Exercise 8.9

Question

Exercise 9: 
    (a) Put $s_N=1+(1/2)+\cdots+(1/N)$. Prove that $$\lim_{N\rightarrow\infty}(s_N-\log N)$$ exists.

(b) Roughly how large must $m$ be so that $N=10^m$ satisfies $s_N>100$.

analambanomenos · Accepted Answer

(a) Since the minimum and maximum values of $1/x$ in the interval $[n,n+1]$ are $1/(n+1)$ and $1/n$, respectively, we have $1/(n+1)\le\int_n^{n+1}(1/x)\,dx=\log(n+1)-\log(1)\le 1/n$. Hence,
    $$\big(s_{N+1}-\log(N+1)\big)-(s_N-\log N)=\frac{1}{N}-\int_N^{N+1}\frac{1}{x}\,dx\ge 0,$$ so that the sequence $s_N-\log N$ is nondecreasing. Also, since $\log(1)=0$, we have
    \begin{align*}
        \frac{1}{2}+\cdots+\frac{1}{N} &\le \sum_{n=0}^{N-1}\big(\log(n+1)-\log(n)\big) = \log N \
        -\log N &\le -\frac{1}{2}-\cdots-\frac{1}{N} \
        s_N-\log N &\le 1.
    \end{align*}
    so that the sequence is also bounded above. Hence by Theorem 3.14 the limit of the sequence exists.

(b) From part (a), $s_N\approx\gamma+\log N$. Using the estimate $\gamma\approx 0.577$ and solving for $100\approx\gamma+\log 10^m$, we get the approximate solution $m=43$.

Exercise 8.9

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Exercise 8.9

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