Finding lower bound of summation in inequality

Question

I'm struggling with the following calculation, for which I don't even have an idea on how to start.

$$\sum_{j=x}^\infty \frac{1}{(j!)^2} \leq \epsilon $$

The problem is to find a $x$ such that for a given $\epsilon$, the previous inequality is true.

What kind of knowledge do I need to have in order to solve this?

I'd like to solve other slightly different inequalities, so I'd really like to learn how to solve this, instead of having the solution.

What have is that: $$\sum_{j=x}^\infty \frac{1}{(j!)^2} - \sum_{j=x-1}^\infty \frac{1}{(j!)^2} = \frac{1}{x!}$$

Answer 1

From Taylor expansion you have $$ e^x = \sum_{n=0}^\infty \frac{x^n}{n!} \implies \sum_{n=0}^\infty \frac{1}{n!} = e. $$ Certainly, $$ \sum_{n=0}^\infty \frac{1}{(n!)^2} < \sum_{n=0}^\infty \frac{1}{n!} = e $$

Answer 2

Note that for $j\ge x$, we have $j!\ge x!\cdot x^{j-x}$. Hence (using the formula for geometric series in the end) $$\sum_{j=x}^\infty\frac1{j!^2}\le\sum_{j=x}^\infty\frac1{x!^2x^{2(j-x)}} =\frac1{x!^2}\sum_{j=x}^\infty\frac1{x^{2(j-x)}}=\frac1{x!^2}\sum_{k=0}^\infty\frac1{(x^2)^k}=\frac1{x!^2}\cdot \frac {x^2}{x^2-1}=\frac1{(x-1)!^2(x^2-1)}.$$ Thus it is sufficient to find $x$ with $$ (x-1)!^2(x^2-1)>\frac1\epsilon.$$