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Find the limit $$\lim_{n\to\infty}\dfrac{1+\frac13+\frac{1}{3^2}+...+\frac{1}{3^n}}{1+\frac15+\frac{1}{5^2}+...+\frac{1}{5^n}}$$

For the numerator we have $1+\frac13+\frac{1}{3^2}+...+\frac{1}{3^n}=\dfrac{1-\left(\frac{1}{3}\right)^{n+1}}{1-\frac13}=\dfrac32-\dfrac12\dfrac{1}{3^n}.$

By analogy, $1+\frac15+\frac{1}{5^2}+...+\frac{1}{5^n}=\dfrac54-\dfrac14.\dfrac{1}{5^n}$

So we have $$\lim_{n\to\infty}\dfrac{1+\frac13+\frac{1}{3^2}+...+\frac{1}{3^n}}{1+\frac15+\frac{1}{5^2}+...+\frac{1}{5^n}}=\lim_{n\to\infty}{\dfrac{\frac32-\frac12.\frac{1}{3^n}}{\frac54-\frac14.\frac{1}{5^n}}}=\lim_{n\to\infty}{\left(\dfrac{3^{n+1}-1}{2.3^n}\cdot\dfrac{4.5^n}{5^{n+1}-1}\right)}$$ What am I supposed to do next?

My initial mistake was that I thought $1+\frac13+\frac{1}{3^2}+...+\frac{1}{3^n}=\dfrac{1-\left(\frac13\right)^n}{1-\frac13}$. How could we actually see (and show) that the terms in the sum(s) are not $n$, but $n+1$?

kormoran
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    Nice work. Note that, $$\lim_{n\to\infty}\frac 1{3^n}=\lim_{n\to\infty}\frac 1{5^n}=0$$ Finally, you have $$\frac{3/2}{5/4} $$ – User Aug 10 '22 at 15:08

3 Answers3

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You were very close:

$$\lim_{n\to\infty}\dfrac{1+\frac13+\frac{1}{3^2}+...+\frac{1}{3^n}}{1+\frac15+\frac{1}{5^2}+...+\frac{1}{5^n}}=\lim_{n\to\infty}{\dfrac{\frac32-\frac12.\frac{1}{3^n}}{\frac54-\frac14.\frac{1}{5^n}}}$$

as you correctly derived.

If you simply use $\lim_{n\to\infty}\frac{1}{3^n}=\lim_{n\to\infty}\frac{1}{5^n}=0$ in that expression, you are left with

$$\lim_{n\to\infty}{\dfrac{\frac32-\frac12.\frac{1}{3^n}}{\frac54-\frac14.\frac{1}{5^n}}} = \frac{\frac32-0}{\frac54-0} = \frac65$$

Andreas Tsevas
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    Appreciate it! May I ask why $\lim_{n\to\infty}\dfrac{1}{3^n}=0$? Maybe as when $n\to\infty \Rightarrow 3^n\to\infty$ as well, and we arrive at $\dfrac{1}{\infty}$? – kormoran Aug 10 '22 at 15:12
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    @Kaloyan Yes, exactly. $3^n$ becomes arbitrarily large for large $n$, so $\frac{1}{3^n}$ approaches $0$. – Andreas Tsevas Aug 10 '22 at 15:17
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How could we actually see (and show) that the terms in the sum(s) are not $n$, but $+1$

The sum is $$1+\frac13+…\frac{1}{3^n}=\frac{1}{3^0}+\frac{1}{3^1}+\frac{1}{3^2}+…+\frac{1}{3^n}$$$$=\sum_{i=0}^n 3^{-n}.$$ The index $i$ starts from the value $\textbf 0$ (and not $1$) and ends at $n$. So the number of terms is $n+1$.

What am I supposed to do next?

From $$\lim_{n\to\infty}\left(\dfrac{3^{n+1}-1}{2\cdot 3^n}\cdot\dfrac{4.5^n}{5^{n+1}-1}\right)$$ you divide the numerator and denominator of the first part by $3^n$ and the numerator and denominator of the second part by $5^n$. $$\lim_{n\to \infty}\left(\frac{3-\frac{1}{3^n}}{2}\cdot\frac{4}{5-\frac{1}{5^n}}\right)$$ Note that $$\lim_{n\to \infty}\frac{1}{3^n}= \lim_{n\to \infty}\frac{1}{5^n}=0$$ so the limit is $$\frac{3}{2}\cdot\frac{4}{5}=\frac65.$$

insipidintegrator
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Sum of IGP is $$S_\infty=1+r+r^2+r^3+\ldots=\frac{1}{1-r}.$$ So the required limit $$=\frac{1/(1-1/3)}{1/(1-1/5)}=\frac{6}{5}.$$

ViktorStein
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MathDona
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