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

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

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.

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

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).
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I tried to solve it like below:
S= 1+ 1/2^2 + 1/3^2 + 1/4^2 + ...
S= 1 + {(1/6^2) * 9} [considering 1/6^2 as the the middle term in the remaining 9 terms. And multiplying it by number of terms. To get an approx value]
S= 1 + 1/4
S= 1.25
Hence (E)

Posted from my mobile device

[quote="Bunuel"]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)
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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

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

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
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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:
\(S < 1 + \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} ... + \frac{1}{(n-1)} - \frac{1}{n} = 2 - \frac{1}{n} < 2\)

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.
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This can be solved in two ways:
1. This is actually a famous Basel equation :
i.e : 1/1^2 + 1/2^2+1/3^2+... infinity = (3.14(pie))^2/6 = 1.6432 which is <2.

2. summation :
This can be written as :
1/n^2 = 1+ 1/n(n-1) , here if we out n = 10 then also it is less than 2.

Correct option : E (S<2)

- the greater the value in denominators, less is the value of fraction, which makes 1/\(2^{2}\) higher value of given fraction
- simple 1 + 0.25 + (all small value less than 0.11)
- 1.45*** something will be the value
- (, which ) is less than 2

The given sequence is:
1, 1/4, 1/9, 1/16, ...

Compare the given sequence to the following geometric sequence:
1, 1/2, 1/4, 1/8, ...

It's clear that every term in this geometric sequence is MORE than the corresponding term of the given sequence.

We can easily find the sum of ALL infinite terms of the geometric sequence using the formula
S = a * (1/(1-r)),
where
a= first term = 1, and
r = common ratio = 1/2
The sum is = 2

Therefore the sum of the given sequence must be less than 2.

Answer choice: E

IanStewart Any easy way to solve this?

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.
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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

Answer: Option E

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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 \(3.5 > S > 1.09\)
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How would you come to a definite answer using this range ?
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