m-th digit approximation of the limit of a convergent serie I am trying to brush up on calculus and picked up Peter Lax's Calculus with Applications and Computing Vol 1 (1976) and I am trying to solve exercise 5.2 a) in the first chapter (page 29):

How large does $n$ have to be in order for
$$ S_n = \sum_{j = 1}^n \frac{1}{j^2}$$
to be within $\frac{1}{10}$ of the infinite sum? within $\frac{1}{100}$? within $\frac{1}{1000}$? Calculate the first, second and third digit after the decimal point of $ \sum_{j = 1}^\infty \frac{1}{j^2}$

Ok so the first part is easy and is derived from the chapter's text:
$ \forall j \geq 1 $ we have $\frac{1}{j^2} \leq \frac{2}{j(j+1)}$ and therefore:
\begin{equation}
\begin{aligned}
\forall n \geq 1,\quad \forall N \geq n +1 \quad S_N - S_n &\leq 2\sum_{k = n+1}^N \frac{1}{k(k+1)}\\
&= 2\sum_{k = n+1}^N \left\{ \frac{1}{k}- \frac{1}{k+1}\right\}\\
&= 2 \left[ \frac{1}{n+1} - \frac{1}{N+1}\right]
\end{aligned}
\end{equation}
Now because we know $S_N$ converges to a limit $l$ from below and by the rules of arithmetic for convergent sequences we have:
$$ 0 \leq S - S_n \leq \frac{2}{n+1}$$
So if we want $S_n$ to be within $\frac{1}{10^k}$ of $S$ it suffices to have:
$$ n \geq N_{k} = 2\times10^k -1$$
But the second part of the question puzzles me. I would like to say that computing $S_{N_{k}}$ is enough to have the first $k$ decimal points of $S$. But earlier in the chapter (on page 9), there is a theorem that states:

if $a$ and $b$ have the same integer parts and the same digits up to the $m$-th, then they differ by less than $10^{-m}$,
$$ |a - b | < 10^{-m}$$
and the converse is not true.

And the example of $a = 0.29999...$ and $b = 0.30000...$ indeed shows that two numbers can differ by less than $2\times 10^{-5}$ and yet have all different first digits.
So I think there is something missing in my "demonstration" above. How to show that I indeed "catch" the first $k$ digits of $S$ by computing $S_{N_k}$?
Thanks!
 A: I shall cheating assuming that you know the value of the infinite sum.
So for your example
$$\sum_{n=1}^{p}\frac1{n^2}=H_p^{(2)}$$ and you want to know $p$ such that
$$\frac {\pi^2}6-H_{p+1}^{(2)} \leq 10^{-k}$$
Using the asymptotics of the generalized harmonic numbers, this will write
$$\frac 1{(p+1)}-\frac 1{2(p+1)^2}+\frac 1{6(p+1)^3}+\cdots\leq 10^{-k}$$ For sure, larger will be $k$ and less terms we shall need.
To stay with equations we know how to solve, let us stay with this cubic in $x=\frac 1{(p+1)} $ and solve
$$x-\frac 1 2 x^2+\frac 16 x^3 - \epsilon=0 \qquad \text{where} \qquad \epsilon=10^{-k}$$
We have
$$\Delta =-\frac{5}{12}+\epsilon -\frac{3 }{4}\epsilon ^2$$ which is very qucikly negative so only one real root. Using the hyperbolic method, we shall find
$$x=1-2 \sinh \left(\frac{1}{3} \sinh ^{-1}(2-3 \epsilon )\right)$$ that is to say
$$p=-\frac{2 \sinh \left(\frac{1}{3} \sinh ^{-1}(2-3 \epsilon )\right)}{1-2 \sinh
   \left(\frac{1}{3} \sinh ^{-1}(2-3 \epsilon )\right)}$$ Since $\epsilon$ is small, a Taylor expansion will give
$$p=\frac{1}{\epsilon }-\frac{3}{2}-\frac{\epsilon }{12}+O\left(\epsilon ^3\right)$$ Back to $k$
$$p \sim 10^k-\frac 32$$ If you want a difference of $10^{-6}$, you will need "almost" one million of terms.
If we make $p=10^6$, the difference is $0.99999950 \times 10^{-6}$
A: If you get an answer like $1.29999$, you'll simply have to compute more terms of the series, but in all likelihood, you'll be able to make a definite statement.  Try to compute the first four digits after the decimal point.  You may be doubtful about the fourth digit, but in all likelihood, you'll be able to make a definite statement about the first three.  With $n=20000$ I got $$\sum_{k=1}^{20000}\frac1{n^2}=1.6448840680982086$$
We can't be sure if the value is $1.6448\dots$ or $1.6449\dots$ but we know it's $1.644\dots$.
(In point of fact, the value is $\frac{\pi^2}6\approx1.6449340668482264$).
