I know the Taylor series for $\tan x$ is,
$$\tan x = \sum_{n\,=\,1}^\infty \frac {(-1)^{n-1}2^{2n} (2^{2n}-1) B_{2n}} {(2n)!} x^{2n - 1} $$
I am trying to find a value for $n$ such that $|\tan 1 -\sum_0^na_n|< 10^{-1000}$. At first, I thought this was an error question where I calculate smallest $n$ such that $a_n < 10^{-1000}$, then $n+1$ is what I am looking for. This requires computing and I realized there is no way that is what I am supposed to be doing for an analysis problem.
I am now trying to do it analytically and could use some help. I am trying to work with
$$\left |\tan 1 - \sum_{n\,=\,1}^\infty \frac {(-1)^{n-1}2^{2n} (2^{2n}-1) B_{2n}} {(2n)!} \right | $$
but I am stuck.
I may be totally off topic, so forgive me if this is the case.
Since this is an alternate expansion, would it be sufficient to find $n$ such that $$A=\frac {2^{2n} (2^{2n}-1) B_{2n}} {(2n)!}<\epsilon $$
If this is the case, you would find here that, for large values of $n$ $$B_{2n}\approx 4\,\sqrt{\pi\, n} \,\left(\frac n {\pi\, e}\right)^{2n}$$ Using Stirling approximation for the factorial, this gives $$A \approx 2 \left(2^{2 n}-1\right) \pi ^{-2 n}\approx 2\left(\frac 2 {\pi}\right)^{2n}$$ from which the problem becomes simple $$n=\frac{\log \left(\frac{\epsilon }{2}\right)}{2\log \left(\frac{2}{\pi }\right)}$$
For $\epsilon=10^{-1000}$, the approximate solution is then $n=2550$.
Edit
Using illimited precision and exact values of Bernoulli numbers
Edit
In fact, we do not need to use Stirling approximation for the factorial since Bernoulli numbers can be expressed in terms of the Riemann zeta function as $$B_{2n}=(-1)^{n+1} \frac {2 (2n)!}{(2\pi)^{2n}}\sum_{i=1}^\infty \frac 1 {i^{2n}}$$ Then $$\frac {(-1)^{n-1}2^{2n} (2^{2n}-1) B_{2n}} {(2n)!}=2 \frac{2^{2 n}-1}{{\pi }^{2 n}}\sum_{i=1}^\infty \frac 1 {i^{2n}}$$ and, using the first term of the summation, we arrive to the previous approximation for $A$.
But take care : this is not an alternating series and all terms are positive. So, what we got is a lower bound for the number of terms to be added together.
So, if we define the approximate coefficient as $$u_n=2^{2 n+1} \pi ^{-2 n}$$ we have $$\sum_{n=1}^\infty u_n=\frac{8}{\pi ^2-4}$$ $$\sum_{n=1}^p u_n=\frac{8}{\pi ^2-4}-\frac{2^{2 p+3} \pi ^{-2 p}}{\pi ^2-4}$$ and so we want $$\frac{2^{2 p+3} \pi ^{-2 p}}{\pi ^2-4}< \epsilon$$ $$p>\frac{\log \left(\frac{1}{8} \left(\pi ^2-4\right) \epsilon \right)}{\log \left(\frac{4}{\pi ^2}\right)}$$ which leads to the same answer. However, this last approach seems to be more rigorous than the previous one.
Edit
Without any approximation, we can define the error as $$\Delta=\sum_{n=p+1}^\infty 2 \left(2^{2 n}-1\right) \pi ^{-2 n} \zeta (2 n)$$ Generating values for $p=100$ to $p=5000$ by steps of $100$, a linear regression gives a perfect fit (standard errors smaller than $10^{-14}$) $$\log(\Delta)=0.309654 -0.903165 \, p$$ which then gives for the number $p$ of terms to be retained $$p=0.342854 -1.10722 \log (\epsilon )$$ while the first approximation which was given here $$p=\frac{\log \left(\frac{\epsilon }{2}\right)}{2\log \left(\frac{2}{\pi }\right)}$$ write $$p=0.767464 -1.10722 \log (\epsilon )$$
Edit
If we define $$a_n=\frac {(-1)^{n-1}2^{2n} (2^{2n}-1) B_{2n}} {(2n)!}$$,we can notice that $\frac{a_{n+1}}{a_n}$ becomes very quickly very close to $\frac {4} {\pi^2}$. So,let us write $$\tan(1)=\sum_{i=1}^n a_i+a_n \sum_{i=n+1}^p \big(\frac {4} {\pi^2}\big)^{i-n}+a_n \sum_{i=p+1}^\infty \big(\frac {4} {\pi^2}\big)^{i-n}$$ and so, the error for stopping after $p$ terms is given by $$\Delta_n=a_n \sum_{i=p+1}^\infty \big(\frac {4} {\pi^2}\big)^{i-n}=a_n\,\frac{4^{-n+p+1} \pi ^{2 n-2 p}}{\pi ^2-4}$$ and for $\Delta_n < \epsilon$, this gives $$p_n=n+\frac{\log \left(\frac{\left(\pi ^2-4\right) \epsilon }{4 a_n}\right)}{\log \left(\frac{4}{\pi ^2}\right)}$$ Using $\epsilon=10^{-1000}$, this gives $$p_1=2550.0365$$ $$p_2=2549.8201$$ $$p_3=2549.8056$$ $$p_4=2549.8041$$ $$p_5=2549.8040$$ These justify that we can safely use $$p\approx p_1=1+\frac{\log \left(\frac{\left(\pi ^2-4\right) \epsilon }{4 }\right)}{\log \left(\frac{4}{\pi ^2}\right)}$$