Big O and approximation to a specific number

Question

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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Answer 1

The since for $0<r<1$, $$\sum_{k=0}^\infty r^k = \frac{1}{1-r}$$ and in general $$\sum_{k=0}^nr^k=\frac{1-r^{n+1}}{1-r},$$ the error in the series truncated at term $n$ is $$\left|\sum_{k=0}^\infty r^k-\sum_{k=0}^n r^k\right|= \frac{1}{1-r} - \frac{1-r^{n+1}}{1-r} = \frac{r^{n+1}}{1-r}$$ so in order to have the error be less than $\epsilon,$ we need $$ \frac{r^{n+1}}{1-r}< \epsilon,$$ which can be rewritten $$ e^{(n+1)\ln(r)} <(1-r)\epsilon,$$ and taking natural log of both sides gives $$(n+1)\ln(r) <\ln((1-r)\epsilon).$$ Solving the inequality for $n$ gives $$ n > \frac{\log(\epsilon(1-r))}{\log(r)}-1$$ (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 $\epsilon = 10^{-4}.$

Answer 2

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?