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I need help with Question 7. I'm confused with this question. I'm not sure what I'm supposed to compare or as a matter of fact, how to even approach this question.

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    Welcome to the site. Please edit your question so that all relevant information is in the question itself, here on the site. It will get more exposure and get you more help that way. – 2017-01-21

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The since for $0 \frac{\log(\epsilon(1-r))}{\log(r)}-1$$ (remember: $log(r)$ is negative for $r<1$).

The series in question is $$ \sum_{k=1}^n e^{-k}$$ corresponding to $r = \frac{1}{e}.$

Plugging in $\epsilon =10^{-4}$ and $r=1/e$ gives $n>8.67.$ This is the threshold above which the error becomes less than $\epsilon = 10^{-4}.$

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    I don't get your first and second step. Can you explain a bit more? – 2017-01-21
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    @john sure, edited – 2017-01-22
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Hint: You can write

$$\sum^{\infty}_{k=0} e^{-k}$$

as

$$\sum^{\infty}_{k=0} (\frac{1}{e})^{k}$$

since $r = \frac{1}{3}$. We have that the partial sum

$$\sum^{n}_{k=0} (\frac{1}{e})^{k} = \frac{1}{1 - \frac{1}{e}} + \mathcal{O}((\frac{1}{e})^{n+1})$$

Now find the value of $(\frac{1}{e})^{10}$. At which place does the first non-zero decimal occur?

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    So, the value is 0.167..x10^-4, since is 10^-4, this proves that is accurate to 10^-4 absolute accuracy, right? and why did you pick 11 as exponent, I don't get that part – 2017-01-21
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    I misread the question – that is what happens when people do not post a self-contained question but resort to scans of problem sheets. – 2017-01-21
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    But the answer is still correct because of the 10^-4 in the answer, right? – 2017-01-21