The question
This question was taken from an International Junior Mathematical Olympiad exam for 9th graders. The question is as follows:
Let $x$ satisfies the equation $\dfrac{1}{x}=\dfrac{1}{2017^2}+\dfrac{1}{2018^2}+\ldots+\dfrac{1}{4030^2}$. Which of the following numbers is the nearest to $x$?
A. $2016$
B. $2017$
C. $3024$
D. $4035$
E. $4037$
D. $4035$
My work
First of all, let's simplify the problem:
$\dfrac{1}{x}=\displaystyle\sum_{k=2017}^{4030}\frac{1}{k^2}$
When searching for algebraic resources, I first used the following two identities to obtain a lower and an upper bound:
- $\dfrac{1}{n}-\dfrac{1}{n+1} = \dfrac{1}{n\left(n+1\right)} < \dfrac{1}{n^2}$
- $\dfrac{1}{n-1}-\dfrac{1}{n} = \dfrac{1}{n\left(n-1\right)} > \dfrac{1}{n^2}$
After a few manipulations (note the application of telescopic sums), this is the result obtained:
$\displaystyle\sum_{n=2017}^{4030}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right) < \dfrac{1}{x} < \displaystyle\sum_{n=2017}^{4030}\left(\dfrac{1}{n-1}-\dfrac{1}{n}\right)\ \ \ \rightarrow\ \ \ \dfrac{1}{2017} - \dfrac{1}{4031} < \dfrac{1}{x} < \dfrac{1}{2016} - \dfrac{1}{4030}$
From this point on, I realized that this approach was inefficient for an Olympiad like IJMO, after all, I had already taken part in it and solved all the questions with solutions that were easier to calculate. Anyway, that's the approximate result:
$4034 < \dfrac{1}{x} < 4037$
As this was not enough to come up with an answer, I looked for other alternatives and used the following identity:
$\dfrac{1}{2}\left(\dfrac{1}{n-1}-\dfrac{1}{n+1}\right) = \dfrac{1}{n^2-1} > \dfrac{1}{n^2} \ \ \ \rightarrow\ \ \ \dfrac{1}{x} < \dfrac{1}{2}\displaystyle\sum_{n=2017}^{4030}\left(\dfrac{1}{n-1}-\dfrac{1}{n+1}\right) \ \ \ \rightarrow$
$\rightarrow \ \ \ \dfrac{1}{x} < \dfrac{1}{2016} + \dfrac{1}{2017} - \dfrac{1}{4030} - \dfrac{1}{4031}\rightarrow \ \ \ 4035.5 < x$
In the end, my work only determined that $4035.5 < x < 4037$ (approximately).
How do I determine the answer? Is there a more efficient solution?
The first approach is not needed; the second approach using the identity $$\frac{1}{n-1} - \frac{1}{n+1} = \frac{2}{n^2-1}$$ yields a tighter bound. With it, we can eliminate choices (A) through (C).
The next thing to do is to estimate the error: the term-wise error is approximately $$\frac{1}{n^2 - 1} - \frac{1}{n^2} = \frac{1}{n^2(n^2-1)} = \frac{1}{n^4 - n^2 + \frac{1}{4} - \frac{1}{4}} = \frac{1}{(n^2-\frac{1}{2})^2 - \frac{1}{4}} \approx \frac{1}{(n^2-\frac{1}{2})^2}.$$ Certainly, $$\frac{1}{n^2-1} - \frac{1}{n^2} < \frac{1}{(n^2-1)^2}$$ for $n > 1$. This error is largest when $n = 2017$, so the total error is $$\sum_{n=2017}^{4030} \frac{1}{n^2-1} - \frac{1}{x} < \frac{4030 - 2017 + 1}{(2017^2 - 1)^2} < \frac{2014}{2016^4} < \frac{1}{2016^3}.$$ This tells us that the given sum is extremely close to $\frac{1}{4035.5}$, namely $$\frac{1}{4035.5} - \frac{1}{x} < \frac{1}{2016^3},$$ and we easily have $x < 4036$: $$\frac{1}{4035.5} - \frac{1}{4036} = \frac{1}{4036(2(4036)+1)} > \frac{1}{2016^3}.$$