# Calculate the limit [closed]

Please help me find $$\lim_{n\rightarrow\infty}{\frac{1+\frac{1}{3}+\frac{1}{9}+\cdots+\frac{1}{3^n}}{1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^n}}}$$

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## closed as off-topic by Jonas Meyer, Eric Stucky, Yes, Adam Hughes, Erick WongFeb 3 at 7:06

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This is really a very easy problem; you should have been able at least to recognize the geometric series in the numerator and denominator and express them in closed form. You’ll get much more benefit from the answers that you get here if you’ve put some serious thought into solving the problem first. –  Brian M. Scott Oct 7 '12 at 11:42

Sequences $1,\frac{1}{3},\frac{1}{9},\ldots\frac{1}{3^n}$ and $1,\frac{1}{2},\frac{1}{4},\ldots\frac{1}{2^n}$ are geometric sequences.

The sum of a geometric sequence $a_1, a_2, \ldots, a_{n-1}, a_n$ is

$$S_n=a_1\frac{1-q^n}{1-q}.$$

For the given sequence $1,\frac{1}{3},\frac{1}{9},\ldots\frac{1}{3^n}$ we have $a_1=1$, $q=\frac{a_2}{a_1}=\frac{1}{3}$, and for the $1,\frac{1}{2},\frac{1}{4},\ldots\frac{1}{2^n}$ we have $a_1=1$, $q=\frac{1}{2}$.

Therefore

$$1+\frac{1}{3}+\frac{1}{9}+\cdots+\frac{1}{3^n}=\frac{1-\left(\frac{1}{3}\right)^n}{1-\frac{1}{3}}\Rightarrow 1+\frac{1}{3}+\frac{1}{9}+\cdots+\frac{1}{3^n}=\frac{3}{2}\left[1-\left(\frac{1}{3}\right)^n\right],$$

$$1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^n}=\frac{1-\left(\frac{1}{2}\right)^n}{1-\frac{1}{2}}\Rightarrow1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^n}=2\left[1-\left(\frac{1}{2}\right)^n\right].$$

Definitely

$$\lim_{n\rightarrow\infty}{\frac{1+\frac{1}{3}+\frac{1}{9}+\cdots+\frac{1}{3^n}}{1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^n}}}=\lim_{n \to \infty}\frac{\frac{3}{2}[1-(\frac{1}{3})^n]}{2[1-(\frac{1}{2})^n]}=\frac{3}{4}\lim_{n \to \infty} \frac{[1-(\frac{1}{3})^n]}{[1-(\frac{1}{2})^n]}=\frac{3}{4},$$

because if $n \to \infty$, $(\frac{1}{3})^n, (\frac{1}{2})^n \to 0$.

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This does not change the result, but to be correct $1+q+q^2+\ldots+q^n=(1-q^{n+1})/(1-q)$ (note the $n+1$). –  enzotib Oct 7 '12 at 11:58

If you see a series sum at the denominator maybe stolz can be applied. $$\lim_{n\rightarrow\infty}{\frac{1+\frac{1}{3}+\frac{1}{9}+\cdots+\frac{1}{3^n}}{1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^n}}}=\lim_{n\rightarrow\infty}{ \frac{2^n \left (1+\frac{1}{3}+\frac{1}{9}+\cdots+\frac{1}{3^n} \right )}{1+2+4+\cdots+2^n}}{\mathrm{stoltz} \atop =}$$ $$\lim_{n\rightarrow\infty}{ \frac{2^{n+1} \left (1+\frac{1}{3}+\frac{1}{9}+\cdots+\frac{1}{3^{n+1}} \right )-{2^{n} \left (1+\frac{1}{3}+\frac{1}{9}+\cdots+\frac{1}{3^{n}} \right )}}{1+2+4+\cdots+2^{n+1}-(1+2+4+\cdots+2^{n})}}$$ $$=\lim_{n\rightarrow\infty}{ \frac{2\cdot2^{n} \left (1+\frac{1}{3}+\frac{1}{9}+\cdots+\frac{1}{3^{n} } \right )+2\cdot2^n\frac{1}{3^{n+1}}-{2^{n} \left (1+\frac{1}{3}+\frac{1}{9}+\cdots+\frac{1}{3^{n}} \right )}}{2^{n+1}}}$$ $$=\lim_{n\rightarrow\infty}{ \frac{2^{n} \left (1+\frac{1}{3}+\frac{1}{9}+\cdots+\frac{1}{3^{n} } \right )+2\cdot2^n\frac{1}{3^{n+1}}}{2^{n+1}}}=\lim_{n\rightarrow\infty} { \frac{ 1+\frac{1}{3}+\frac{1}{9}+\cdots+\frac{1}{3^{n} } }{2}}$$ $$=\lim_{n\rightarrow\infty} { \frac{ 1+3+9+\cdots+3^n }{2\cdot 3^n}}= \frac{1}{2}\lim_{n\rightarrow\infty} { \frac{ 1+3+9+\cdots+3^n }{ 3^n}}{\mathrm{Stoltz \atop =}}$$ $$\frac{1}{2}\lim_{n\rightarrow\infty} { \frac{ 1+3+9+\cdots+3^{n+1}- ( 1+3+9+\cdots+3^n )}{ 3^{n+1}-3^n}}=\frac{1}{2}\lim_{n\rightarrow\infty}\frac{3^{n+1}}{2\cdot 3^n}=\frac{3}{4}$$

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I think about how you could type this neat answer. :) –  Nancy Rutkowskie Oct 7 '12 at 13:05