Find the largest $n$ so that $f(2) \cdot f(3) \cdot f(4) \cdots f(n-1) \cdot f(n) < 1.98.$
what the answer?
98
We can factor $f(x)$ as
$$f(x) = \frac{x \cdot x}{(x-1)(x+1)}.$$
So if we write out the product, we see that $f(2) \cdot f(3) \cdot f(4) \cdots f(n-1) \cdot f(n)$ is
$$\frac{2 \cdot 2}{1 \cdot 3} \frac{3 \cdot 3}{2 \cdot 4} \frac{4 \cdot 4}{3 \cdot 5} \cdots \frac{(n-1) \cdot (n-1)}{(n-2) \cdot n} \frac{n \cdot n}{(n-1) \cdot (n+1)}.$$
We count that there are two of each factor from 2 to $n,$ inclusive, in the numerator. In the denominator, we count two of each factor from 3 to $n-1,$ inclusive, and one each of 1, 2, $n,$ and $n+1.$ Most of these cancel, leaving
$$\frac{2 \cdot n}{1 \cdot (n+1)} = \frac{2n}{n+1}.$$
We observe that $1.98 = \frac{198}{100} = \frac{2 \cdot 99}{99 + 1}.$ So $n = 99$ is too high. Also, $\frac{2 \cdot 98}{99} < \frac{2 \cdot 99}{100}.$ (We can check this by cross-multiplying to clear the fractions.) So the answer is $n = \boxed{98}.$
To find the largest $n$ that satisfies the given inequality, we need to determine the value of $f(k)$, where $k$ is a positive integer. However, since the expression $f(2) \cdot f(3) \cdot f(4) \cdots f(n-1) \cdot f(n)$ is given without any information about the nature of $f(k)$, we cannot determine the specific value for each term $f(k)$.
Hence, we need to employ a general strategy to solve this problem. We know that the product of all these terms is smaller than $1.98$. This implies that at least one of the terms should be less than $\sqrt[n]{1.98} \approx 1.122$. Thus, the largest possible value for $f(k)$ is $2$.
In order to avoid a larger value of $n$, we should choose the smallest value for $f(k)$ that is less than or equal to $\sqrt[n]{1.98}$, which in this case is $1$. Therefore, we have $f(k) = 1$ for all $k \geq 2$. Consequently, the inequality becomes $1 < 1.98$.
Since $1$ is less than $1.98$, any value of $n$ satisfies the inequality. Therefore, the largest possible value for $n$ is infinity.