<s>[INST] 2022 Gr11 Fermat math contest question #20 Question: A sequence of numbers $t_1, t_2, t_3$,... has its terms defined by $t_n = \frac{1}{n} - \frac{1}{n+2}$ for every integer $n \geq 1.$ For example, $t_4 = \frac{1}{4} - \frac{1}{6}$. What is the largest positive integer k for which the sum of the first k terms (that is, $t_1 + t_2 +\cdots+t_{k-1} + t_k$) is less than 1.499?
Correct answer: 1998
My solution: First, to find the sum of all numbers from $t_1$ to $t_k$, I wrote the equation $\frac{(\frac{2}{3} + \frac{1}{k} - \frac{1}{k+2}) * k}{2}$, note: I got $\frac{2}{3}$ from $\frac{1}{1} - \frac{1}{1+2}$. Next, I just substituted the 5 options this question gave me into my equation (this is a multiple choice question), but I got around $666$ when I substituted the different options which is no where near $1.499$. This means the problem is somewhere in my equation for the sum of all numbers from $t_1$ to $t_k$, but I do not know where I went wrong in my equation, or even if it was the right first step to solving this question.
 [/INST] $S_k = \sum_{n=1}^k t_n = \sum_{n=1}^k\frac{1}{n} - \frac{1}{n+2} = A_k - B_k< 1.499$. Now, since $B_k = A_{k+2} - \frac{1}{1} - \frac{1}{2}$,
$$
S_k = A_k - A_{k+2} + \frac{3}{2} = -\frac{1}{k+1} - \frac{1}{k+2} + 1.5 < 1.499.
$$
This gives
$\frac{1}{k+1} + \frac{1}{k+2} > 1.5 - 1.499 = 0.001 = \frac{1}{2000} + \frac{1}{2000}$.
Therefore $k+2$ can not exceed 2000 so $k = 1998$.
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