Find the limit of $\frac{1+\frac13+\frac{1}{3^2}+...+\frac{1}{3^n}}{1+\frac15+\frac{1}{5^2}+...+\frac{1}{5^n}}$ 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 $n+1$?
 A: 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
$$\lim_{n\to\infty}{\dfrac{\frac32-\frac12.\frac{1}{3^n}}{\frac54-\frac14.\frac{1}{5^n}}} = \frac{\frac32-0}{\frac54-0} = \frac65$$
A: 
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 $$\frac{3}{2}\cdot\frac{4}{5}=\frac65.$$
A: Sum of IGP is $$S_\infty=1+r+r^2+r^3+\ldots=\frac{1}{1-r}.$$
So the required limit $$=\frac{1/(1-1/3)}{1/(1-1/5)}=\frac{6}{5}.$$
