Why Does the Alternating Test Estimation Theorem Not Give The Correct Solution Here? We were given a question in one our Calculus II class concerning estimating definite integrals using the integrand's Maclaurin Series. Specifically, this was the problem at hand:

If we estimated the definite integral $\displaystyle \int_0^1 \frac{\arctan{x}}{x} \, dx$ using a Maclaurin polynomial, what order should the polynomial at least be so that the error of our estimation is less than $0.01$.

Over here, I found the Maclaurin Series of the integrand, which is: $\displaystyle \sum_{n = 0}^{\infty}(-1)^n \frac{x^{2n}}{2n + 1}$. Integrating the resulting series would yield an alternating series whose value equals the definite integral. Namely,
$$\int_0^1\frac{\arctan{x}}{x} \, dx = \sum_{n = 0}^{\infty}(-1)^n \frac{1}{(2n+1)^2}$$
Now, the question shifts focus to what value $k$ should we choose so that the sum of the first $k + 1$ terms of the series yields an error less than $0.01$. To answer this question, we were given the hint of using the Alternating Series Remainder Theorem ($\lvert L - s_n \rvert < \lvert a_{n + 1}\rvert$). I applied this theorem in the wrong manner in the beginning.
$$\frac{1}{(2(n+1) + 1)^2} > 0.01 \implies 2n + 3 < 10 \implies 2n < 7 \implies n < \frac{7}{2} \text{. So choose } n = 3.$$
However, this actually works. If we plugged $3$ into the series $\displaystyle \sum_{n=0}^{3}(-1)^n \frac{1}{(2n + 1)^2}$ we get a value that is off by the actual definite integral by less than $0.01$. So, the Maclaurin polynomial would be of order $6$ in this case.
However, I later learnt that the way I applied the theorem was wrong, so I revised my work to become:
$$\frac{1}{(2(n+1) + 1)^2} < 0.01 \implies 2n + 3 > 10 \implies 2n > 7 \implies n > \frac{7}{2} \text{. So choose } n = 4.$$
This also gives a value that is within the desired range; however, it is not the least value. This answer would suggest that we need the Maclaurin polynomial to be at least 8, which is not true as we've seen before.
Basically, why is the test not leading us to the right value for $k$?
 A: You are correct that if $n > \frac{7}{2}$, then the Alternating Series Remainder Theorem (ASRT) guarantees that $\sum_{k=0}^n (-1)^n \frac{1}{(2n+1)^2}$ is within $0.01$ of the true value of the integral, so $n=4$ is certainly large enough.  But if $n < \frac{7}{2}$, the ASRT makes no promises either way about whether the $n$-th partial sum is close enough.  It happens to be false for $n=0, 1, 2$ and true for $n=3$, but this can't be deduced (at least not immediately) from the ASRT.
To see why, let's recall why the ASRT works to begin with.  If $\sum_n a_n$ is a series satisfying the hypotheses of the alternating series test, then its partial sums $s_n$ "zigzag inwards" towards the sum:  they alternate between increasing and decreasing, and each of these increases and decreases overshoots the actual sum.  Therefore, any given partial sum $s_n$ must be within distance $|a_{n+1}|$ of the actual sum, in order for the next jump to overshoot the sum.  (Moreover, the error has the same sign as $a_{n+1}$.)  This is essentially what the ASRT says.
But without having some more specific information about our series, we can't say exactly how big the error is--it might be very close to $a_{n+1}$, or very close to $0$, or anywhere in between.  In the case of $n=3$ in your series, the next term $a_{n+1}$ is $1/9^2 \approx 0.0123$, so according to the ASRT the error could be anywhere in the interval $(0, 1/9^2)$--which does not guarantee that the error is less than $0.01$.  But the error is in fact only about $0.0075$, which is good enough.
