ISI2004-MIII: 13

Question

Let $X =\frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+\ldots+\frac{1}{3001}$. Then

• A. $X< 1$
• B. $X>\frac{3}{2}$
• C. $1< X< \frac{3}{2}$
• D. none of the above

Comments

Is option B correct?

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I am getting C and answer is not given

Answer 1

Take the series as $\dfrac{1}{x}$ 

and use  Riemann integral to evaluate the series 

integrate $\dfrac{1}{x} {dx}$   from $x=1001$ to $3001$ then we get $1.0979$ which is in between $1$ and $\frac{3}{2}.$

So $C$ is correct option.

$1<X<\frac{3}{2}$

Comments

@Golam Murtuza

$\int dx/x = ln\left | x \right |+C$   not   $logx, log\left | x \right | ,or, log⁡\left | x \right |+C$.

https://math.stackexchange.com/questions/206032/what-is-the-integral-of-1-x 

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The idea of using integration while summing the terms comes when there is an infinite no. of terms in the series, that’s basically evaluating the definite integral using the first principle. Where in the world did the idea of using integral whilst evaluating finite series strike you??  

P.S. I am not questioning your answer if it is right or nt. I just to know where u got this idea

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can we write it in difference of harmonic sums and approximate it ?

Answer 2

If you make the gap equal to 250, then from 1/1001 to 1/1250, you have
250 numbers, each greater than 1/1251.

So, 1/1001 + ... + 1/1250 > 250*1/1251.

Similarly, for the next gap of 250 numbers,

So, 1/1251 + ... + 1/1500 > 250*1/1501.

Continuing like this,

x = 1/1001 + ... + 1/3000 (+ 1/3001)  > 250(1/1251 + 1/1501 + 1/1751 + 1/2001 + 1/2251 + 1/2501 + 1/2751 + 1/3001)+1/3001
which comes to 1.02, to two decimal places.
Therefore, x > 1.

For the next part of the question, it was necessary to only divide the 2001 numbers
into 4 parts, mostly of 500 numbers apiece.

From 1/1001 to 1/1500, there are 500 numbers, each less than 1/1000.
Therefore, 1/1001 + ... + 1/1500 < 500 * 1/1000 (i.e. < 1/2)

From 1/1501 to 1/2000, there are 500 numbers, each less than 1/1500.
Therefore, 1/1501 + ... + 1/2000 < 500 * 1/1500 (i.e. < 1/3)

From 1/2001 to 1/2500, there are 500 numbers, each less than 1/2000.
Therefore, 1/2001 + ... + 1/2500 < 500 * 1/2000 (i.e. < 1/4)

From 1/2501 to 1/3001, there are 501 numbers, each less than 1/2500.
Therefore, 1/2501 + ... + 1/3001 < 501 * 1/2500 (i.e < 1/5 + 1/2500)

Adding them all :
1/1001 + ... + 1/3001  <  1/2 + 1/3 + 1/4 + 1/5 + 1/2500 (i.e < ≈1.2837)
Therefore, x < 3/2.

Thus, (C) 1 < x < 3/2 is the correct answer

Comments

When to use ln and log.in this case ln is used log gives different  ans. Please explain.

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@CHIRAG CHAWLA You should have used ln (natural logarithm) not log ( which is base 10).

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In case of decibel, Richter scale (earth quake) we use log base 10, which is common logarithm.

Generally in case of calculus we use log base e= ln, which is natural logarithm.

In case of Complexity of algorithm we use log base 2.

Answer 3

The $n^{th}$ Harmonic number is defined as :

$$H_n = 1 + \frac{1}{2} + \frac{1}{3} + ...+\frac{1}{n}$$

This is a divergent series and can be proved easily and there is no nice closed form for it but we can approximate it as:

$H_n \approx \ln n + \gamma + \frac{1}{2n} – \frac{1}{12n^2}$

More precisely,

$H_n = \ln n + \gamma + \frac{1}{2n} – \frac{1}{12n^2} + \frac{\epsilon_n}{120 n^4}$ for $ 0< \epsilon_n < 1 $

where $\gamma$ is known as Euler-Mascheroni Constant or simply Euler’s Constant and it’s value is $0.5772156649…$

 I will approximate it by writing $H_n \approx \ln n + \gamma $

So, here, we can write $X$ as:

$X = \left(1 + \frac{1}{2} + \frac{1}{3} + ...+\frac{1}{3001}\right) – \left(1 + \frac{1}{2} + \frac{1}{3} + ...+\frac{1}{1000}\right)$

$X = H_{3001} – H_{1000}$

$X \approx \ln 3001 – \ln 1000 = 1.099$

So, (C) is correct.

There are many interesting things to know about Harmonic Numbers.

$1)$ Nobody knows whether Euler’s Constant $\gamma$ is rational or irrational. This is an unsloved problem.

$2)$ We call it harmonics because a tone of the wavelength $\frac{1}{n}$ is called the $n^{th}$ harmonic of a tone whose wavelength is 1.

$3) $ There are some inequalities for Harmonic Numbers:

$A) $ $\frac{\lfloor (\lg n) \rfloor + 1} {2} < H_n \leq \lfloor (\lg n) \rfloor + 1$

$B) $ $\ln n < H_n < ln + 1$ etc.

$4) $ $\lim_{n \rightarrow \infty} (H_n - \ln n) = \gamma$