In an alternating series remainder where the 1st term in remainder is a negative, why is the approximate series an overestimate?

Question

Saw this answer but it doesn't go deep enough to help understanding this. I need help identifying my knowledge gap as I struggle to understand why it is an overestimate. My thinking below results in the approximation being an underestimate, rather than overestimate (which is stated in the answer).

When taking the values of the series:

$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}$ is approximated using the partial sum $\sum_{n=1}^{99} \frac{(-1)^{n+1}}{n}$

Setting up:

1. $\sum^{99}_{n=1}\frac{(-1)^{n+1}}{n} = S_{99}$ , is the partial sum
2. $\sum^{\infty}_{n=99}\frac{(-1)^{n+1}}{n} = R_{99}$ , where $S_{99} +R_{99} = S$ (Sum of series)
3. First term of $R_{99} = \frac{(-1)^{101}}{100} = \frac{-1}{100} = Rt_{100}$
4. $S_{100}+Rt_{100}$ is the approximation
5. Thus: $S_{99}+Rt_{100} < S < S_{99}$

Since $Rt_{100}$ is the smallest number of the $R_{99}$: $S_{99}+Rt_{100}(aka \frac{-1}{100})$ < S

This is verified using wolfram alpha:

$S_{99}+Rt_{100} = \sum^{99}_{n=1}\frac{(-1)^{n+1}}{n} + (\frac{-1}{100})\approx 0.688172$ (link)

$S = \sum^{\infty}_{n=1}\frac{(-1)^{n+1}}{n} =\log{2}\approx 0.69315$ (link)

$S_{99} = \sum^{99}_{n=1}\frac{(-1)^{n+1}}{n} \approx 0.698172$ (link)

Every thing seem to point to the approximation being an underestimate. Why does Khan academy say it is an over estimate? (Ref below)

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The part of the question about over or under estimate (which I got the wrong answer on):

My intuition:

• I suspect it's got something to do with error bound/error estimates, but even then the math should make sense that the approximation should be smaller than the full alternating series thus an under estimate?(even for a negative 1st term of remainder, $Rt_{100} = \frac{-1}{100}$)

• I'm at this part of my journey (quiz) and Khan hasn't really explained error bounds yet, ideally I want to unknot the conflicting math vs my results. Even from reading, error bound seem to just mean the difference between "S and the approximation".

Answer 1

If we visualize $s$, $S_k$, and $R_k$ as points on the real line, where the negative numbers are off to the left and the positive ones off to the right, you can think of the error $R_k$ as the distance you need to walk to get from the partial sum $S_k$ to the actual sum $s$. This intuition works because of the definition of $R_k$.

$$R_k=s-S_k$$

This says the same thing as $s=S_k+R_k$, implying that if the partial sum $S_k$ is greater than $s$, you'll need to walk in the negative direction, aka to the left, to get to $s$. In symbols, this means that $R_k<0$ (similar reasoning holds when $R_k>0$). We can also easily reach this conclusion analytically:

$$S_k>s\iff s-S_k<0\iff R_k<0$$

Let's now consider your situation. You and Khan Academy correctly found the error to be approximately $-0.01$, and hence negative. Applying our intuition here, this means that we need to walk approximately $0.01$ units in the negative direction (to the left) to get from $S_{99}$ to the sum $s$. But if we need to walk to the left to get to $s$, then clearly we must've started our walk to the right of $s$! Thus, the partial sum $S_{99}$ must've been greater than the actual sum.

Likewise, if we know that the error is positive, then you need to walk to the right to get to the actual sum $s$, so our partial sum must've been to the left of $s$, i.e. $S_k<s$.