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.
Given, $$t_n=\frac1n-\frac{1}{n+2}$$ we get $$S_k=\sum_{n=1}^k t_n=\sum_{n=1}^k \frac1n-\frac{1}{n+2}$$$$= \sum_{n=1}^k\frac1n-\sum_{n=1}^k\frac{1}{n+2}$$ Reindexing the second sum and replacing $n+2=i$, we get $$S_k= \sum_{n=1}^k\frac1n- \sum_{i=3}^{k+2}\frac1k$$$$=1+\frac12-\frac{1}{k+1}-\frac{1}{k+2}=\frac{k(3k+5)}{2(k+1)(k+2)}$$ Now, given that $S_k<1.499$ we infer that $$1.5-\frac{1}{k+1}-\frac{1}{k+2}<1.499$$ or that $$\frac{1}{k+1}+\frac{1}{k+2}>0.001$$
To better visualise the reindexing, we can write $$S_k=\left(\color{red}{1+\frac12}+ \frac13+ …+ \frac{1}{k}\right)-\left(\frac13+ \frac14+…+ \frac{1}{k}+\color{red}{\frac{1}{k+1}+\frac{1}{k+2}}\right)$$ where the $\displaystyle \frac13+ …+ \frac{1}{k}$ gets cancelled.