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# 1+1/2+1/3+.....+1/16 is

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29 Apr 2014, 23:19
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Qs. $$1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{16}$$ is

A) between 1 and 2
B) between 2 and 3
C) between 3 and 4
D) between 4 and 5
E) between 5 and 6

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30 Apr 2014, 00:03
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bagdbmba wrote:
Qs. $$1+1/2+1/3+.....+1/16$$ is

A) between 1 and 2
B) between 2 and 3
C) between 3 and 4
D) between 4 and 5
E) between 5 and 6

I don't think the question is appropriate for GMAT until and unless they want you to do lots of calculations and then approximate. You are not expected to find the sum of harmonic mean.

The sum of harmonic series is approximated by comparison.

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

$$T = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{16} + ...$$

Every term of S is greater than or equal to corresponding term of T.

$$T = 1 + \frac{1}{2} + [\frac{1}{4} + \frac{1}{4}] + [\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}] + [\frac{1}{16} + ...]$$

$$T = 1 + \frac{1}{2} + [\frac{1}{2}] + [\frac{1}{2}] + [\frac{1}{2}] = 3$$

So S must be more than 3.

Now note that max diff between two corresponding terms of S and T is 1/3 - 1/4 = 1/12. There are only 11 terms in S which are more than their corresponding terms in T. So even if we account for all the difference as 11*(1/12) = 11/12 and add it to 3, we will still not get 4.

Hence S must lie between 3 and 4.
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##### General Discussion
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29 Apr 2014, 23:22
Bunuel - MGMAT has a solution, but I'd like to know whether you'd recommend any other way to solve this ?

Thank you.
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03 May 2014, 00:10
VeritasPrepKarishma wrote:
bagdbmba wrote:
Qs. $$1+1/2+1/3+.....+1/16$$ is

A) between 1 and 2
B) between 2 and 3
C) between 3 and 4
D) between 4 and 5
E) between 5 and 6

I don't think the question is appropriate for GMAT until and unless they want you to do lots of calculations and then approximate. You are not expected to find the sum of harmonic mean.

The sum of harmonic series is approximated by comparison.

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

$$T = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{16} + ...$$

Every term of S is greater than or equal to corresponding term of T.

$$T = 1 + \frac{1}{2} + [\frac{1}{4} + \frac{1}{4}] + [\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}] + [\frac{1}{16} + ...]$$

$$T = 1 + \frac{1}{2} + [\frac{1}{2}] + [\frac{1}{2}] + [\frac{1}{2}] = 3$$

So S must be more than 3.

Now note that max diff between two corresponding terms of S and T is 1/3 - 1/4 = 1/12. There are only 11 terms in S which are more than their corresponding terms in T. So even if we account for all the difference as 11*(1/12) = 11/12 and add it to 3, we will still not get 4.

Hence S must lie between 3 and 4.

Hi I did not get, how did you write T terms??
Can anyone please explain? Or is there another way to do this problem???
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03 May 2014, 00:21
1
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bagdbmba wrote:
Qs. $$1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{16}$$ is

A) between 1 and 2
B) between 2 and 3
C) between 3 and 4
D) between 4 and 5
E) between 5 and 6

Sum = $$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}...........$$

Sum = 1 + 0.5 + 0.33 + 0.25 + 0.20 + 0.16 + 0.14 + 0.12 + 0.11 + ..............

Sum = 2.81 + .........

Each of 1/10 and 1/11 would add approximately 0.10 to the above sum, so we can infer that the sum will definitely be greater than 3 but would not go beyond 4

Choice C
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04 May 2014, 22:50
drkomal2000 wrote:
VeritasPrepKarishma wrote:
bagdbmba wrote:
Qs. $$1+1/2+1/3+.....+1/16$$ is

A) between 1 and 2
B) between 2 and 3
C) between 3 and 4
D) between 4 and 5
E) between 5 and 6

I don't think the question is appropriate for GMAT until and unless they want you to do lots of calculations and then approximate. You are not expected to find the sum of harmonic mean.

The sum of harmonic series is approximated by comparison.

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

$$T = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{16} + ...$$

Every term of S is greater than or equal to corresponding term of T.

$$T = 1 + \frac{1}{2} + [\frac{1}{4} + \frac{1}{4}] + [\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}] + [\frac{1}{16} + ...]$$

$$T = 1 + \frac{1}{2} + [\frac{1}{2}] + [\frac{1}{2}] + [\frac{1}{2}] = 3$$

So S must be more than 3.

Now note that max diff between two corresponding terms of S and T is 1/3 - 1/4 = 1/12. There are only 11 terms in S which are more than their corresponding terms in T. So even if we account for all the difference as 11*(1/12) = 11/12 and add it to 3, we will still not get 4.

Hence S must lie between 3 and 4.

Hi I did not get, how did you write T terms??
Can anyone please explain? Or is there another way to do this problem???

This is a standard method of comparing the sum of harmonic series. We compare the sum with another series in which the terms are easy to add and the sum is easy to find. That other series is T. Compare the corresponding terms of S and T - i.e. first term of S with first term of T, second term of S with second term of T and so on. Every term of S will be greater than or equal to the corresponding term of T.

From where do we get T? It has been obtained by mathematicians using the properties of numbers and their powers. If you don't know it, it's highly unlikely that you will be able to think of it in the exam. GMAT doesn't expect you to. The other method is using approximate values of the fractions and adding them but that's way too cumbersome in my opinion. GMAT doesn't do cumbersome.

But if you do understand the method used by me above, as an intellectual exercise, try solving for S = 1 + 1/3 + 1/5 + 1/7 + 1/9 + 1/11 + ... + 1/27
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25 Aug 2014, 03:02
To find : S = 1 + 1/3 + 1/5 + 1/7 + 1/9 + 1/11 + ... + 1/27

Let $$S' = (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} +\frac{1}{6} +\frac{1}{7} +\frac{1}{8}+.....+ \frac{1}{32} )$$

Or, $$S' = ( 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \frac{1}{10} + ... + \frac{1}{32}) + (\frac{1}{3} +\frac{1}{5} + \frac{1}{7} + \frac{1}{9} + \frac{1}{11} + ... + \frac{1}{27}+\frac{1}{29}+\frac{1}{31})$$

Or, $$S' = 1 + (\frac{1}{2} (1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} +\frac{1}{10} + ... + \frac{1}{16})) + (\frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{9} + \frac{1}{11} + ... + \frac{1}{27})+(\frac{1}{29}+\frac{1}{31})$$

Or,$$S' > ( \frac{1}{2}( 1+ 4(\frac{1}{2})) + (1 + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{9} + \frac{1}{11} + ... + \frac{1}{27})+(\frac{1}{29}+\frac{1}{31})$$

Or, $$S' > \frac{3}{2} + S + (\frac{1}{29}+\frac{1}{31})$$

Or, $$S' > \frac{3}{2} + S + \frac{1}{15}$$ [Since, $$\frac{1}{29} +\frac{1}{31} >\frac{2}{30} or \frac{1}{15}$$]

Or, $$S < S' - \frac{3}{2} -\frac{1}{15}$$

Or, $$S < 1 + \frac{5}{2}-\frac{3}{2}-\frac{1}{15}$$ [ Since, $$S ' > 1 + 5(\frac{1}{2})$$ ]

Or. $$S < 2 - \frac{1}{15}$$

Or, $$S < \frac{29}{15}$$

This calculated value is incorrect, as S > 2. Where have I gone wrong?
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27 Aug 2014, 11:30
If this comes on the GMAT, I'd prefer going the decimal addition way. Decimals till 1/10 we should know for the GMAT in my opinion. Quick tenths digits addition till 1/10 should give the required range.
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28 Aug 2014, 22:27
bagdbmba wrote:
Bunuel - MGMAT has a solution, but I'd like to know whether you'd recommend any other way to solve this ?

Thank you.

Could you post the Bunuel's solution? I really want to see it
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29 Nov 2016, 21:33
Outstanding Question.
I picked the wrong Answer as a result of which this is going in my error log.

Here is what i will do the next time i see this Question =>

Let n=1+1/2+1/3+...

No need to know the complicated harmonic sum.
1
0.5
0.33
0.25
0.20
0.16
0.14
0.12
0.11
0.10
0.09
and some more terms

Add them up we get 3
So the sum must be greater than 3.

Abhishek009 You have any tricks for this one ?

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30 Nov 2016, 00:33
bagdbmba wrote:
Qs. $$1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{16}$$ is

A) between 1 and 2
B) between 2 and 3
C) between 3 and 4
D) between 4 and 5
E) between 5 and 6

Here is my solution for this one.

$$\begin{split} A&=1+(\frac{1}{2}+\frac{1}{3})+(\frac{1}{4}+\frac{1}{5})+...+(\frac{1}{14}+\frac{1}{15})+\frac{1}{16}\\ &<1+(\frac{1}{2}+\frac{1}{2})+(\frac{1}{4}+\frac{1}{4})+...+(\frac{1}{14}+\frac{1}{14})+\frac{1}{16}\\ &=1+\frac{2}{2}+\frac{2}{4} +\frac{2}{6} +...+ \frac{2}{14}+\frac{1}{16}\\ &=1+1+(\frac{1}{2}+\frac{1}{3})+(\frac{1}{4}+\frac{1}{5})+(\frac{1}{6}+\frac{1}{7})+\frac{1}{16}\\ &<1+1+(\frac{1}{2}+\frac{1}{2})+(\frac{1}{4}+\frac{1}{4})+(\frac{1}{6}+\frac{1}{6})+\frac{1}{16}\\ &=2+\frac{2}{2}+\frac{2}{4}+\frac{2}{6}+\frac{1}{16}\\ &=2+1+\frac{1}{2}+\frac{1}{3}+\frac{1}{16}\\ &=3 + \frac{43}{48}<4 \end{split}$$

$$\begin{split} A&=1+\frac{1}{2}+(\frac{1}{3}+\frac{1}{4})+...+(\frac{1}{15}+\frac{1}{16})\\ &>1+\frac{1}{2}+(\frac{1}{4}+\frac{1}{4})+...+(\frac{1}{16}+\frac{1}{16})\\ &=1+\frac{1}{2}+\frac{2}{4}+...+\frac{2}{14}+\frac{2}{16}\\ &=(1+\frac{1}{2}+\frac{1}{2})+(\frac{1}{3}+\frac{1}{4})+...+(\frac{1}{7}+\frac{1}{8})\\ &>2+(\frac{1}{4}+\frac{1}{4})+...+(\frac{1}{8}+\frac{1}{8})\\ &=2+\frac{2}{4}+\frac{2}{6}+\frac{2}{8}\\ &=2+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\\ &=2+\frac{13}{12}>3 \end{split}$$

Hence $$3 < A < 4$$, the answer is C

The solution for this type of question requires a lot of time, so it isn't the best way to solve this one under 2 minutes during take actual GMAT test. We could solve for the general problem like this one:
$$B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$$
also, calculating approximate value of B is rather tough.
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30 Nov 2016, 01:03
Here is the solution for the general problem. This one is out of scope of GMAT, but I still post here for anyone who wants to understand the concept.
$A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^n} \quad with \: n>2$

We have
$\begin{split} A&=1+\frac{1}{2}+\frac{1}{3}+(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7})+...+(\frac{1}{2^k}+\frac{1}{2^k+1}+\frac{1}{2^k+2}+...+\frac{1}{2^{k+1}-1})+...+\frac{1}{2^n}\\ &<1+\frac{1}{2}+\frac{1}{3}+(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4})+...+(\frac{1}{2^k}+\frac{1}{2^k}+\frac{1}{2^k}+...+\frac{1}{2^k})+...+\frac{1}{2^n}\\ &=1+\frac{1}{2}+\frac{1}{3}+\frac{4}{4}+...+\frac{2^k}{2^k}+...+\frac{2^{n-1}}{2^{n-1}}+\frac{1}{2^n}\\ &=1+\frac{1}{2}+\frac{1}{3}+1+1+...+1+\frac{1}{2^n}\\ &=1+\frac{1}{2}+\frac{1}{3}+n-2+\frac{1}{2^n}\\ &=n-1+\frac{1}{2}+\frac{1}{3}+\frac{1}{2^n}\\ &=n-1+\frac{5\times 2^n+6}{6\times 2^n} \end{split}$

Since $$\frac{5\times 2^n+6}{6\times 2^n} < 1 \iff 6 \leq 2^n$$, so we have $$A<n$$.

$\begin{split} A&=1+\frac{1}{2}+(\frac{1}{3}+\frac{1}{4})+(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8})+...+(\frac{1}{2^{k-1}+1}+\frac{1}{2^{k-1}+2}+...+\frac{1}{2^{k}})+...+\frac{1}{2^n}\\ &>1+\frac{1}{2}+(\frac{1}{4}+\frac{1}{4})+(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8})+...+(\frac{1}{2^k}+\frac{1}{2^k}+...+\frac{1}{2^k})+...+\frac{1}{2^n}\\ &=1+\frac{1}{2}+\frac{2}{4}+\frac{4}{8}+...+\frac{2^{k-1}}{2^k}+...+\frac{2^{n-1}}{2^n}\\ &=1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+...+\frac{1}{2}\\ &=1+\frac{n}{2}\\ \end{split}$

Finally, we have $$1+\frac{n}{2} < A < n$$

With $$n=4 \implies 3 < A< 4$$

Phew, quite hard one.
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11 Feb 2017, 08:19
Using Narenn approach, I solved it in 2:25! Is it GMAT like question?
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18 Mar 2017, 12:53
I got there just by calculations: we can estimate that the number is more than 3
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22 Apr 2018, 12:44
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