Let $$f(x) = \frac{x^2}{x^2 - 1}.$$Find the largest integer $n$ so that $f(2) \cdot f(3) \cdot f(4) \cdots f(n-1) \cdot f(n) < 1.98.$
f(x) = x^2/(x^2-1)
It might help to visit
https://math.stackexchange.com/questions/312381/proving-prodk2-1-k2-n1-2n-by-induction
to see that
n
Π (k^2-1)/k^2 = (n+1)/(2n)
k=2
so, you want n such that
2n/(n+1) < 1.98
To find the largest integer $n$ that satisfies the given condition, we need to evaluate the expression $f(2) \cdot f(3) \cdot f(4) \cdots f(n-1) \cdot f(n)$ and find where it is less than $1.98$. Let's start by finding the expression for $f(x)$.
Given that $$f(x) = \frac{x^2}{x^2 - 1},$$ we can substitute $x$ with $n$ in the expression to get $f(n)$. Thus, $$f(n) = \frac{n^2}{n^2 - 1}.$$
To simplify the expression, we can factor $n^2 - 1$ as $(n-1)(n+1)$. This allows us to rewrite $f(n)$ as $$f(n) = \frac{n^2}{(n-1)(n+1)}.$$
Now that we have an expression for $f(n)$, we can write out the expression we need to evaluate as:
$$f(2) \cdot f(3) \cdot f(4) \cdots f(n-1) \cdot f(n) = \frac{2^2}{(2-1)(2+1)} \cdot \frac{3^2}{(3-1)(3+1)} \cdot \frac{4^2}{(4-1)(4+1)} \cdots \frac{n^2}{(n-1)(n+1)}.$$
Notice that many terms in the numerator and denominator of each fraction cancel out, simplifying the expression significantly.
We can rewrite this product as:
$$\frac{2^2 \cdot 3^2 \cdot 4^2 \cdots n^2}{(2-1) \cdot (3-1) \cdot (4-1) \cdots (n-1)(n+1)}.$$
Next, we can simplify it further by cancelling out common terms in the numerator and denominator:
$$\frac{2^2 \cdot 3^2 \cdot 4^2 \cdots n^2}{1 \cdot 2 \cdot 3 \cdot 4 \cdots (n-1) \cdot (n+1)}.$$
Now, we can express the entire expression as a single fraction:
$$f(2) \cdot f(3) \cdot f(4) \cdots f(n-1) \cdot f(n) = \frac{2^2 \cdot 3^2 \cdot 4^2 \cdots n^2}{1 \cdot 2 \cdot 3 \cdot 4 \cdots (n-1) \cdot (n+1)}.$$
We want to find the largest integer $n$ such that this expression is less than $1.98$.
To solve this, we can start evaluating the expression for different values of $n$ until we find the largest integer that satisfies the condition. We can start from $n=2$ and increment $n$.
For example, when $n = 2$, the expression becomes $\frac{2^2}{1 \cdot 2}$, which is equal to $2$.
When $n = 3$, the expression becomes $\frac{2^2 \cdot 3^2}{1 \cdot 2 \cdot 3 \cdot 4}$, which is equal to $\frac{9}{4}$.
Continuing this process, we find that when $n=5$, the expression becomes $\frac{2^2 \cdot 3^2 \cdot 4^2 \cdot 5^2}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6}$, which is equal to $\frac{100}{72} = \frac{25}{18} \approx 1.3889$.
We can see that for $n \geq 5$, the expression becomes less than $1.98$.
Therefore, the largest integer $n$ that satisfies $f(2) \cdot f(3) \cdot f(4) \cdots f(n-1) \cdot f(n) < 1.98$ is $\boxed{5}$.
By evaluating the expression for $n=5$ and comparing it to $1.98$, we can conclude that any larger value of $n$ will also satisfy the inequality.