Harmonic At It's Best

Calculus Level 4

Let

$\text{A}$ = $\frac{1}{1001} + \frac{1}{1002} + \frac{1}{1003} + \dots + \frac{1}{3001}$ .

Then $x$ $\lt$ $\text{A}$ $\lt$ $\frac{y}{3}$ .

Where $x$ is the greatest integer possible in its range and $y$ is the smallest integer possible in its range.

Find the value of $x + y$ .

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1 solution

Nathanael Case
Feb 13, 2015

I used the continuous form 1/x to find a lower bound and upper bound for A

So if we call the sum "A" like in the problem then we have:

$\int\limits_{1001}^{3002}\frac{dx}{x}<A<\int\limits_{1000}^{3001}\frac{dx}{x}$

$\ln(\frac{3002}{1001})<A<\ln(\frac{3001}{1000})$

$1.0983<A<1.0989$

That puts a pretty good restriction on A! Therefore $1<A<\frac{4}{3}$

Here is a picture to explain the idea:

There is a typo.

It should be $\ln{(\frac{3001}{1001})}$ not $\ln{(\frac{3001}{10001})}$

Rajdeep Dhingra - 6 years, 4 months ago

Thank you for pointing out the typo. I don't have a link or anything, but I think I can make it more clear if I draw a picture. Give me some time I will add a picture to the solution.

Nathanael Case - 6 years, 4 months ago

Could you explain this continuous approximation of $\frac{1}{x}$ ?

Could you give me a link?

Rajdeep Dhingra - 6 years, 4 months ago

Ok, I've added the picture. If you have any more questions after studying the picture, I will try to answer them.

Also, making the picture made me realize my first edit was slightly off! I corrected the limits and now the restriction on A is even better ! :)

Nathanael Case - 6 years, 4 months ago

Thanks for the explanation

Rajdeep Dhingra - 6 years, 4 months ago

how did you get the picture

Rajdeep Dhingra - 6 years, 4 months ago

@Rajdeep Dhingra – I drew the picture on my computer... Do you mean how did I think of it?

I thought of it because a series, (like $\Sigma \frac{1}{n}$ ), is just an approximation of the area of the continuous case, (like $\int\frac{1}{x}dx$ ). Instead of $dx$ , you have $\Delta x=1$ .

But there are two ways to approximate an integral; by too much (rectangles above) or by too little (rectangles below). I just used this idea to constrain the sum.

This idea (of approximating integrals with $\Delta x$ instead of $dx$ ) is pretty fundamental to calculus. Calculus is built out of the limit of $\Delta x$ → $dx$

To be honest, I don't know any 13 year olds who know calculus, but if you're on this website, you must be pretty smart :)

Nathanael Case - 6 years, 4 months ago

@Nathanael Case – No I meant how did you drew the image on your computer?

Using which software?

Rajdeep Dhingra - 6 years, 4 months ago

@Rajdeep Dhingra – I have a mac, so I used something called "Paintbrush"

Nathanael Case - 6 years, 4 months ago

Side note: If we average the two limits:

$\frac{1}{2}\ln(\frac{3002}{1001})+\frac{1}{2}\ln(\frac{3001}{1000})$

we get a very good approximation of the sum (correct to 6 decimal places)

Nathanael Case - 6 years, 4 months ago

From a question I encountered yesterday, the value of the nth harmonic number is the same as $y+ln(n)$ , where y is the constant, which by manipulating gives us $2y + ln(3001)-ln(1000)-1 = 1.2533$ Is this still a good approximation? I'm talking about when the value of this becomes $0.99$ while the true value is $1.01$ which gives different answers for the question.

Marc Vince Casimiro - 6 years, 4 months ago

Actually, because $H_{n} \sim \gamma+\ln n$ (that's a gamma, by the way), what you get is $A \approx \ln 3.001$ . You made a mistake somewhere in the algebraic manipulation. We can then use our intuition to bound $A$ .

Jake Lai - 6 years, 3 months ago

Harmonic At It's Best

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