# 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 $a>1$

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

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 $$ln(1-x)=-\left(x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+...\right)$$ Nov 28, 2019 at 11:31
• One problem with this is the series does not converge for the values of $a$ we're concerned about $(a>1)$. Nov 28, 2019 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, 2019 at 11:35
• This is almost a particular case, maybe it could help. Nov 28, 2019 at 11:36
• @ArnaudD. thank you! I will analyze it. Nov 28, 2019 at 11:39

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

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

• Is the last limit correct? It would look like the limit should be $1/(a-1)$. But using Stolz–Cesàro is a good idea! Nov 28, 2019 at 11:28
• @user Thank you very much! Is there any literature you could recommend on this topic? Nov 28, 2019 at 11:29
• @Fimpellizieri Opsss yes of course! I fix it
– user
Nov 28, 2019 at 11:29
• @VerkhotsevaKatya Simply refer to Stoltz-Cesaro theorem.
– user
Nov 28, 2019 at 11:31

\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 $$(2)$$.

• @VerkhotsevaKatya: when $a\gt1$, those terms are larger. It's always best to start with the major terms.
– robjohn
Nov 28, 2019 at 14:11
• I don't understand why the domination of (1) by (4) proves (2). Could someone explain me? Nov 29, 2019 at 16:35
• @Jeanba: This follows from the discrete version of the Dominated Convergence Theorem.
– robjohn
Nov 29, 2019 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. Nov 29, 2019 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, 2019 at 21:04