Stairway to heaven

Calculus Level 5

Let $P_{n}$ for any positive integer $n$ describe a particle's "stairway" path as follows:

A particle, starting at the origin of a standard $xy$ -grid, first moves east, (i.e., in the positive $x$ -direction), then up, (i.e., in the positive $y$ -direction), then east, up, and so on for a total of $n$ moves, such that the $k$ -th move has length $\displaystyle \frac{1}{\sqrt{kn}}$ for $1 \le k \le n$ .

Now let $D_{n}$ be the displacement between the starting and finishing points, and let $|P_{n}|$ be the total distance traveled.

If $\displaystyle S = \lim_{n \rightarrow \infty} (|P_{n}| - D_{n})$ , then find $\lfloor 1000S \rfloor$ .

The answer is 585.

1 solution

Brian Charlesworth
May 9, 2015

I'll post a full answer shortly, but for now I'll just note quickly that the keys to the question are the facts that

$\displaystyle\lim_{n \rightarrow \infty} \sum_{k=1}^{n} \dfrac{1}{\sqrt{kn}} = 2$ , and that $\displaystyle\lim_{n \rightarrow \infty} \sum_{k=1}^{\lfloor \frac{n}{2} \rfloor} \dfrac{1}{\sqrt{2kn}} = 1.$

Hint: Use Riemann sums.

Great problem! I never knew the approximation of representing a sum as an integral was called Riemann sums!

Raghav Vaidyanathan - 6 years, 1 month ago

Thanks! Again, I'm relieved that you got the same answer as I did. :)

Brian Charlesworth - 6 years, 1 month ago

I didn't use the $\frac{n}{2}$ upper limit in the second sum(s) ;-;

Jake Lai - 6 years, 1 month ago

That's too bad. :( By the way, this is one of my contributions to the set of questions that you and Trevor will be posting later today, (or at least I think that was the plan).

Brian Charlesworth - 6 years, 1 month ago

Posted the first one! Stay tuned for a 5D variant I'm cooking up ;3

Jake Lai - 6 years, 1 month ago

@Jake Lai – Great! Your first question is a good one, but I think that Dr. Bretscher makes a valid point; you may need to specify that the value you are asking for is the magnitude of the displacement of Zeno at journey's end, and not the actual distance he has traveled.

Brian Charlesworth - 6 years, 1 month ago

Brian Charlesworth Good Problem! I wasn't able to solve it though.What I did was this: $S =\lim_{n\rightarrow \infty}\displaystyle \sum_{k=1}^{n}\dfrac{1}{\sqrt{kn}} -\lim_{n \rightarrow \infty}\left|\sum_{k=1}^{n}\dfrac{e^{ik\pi/2}}{\sqrt{kn}} \right|$ the left summation equals $2$ ,I know it, but the right hand summation is not happening??!!

Kunal Gupta - 5 years, 8 months ago

Can you post a full answer? I don't understand how the upper limit is $\left \lfloor \dfrac n2 \right \rfloor$ .

Pi Han Goh - 5 years, 6 months ago

Stairway to heaven

The answer is 585.

1 solution

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