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$$\underset{n\rightarrow\infty}\lim{\frac{n}{a^{n+1}}\left(a+\frac{a^2}{2}+\frac{a^3}{3}+\cdots+\frac{a^n}{n}\right)}=?, \;\;a>1$$

In Shaum's Mathematical handbook of formulas and tables I've seen: $$\;\;\;\;\;\;\;\;\;\;\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots\;,x\in\langle-1,1]\;\;\;\;\;\;\;$$

$$\frac{1}{2}\ln{\Bigg(\frac{1+x}{1-x}\Bigg)}=1+\frac{x^3}{3}+\frac{x^5}{5}+\frac{x^7}{7}+\cdots\;\;\;,x\in\langle-1,1\rangle$$ The term in parentheses reminded me of the harmonic series. I thought of using the Taylor series. Is that a good idea? It says $a>0$ so I probably can't use these two formulas. On the other hand: $$e^x=x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots\;\;\;\;\;\;,$$ but there are no factorials in the denominators.

Source in Croatian: 2.kolokvij, matematička analiza

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  • $\begingroup$ Shouldn't $$ln(1-x)=-\left(x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+...\right)$$ $\endgroup$
    – Naman Jain
    Nov 28, 2019 at 11:31
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    $\begingroup$ One problem with this is the series does not converge for the values of $a$ we're concerned about $(a>1)$. $\endgroup$ Nov 28, 2019 at 11:33
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    $\begingroup$ Look at $\frac{a^n}n+\frac{a^{n-1}}{n-1}+\cdots$ as $\frac{a^n}n\left(1+\frac{n}{n-1}\frac1a+\frac{n}{n-2}\frac1{a^2}+\cdots\right)$ $\endgroup$
    – robjohn
    Nov 28, 2019 at 11:35
  • $\begingroup$ This is almost a particular case, maybe it could help. $\endgroup$
    – Arnaud D.
    Nov 28, 2019 at 11:36
  • $\begingroup$ @ArnaudD. thank you! I will analyze it. $\endgroup$
    – PinkyWay
    Nov 28, 2019 at 11:39

2 Answers 2

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By Stoltz-Cesaro

$$\frac{n}{a^{n+1}}\left(a+\frac{a^2}{2}+\frac{a^3}{3}+…+\frac{a^n}{n}\right)=\frac{\left(a+\frac{a^2}{2}+\frac{a^3}{3}+…+\frac{a^n}{n}\right)}{\frac{a^{n+1}}{n}}$$

we obtain

$$\frac{\frac{a^{n+1}}{n}}{\frac{a^{n+2}}{n+1}-\frac{a^{n+1}}{n}}=\frac1{\frac{na}{n+1}-1} \to \frac1{a-1}$$

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    $\begingroup$ Is the last limit correct? It would look like the limit should be $1/(a-1)$. But using Stolz–Cesàro is a good idea! $\endgroup$ Nov 28, 2019 at 11:28
  • $\begingroup$ @user Thank you very much! Is there any literature you could recommend on this topic? $\endgroup$
    – PinkyWay
    Nov 28, 2019 at 11:29
  • $\begingroup$ @Fimpellizieri Opsss yes of course! I fix it $\endgroup$
    – user
    Nov 28, 2019 at 11:29
  • $\begingroup$ @VerkhotsevaKatya Simply refer to Stoltz-Cesaro theorem. $\endgroup$
    – user
    Nov 28, 2019 at 11:31
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$$ \begin{align} \lim_{n\to\infty}\frac{n}{a^{n+1}}\left(a+\frac{a^2}2+\cdots+\frac{a^n}n\right) &=\lim_{n\to\infty}\left(\frac1a+\frac{n}{n-1}\frac1{a^2}+\frac{n}{n-2}\frac1{a^3}+\cdots\right)\tag1\\ &=\frac1a+\frac1{a^2}+\frac1{a^3}+\cdots\tag2\\ &=\frac1{a-1}\tag3 \end{align} $$ The series on the right side of $(1)$ is dominated by $$ \frac1a+\frac2{a^2}+\frac3{a^3}+\cdots=\frac{a}{(a-1)^2}\tag4 $$ which validates $(2)$.

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    $\begingroup$ @VerkhotsevaKatya: when $a\gt1$, those terms are larger. It's always best to start with the major terms. $\endgroup$
    – robjohn
    Nov 28, 2019 at 14:11
  • $\begingroup$ I don't understand why the domination of (1) by (4) proves (2). Could someone explain me? $\endgroup$
    – Jeanba
    Nov 29, 2019 at 16:35
  • $\begingroup$ @Jeanba: This follows from the discrete version of the Dominated Convergence Theorem. $\endgroup$
    – robjohn
    Nov 29, 2019 at 17:25
  • $\begingroup$ Thanks for answering! I still don't see from the formula what N and $f_N(n)$ are exactly here. $$\lim_{N\rightarrow\infty}\sum_{n=-\infty}^\infty f_N(n)$$ can $N$ depends on $n$? I'm a bit lost, sorry if this is super easy. $\endgroup$
    – Jeanba
    Nov 29, 2019 at 18:25
  • $\begingroup$ @Jeanba: Since $N$ appears outside of the scope of $n$ (which is only defined inside the summation), $N$ cannot depend on $n$. $\endgroup$
    – robjohn
    Nov 29, 2019 at 21:04

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