How much do I need?

I'll try to provide a sensible solution here:

It is obvious that the sum $S$ is bounded below by $0$ and above by $1$ because the sum must be positive since only positive terms are added. The upper bound is given by comparison.

$0\lt S\lt \sum_{n=1}^\infty \frac{1}{2^n}=1$

So, let's consider that the sum $S$ is of the form $S=\overline{0.abcdefgh\cdots}$ , where $a,b,c,d,e,f,g,\cdots\in\{0,1,2,3,\ldots,9\}$ . Then, we have,

$\lfloor 10^5\times S\rfloor = \lfloor \overline{abcde.fgh\cdots}\rfloor=\overline{abcde}$

We need to find $\overline{abcde}$ . Now, comes the calculator part. We test how many terms we need to get the correct result. From the above result, note that the insignificant terms that doesn't change our answer are of the order of magnitude $\leq (-6)$ since those terms only contribute to the fractional part of $10^5S$ and can never affect the integral part as subsequent terms rapidly decrease .

Hence, we use only those terms which are of the order of magnitude $\geq (-5)$ . Now, well, let's just say that the calculator works here more than you do.

The last term in the series that has order of magnitude $\geq (-5)$ is $\dfrac{1}{^42}$ .

This is because $\dfrac{1}{^52}\lt \dfrac{1}{^42\times 2}$ and the latter has order of magnitude $\leq (-6)$ .

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So, it all boils down to this:

$\lfloor 10^5\times S\rfloor=\left\lfloor10^5\times \sum_{n=1}^4\frac{1}{^n2}\right\rfloor=81251$

Add up the first four terms because 1/2^65536 is very tiny