I know that a sequence of real numbers $\{x_n\}_{n=0}^\infty$ converges to a limit $L$ which is an irrational number that is only known as the limit of the above sequence (ie there is no other way to have access to an approximate value of $L$). Is there a way to find after which number of terms $n$ in the above sequence, a given number of digits of $L$ are exact?
As an example, the sequence is $x_n=\sum_{i=1}^n\frac{1}{i^2}$ but my question is general. I should also mention that relying on an algorithm to answer the question is acceptable. Is there a systematic way to check that the $k$ first numerically found digits are exact?
No, and there are three different problems that stop it.
One is that there are breakpoints where the number changes from $2.9999$ to $3.0000$ or something like that. A tiny change in the sum can change a lot of digits. We can eliminate this one by asking that we know the sum better than $\pm 5\cdot 10^{-k}$ instead of asking for $k$ accurate digits.
The second is that there might be large numbers far out in the sum. The sequence $x_n$ may look well behaved, but unknown to you $x_{10^{1000}}=1,000,000$. You can't get any decimals right until you add more terms than that.
The third is that some nicely specified series converge very slowly. We know that $\sum \frac 1n$ diverges like $\log n$. Less well known is that $\sum \frac 1{n \log n}$ diverges like $\log(\log(n))$. On the other hand, $\sum \frac 1{n(\log n)^2}$ converges, but very slowly. $\sum \frac 1{n\log n (\log(\log n))^2}$ converges more slowly yet.
Responding to your comment that you have the specific series $x_n=\sum_{i=1}^n 1/i^2$ we can give an answer. You can just look at the partial sum and see it does not have a string of $9$s at the end. We know there are not stray terms that are very large because you have a nice formula. We can improve the convergence by seeing it as a Riemann sum for the integral $\int_1^\infty \frac 1{x^2}dx=1$ and work on the error term. The error term will be of order $\frac 1{x^3}$ so is more manageable. When the series has a nice formula you can often bound the truncation error to give a number of terms to get a number of decimals. There is not a general procedure.