Round to n decimal places

Solution 1:

(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.