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After asking a question in another thread regarding a specific series, I was directed towards Raabe's test. However, I have some doubts regarding the proof at https://en.wikipedia.org/wiki/Ratio_test.

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Questions:

1) How does the author deduce $ca_Ne^{-R\log(n)}$ from $a_Ne^{-R(\frac{1}{N}+...+\frac{1}{n})}$? I mean specifically $\log(n)$. I tried to use Maclaurin expansion, but I could not see.

2) In the same expression $a_Ne^{-R(\frac{1}{N}+...+\frac{1}{n})}\geqslant ca_Ne^{-R\log(n)}$. What is $c$? Why use $c$?

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    $\begingroup$ There is a known limit: $\lim_{n\to \infty} (\frac{1}{1}+\frac{1}{2}+\dots+\frac{1}{n}-\log(n)) = \gamma$, where $\gamma$ is the Euler-Mascheroni constant. I believe it was used here. Or some other similar approximation. $\endgroup$
    – Jakobian
    Jul 1, 2018 at 16:08
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    $\begingroup$ Bound $\ln(n)=\int_{1}^{n}\frac{dx}{x}$ by its upper Riemann and lower Riemann sum for the partition at the integer points. Constants like $1+\frac{1}{2}+...+\frac{1}{N}$ you can throw into the $O(1)$. $\endgroup$
    – user566930
    Jul 1, 2018 at 16:49
  • $\begingroup$ In the article, the sentence "The proof of the other half is entirely analogous, with all the equalities reversed" is baloney... In particular the inequality $1+R/n \leq e^{R/n}$ can't be reversed. $\endgroup$ Jul 1, 2018 at 18:04
  • $\begingroup$ @DanielWainfleet You might note the sentence after that one... $\endgroup$ Jul 2, 2018 at 0:26

2 Answers 2

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1) I comes from the lower Riemann sums for $\int_{N-1}^n\frac{\mathrm d t}t$ $$\log n-\log(N-1)=\int_{N-1}^n\frac{\mathrm d t}t\ge\sum_{k=N}^n\frac 1k$$ $$\text{so that }\qquad\mathrm e^{-R\bigl(\tfrac 1N+\dots+\tfrac 1n\bigr) }\ge\mathrm e^{-R\bigl(\log n-\log(N-1)\bigr)}=\mathrm e^{-R\log n}\underbrace{\mathrm e^{R\log(N-1)}}_{=c}$$

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There's a common convention in analysis that the letter $c$ denotes some constant, the value of which may change each time it appears. (That is, when one writes $A\le cB$ and then later writes $X\le cY$ the two $c$'s are not necessarily the same constant.)

The author is assuming first that the reader knows that $\log(n)-(1+1/2+\dots+1/n)$ is bounded. This is easy to see:

$$\begin{align}\log(n)-\log(n-1)-\frac1n&=\int_{n-1}^n\left(\frac 1t-\frac1n\right)\,dt \\&=\int_{n-1}^n\frac{n-t}{nt}\,dt.\end{align}$$

Now $t\in[n-1,n]$ implies that $$0\le\frac{n-t}{nt}\le\frac1{n(n-1)},$$ so $$0\le \log(n)-\log(n-1)-\frac1n\le\frac1{n(n-1)}.$$Since $\sum\frac1{n(n-1)}<\infty$ it follows from the triangle inequality that there exists $c$ such that $$\left|\log(n)-(1+\frac12+\dots+\frac1n\right|\le c.$$

Hence if $N$ is fixed there exists $c_N$ such that$$\left|\log(n)-(\frac1N+\frac1{N+1}+\dots+\frac1n\right|\le c_N$$for $n>N$. So if $R>0$ and $n>N$ then $$e^{-R(\frac1N+\dots\frac1n)}\ge e^{-R(\log(n)+c_N)} =e^{-Rc_N}e^{-R\log(n)}=ce^{-R\log(n)},$$where $c=e^{-Rc_N}$.

In case it comes up elsewhere, the same argument shows that$$e^{-R(\frac1N+\dots+\frac1n)}\le e^{Rc_N}e^{-R\log(n)}.$$

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