Finding $\lim_{n\to\infty}{\frac{n}{a^{n+1}}\left(a+\frac{a^2}{2}+\frac{a^3}{3}+\cdots+\frac{a^n}{n}\right)}$ where $a>1$

Question

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

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

Shouldn't $$ln(1-x)=-\left(x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+...\right)$$ — Naman Jain, Nov 28 '19 at 11:31
One problem with this is the series does not converge for the values of $a$ we're concerned about $(a>1)$. — Fimpellizzeri, Nov 28 '19 at 11:33
Look at $\frac{a^n}n+\frac{a^{n-1}}{n-1}+\cdots$ as $\frac{a^n}n\left(1+\frac{n}{n-1}\frac1a+\frac{n}{n-2}\frac1{a^2}+\cdots\right)$ — robjohn, Nov 28 '19 at 11:35

user · Answer 1 · 2019-11-28T11:30:27.997

8

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

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

edited Nov 28 '19 at 11:30

answered Nov 28 '19 at 11:24

user

154,566

2

Is the last limit correct? It would look like the limit should be $1/(a-1)$. But using Stolz–Cesàro is a good idea! – Fimpellizzeri Nov 28 '19 at 11:28
@user Thank you very much! Is there any literature you could recommend on this topic? – PinkyWay Nov 28 '19 at 11:29
@Fimpellizieri Opsss yes of course! I fix it – user Nov 28 '19 at 11:29
@VerkhotsevaKatya Simply refer to Stoltz-Cesaro theorem. – user Nov 28 '19 at 11:31

robjohn · Accepted Answer · 2019-11-28T12:17:19.377

5

$$ \begin{align} \lim_{n\to\infty}\frac{n}{a^{n+1}}\left(a+\frac{a^2}2+\cdots+\frac{a^n}n\right) &=\lim_{n\to\infty}\left(\frac1a+\frac{n}{n-1}\frac1{a^2}+\frac{n}{n-2}\frac1{a^3}+\cdots\right)\tag1\\ &=\frac1a+\frac1{a^2}+\frac1{a^3}+\cdots\tag2\\ &=\frac1{a-1}\tag3 \end{align} $$ 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 $(2)$.

edited Nov 28 '19 at 12:17

answered Nov 28 '19 at 12:01

robjohn

345,667

Thank you for your approach! It is really elegant to look in the opposite direction. – PinkyWay Nov 28 '19 at 13:19
1

@VerkhotsevaKatya: when $a\gt1$, those terms are larger. It's always best to start with the major terms. – robjohn Nov 28 '19 at 14:11
I don't understand why the domination of (1) by (4) proves (2). Could someone explain me? – Jeanba Nov 29 '19 at 16:35
@Jeanba: This follows from the discrete version of the Dominated Convergence Theorem. – robjohn Nov 29 '19 at 17:25
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. – Jeanba Nov 29 '19 at 18:25
@Jeanba: Since $N$ appears outside of the scope of $n$ (which is only defined inside the summation), $N$ cannot depend on $n$. – robjohn Nov 29 '19 at 21:04
@Jeanba: the limit in this question is $\lim\limits_{n\to\infty}\sum\limits_{k\in\mathbb{Z}}\frac{n}{n-k+1}\frac{[1\le k\le n]}{a^k}$ where $[\dots]$ are Iverson brackets. Does that help? – robjohn Nov 30 '19 at 12:43

Finding $\lim_{n\to\infty}{\frac{n}{a^{n+1}}\left(a+\frac{a^2}{2}+\frac{a^3}{3}+\cdots+\frac{a^n}{n}\right)}$ where $a>1$

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