.06 +  0.04  + . . .

Add the first 4 values....
S = 1.42 +  0.04  + . . .

IMPORTANT: Notice that each fraction is  less than  the fraction before it 
So, all of the 5 decimals after  0.04  will be  less than   0.04 
So, if we replace all of those 5 decimals with  0.04 , our new sum will be  greater than  the original sum

So: S < 1.42 +  0.04  +  0.04  +  0.04  +  0.04  +  0.04  +  0.04

Simplify: S < 1.42 +  0.24  
Simplify: S < 1.66

Answer: E

Cheers,
Brent

ganand · Answer

Hi,

For simplicity, let's break the given series in three parts as follows:

S_{1} = 1 ,

S_{2} = \frac{1}{2^2} + \frac{1}{3^2} , and

S_{3} = \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \frac{1}{7^2} + \frac{1}{8^2} + \frac{1}{9^2} + \frac{1}{10^2}

In  S_{2}  larger term is  \frac{1}{2^2} , and there are two terms. Hence,

S_{2} = \frac{1}{2^2} + \frac{1}{3^2} < 2*\frac{1}{4} = \frac{1}{2}  --- (1)

Similarly, in series  S_{3}  the largest term is  \frac{1}{4^2} = \frac{1}{16}  and there are total 7 terms. Hence,

S_{3} = \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \frac{1}{7^2} + \frac{1}{8^2} + \frac{1}{9^2} + \frac{1}{10^2} < 7* \frac{1}{16} < 8*\frac{1}{16} = \frac{1}{2}  --- (2)

S = S_{1} + S_{2} + S_{3} < 1 + \frac{1}{2} + \frac{1}{2} = 2

Hence  S < 2 . Answer (E).

Thanks.

MartyTargetTestPrep · Answer

S = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \frac{1}{7^2} + \frac{1}{8^2} + \frac{1}{9^2} + \frac{1}{10^2}

GMAT quant questions tend to be designed to be answered efficiently via the use of hacks. So, let's start hacking.

Start off with the obvious:  1 + \frac{1}{2^2} = 1 \frac{1}{4}

It becomes apparent that to get to 2 we need 3/4 more, and those other fractions look pretty small. They probably won't get us to 2.

If we can prove that 1 1/4 + (the sum of the rest of those fractions) < 2, then the correct answer will be S < 2 and we'll be done.

The next fraction is  \frac{1}{3^2} .

That is effectively  \frac{1}{9} .

We have 1 1/4 + 1/9.

Without adding, we know that 1 1/4 + 1/9 < 1 1/2.

The next seven are  \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \frac{1}{7^2} + \frac{1}{8^2} + \frac{1}{9^2} + \frac{1}{10^2} .

Since 8 x 1/16 = 1/2, and the first fraction of this set of seven fractions is 1/16 and all the rest are smaller, this set has to add up to less than 1/2.

So, S = 1 1/4 + 1/9 + (a number < 1/2).

S < 2.

The correct answer is (E).

PKN · Answer

If  S = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \frac{1}{7^2} + \frac{1}{8^2} + \frac{1}{9^2} + \frac{1}{10^2} , which of the following is true?

A. S > 3
 
B. S = 3
 
C. 2 < S < 3
 
D. S = 2
 
E. S < 2

We have the terms after  \frac{1}{5^2}  are close to zero. Hence the sum of the terms henceforth  \frac{1}{5^2}  can be approximated to zero.
Hence  S = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \frac{1}{7^2} + \frac{1}{8^2} + \frac{1}{9^2} + \frac{1}{10^2}  reduces to
 S=1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} 
Or,  S=\frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} =1+0.25+0.11+0.0625+0.04=1.4625<2
So S<2.

Ans (E)

DhruvS · Answer

I tried to solve it like below:
S= 1+ 1/2^2 + 1/3^2 + 1/4^2 + ...
S= 1 + {(1/6^2) * 9}  
S= 1 + 1/4
S= 1.25
Hence (E)

Posted from my mobile device

Ved22 · Answer

NO Need of huge calculations done above.

Simple, the entire expression can be generalized as below:

S = 1+ 9*(1/n^2)
 = n^2+9/n^2
so finally S = n^2+9/n^2

now put n=2 so S= 4 equation equals to 1.56 , put n=9 so 1.11

So, merely in all S< 2 so the option E is valid

shreyagupta1401 · Answer

Easy way to solve - multiply both sides by 100 and then figure out rough estimation of S. Mine was ~1.6

Target4bschool · Answer

IanStewart  Any easy way to solve this?

IanStewart · Answer

It's an estimation question, so we don't want to do exact calculations. I think once you notice the later terms in the sum are minuscule compared to the early terms, you can quickly become confident the sum will be less than 2. To be sure of that, I'd do something similar to what others did above: we have seven terms from 1/4^2 onwards, and they're all less than or equal to 1/4^2, so they sum to something less than 7/16 < 1/2. So the last seven terms sum to something less than 0.5, and since 1 + 1/4 + 1/9 is clearly less than 1.5 (since 1 + 1/4 + 1/4 is exactly 1.5) we can be sure the overall sum is less than 2.

abhishekpasricha · Answer

I think we cannot take this approach as the terms are not in AP. The approach of multiplying the middle term by the number of terms can be done only when the terms are in AP.

Bunuel  & Experts please let me know if I am incorrect here

Thanks in advance

IanStewart · Answer

Yes, you are completely correct, and it's easy to see why it's a bad idea, even if only estimating, to try to shoehorn a non-arithmetic sequence into an arithmetic sequence formula. I'm not sure why that solution separated out the first term, but if it hadn't, and if it had simply multiplied the median (the average of 1/5^2 and 1/6^2) by the number of terms, 10, that method would have estimated the sum to be roughly 0.34. That's nowhere close to the right answer -- it's far less than just the first term alone. You really can never reliably use arithmetic sequence formulas except in the lone situation when you know you're dealing with an arithmetic sequence.

GMATinsight · Answer

Answer: Option E

Video solution by  GMATinsight

https://www.youtube.com/watch?v=yXo-33jt9JE

errorlogger · Answer

Hi  Bunuel ,  chetan2u  or any other experts,
Can anyone please confirm if I can use this approach:

S = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \frac{1}{7^2} + \frac{1}{8^2} + \frac{1}{9^2} + \frac{1}{10^2} 
Ignoring the first term(1), we can see that the highest term is  \frac{1}{4} , so max value can be 9/4 or 2.5
Similarly, the lowest value can be 9/100 = 0.09
Adding the first term(1), so the range would be  3.5 > S > 1.09

chetan2u · Answer

Yes, that will be the correct way. However, the range you get contains all the options.

So what do we do?
We narrow down the range.

And how do we do it?
By taking the extremes of last 8 terms.
So  S = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \frac{1}{7^2} + \frac{1}{8^2} + \frac{1}{9^2} + \frac{1}{10^2} 
  S = 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \frac{1}{7^2} + \frac{1}{8^2} + \frac{1}{9^2} + \frac{1}{100} 
So  1+\frac{1}{4}+\frac{8}{9}=1+0.25+0.9=2.15  at higher end and  1+\frac{1}{4}+\frac{8}{100}=1+0.25+0.08=1.33  at lower end

You will have to repeat the step again till you get your answer as a range with the range given in option

Nikenproone · Answer

It can be solved easily without guessing:
 \frac{1}{(2^2)} < \frac{1}{(1*2)} = \frac{1}{1} - \frac{1}{2} 
 \frac{1}{(3^2)} < \frac{1}{(2*3)} = \frac{1}{2} - \frac{1}{3} 
 \frac{1}{(4^2)} < \frac{1}{(3*4)} = \frac{1}{3} - \frac{1}{4} 
......
 \frac{1}{(n^2)} < \frac{1}{((n-1)*n)} = \frac{1}{(n-1)} - \frac{1}{n}

Sum them alltogether we have:
 S < 1 + \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} ... + \frac{1}{(n-1)} - \frac{1}{n} = 2 - \frac{1}{n} < 2

avigutman · Answer

Video solution from Quant Reasoning:

https://www.youtube.com/watch?v=HUq7hxxYucs

Tanish1212 · Answer

If we do this as 1 + (consider all 1/100) i.e it will be 9/100 
then 1 + 0.09 = 1.09 which is less than 2, So S<2

pratik511 · Answer

sum of infinite sum of 1 + (1/2)**2 + (1/3)**2 + .... to infinity is Pi squared by 6 = 1.64
so the given number will be less than that. Hence less then 2.

If S = 1 + 1/2^2 + 1/3^2 + 1/4^2 + 1/5^2 + 1/6^2 + 1/7^2 + 1/8^2 + 1/9

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If S = 1 + 1/2^2 + 1/3^2 + 1/4^2 + 1/5^2 + 1/6^2 + 1/7^2 + 1/8^2 + 1/9

Prep Toolkit

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Perfect 805 Score Debrief by Julia

My Rewards