So, let's say we want to evaluate $\pi$ using $\sum\limits^\infty_{n=0} \frac{4(-1)^n}{2n+1}$ (or any other series) correct to four decimal places.
(i) Do we compare the series result to 3.1415 or 3.1416 or to the exact value of $\pi$ ?
(ii) Let's say based on calculation we find $\pi$ correct to four decimal places to be $\sum\limits^{136115}_{n=0} \frac{4(-1)^n}{2n+1}$ . Now, do we round this answer to four decimal places and compare it to one of the above cases in (i) or do we not round the sum and we look at the first four decimal digits of the sum as the result without rounding, which is 3.1415?
(iii) Which of the following is the correct setup if we apply the alternating series remainder formula $(|R_n|\leq b_{n+1})$ to get the smallest n such that the sum gives $\pi$ correct to four decimal places?
$b_{n+1}<10^{-4}\qquad,\qquad b_{n+1}\leq10^{-4}\qquad,\qquad b_{n+1}<\frac{5}{10^5}\qquad,\qquad b_{n+1}\leq \frac{5}{10^5}$
(iv) Now by using $b_{n+1}\leq10^{-4}$ the smallest number for n is n=39999. Is $\pi$ correct to four decimal places equal to $\pi \approx3.1415$ or $\pi \approx3.1416$ knowing that $\sum\limits^{39999}_{n=0} \frac{4(-1)^n}{2n+1}\pi \approx3.14156765$
(v) Is n=136115 the smallest number that gives $\pi$ correct to four decimal places? because it is the smallest number after which the sum of the series' first four decimal digits never change?
(i) Since the exact value of $\pi$ is unknown (and for a general constant for which, we aim to approximate), the error function is the one determining how many terms of the sequence should be regarded to keep the error below a threshold. Most often, an error bound is achievable instead of an exact error function. In this case $$ {\pi=\sum_{n=0}^\infty 4\frac{(-1)^n}{2n+1} \\ e(N)=\sum_{n=N+1}^\infty 4\frac{(-1)^n}{2n+1} } $$ hence $$ e(2N-1){=\sum_{n=2N}^\infty 4\frac{(-1)^n}{2n+1} \\= \sum_{n=N}^\infty 4\left[\frac{1}{4n+1}-\frac{1}{4n+3}\right] \\< \sum_{n=N}^\infty 4\left[\frac{1}{4n-1}-\frac{1}{4n+3}\right] \\=\frac{4}{4N-1} \\\implies |e(N)|<\frac{4}{2N+1} } $$ In this case, we wish $|e(N)|<5\times 10^{-5}$ which leads to a sufficient condition of $\frac{4}{2N+1}<5\times 10^{-5}$ or $N\ge 40000$.
(ii and iii)
For small errors and large $N$s, such round-ups are seldom important. However, to measure the error, no round-up is used and $b_{n+1}\leq \frac{5}{10^5}$ is preferred among all.