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$$\sum_{n=1}^{\infty} \frac{1}{3^n-2^n}$$

I know this series is convergent and using the ratio test. But I can't conclude the proving.

$$\begin{align}\lim_{n\to\infty} \left| \frac{\frac{1}{3^{n+1}-2^{n+1}}}{ \frac{1}{3^n-2^n}}\right| &= \lim_{n\to\infty} \left| \frac{3^n-2^n}{ 3^{n+1} - 2^{n+1}} \right|\\ &= \lim_{n\to\infty} \left| \frac{1-(\frac{2}{3})^n}{ 3(1-2^{n+1}/3^{n+1})}\right| \end{align}$$

By using calculation, the limit is 0. But I can't compute this limit by myself without using calculation. Any help?

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  • $\begingroup$ MAy be it is better to write in the same form the numerator and the denominator? $\endgroup$ – kmitov Feb 22 '16 at 12:29
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$\lim_{n\to\infty} \left| \frac{1-(\frac{2}{3})^n}{3(1-(\frac{2}{3})^{n+1})}\right| = \frac{1}{3}$ because $\lim_{ n \to \infty} \left(\frac{2}{3}\right)^n=0$

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Compare the general term of your series with $ \frac1{3^n} $: $$ \frac{\frac{1}{3^n-2^n}}{\frac1{3^n}}=\frac{3^n}{3^n-2^n}\stackrel{n\to+\infty}{\longrightarrow}1 $$ thus, being $\sum_n\frac1{3^n}$ convergent, so is the given series.

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You've almost got this one solved

$\lim_{n\to\infty} {(\frac{2}{3})}^n = \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} .... = 0.\dot{6} \times 0.\dot{6} \times 0.\dot{6} ... = 0$

so your solution will come out to $\frac{1}{3}$

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Also old style comparison: $$ n>1\implies 3^n-2^n=2^n((3/2)^n-1)\ge2^n\implies \frac1{3^n-2^n}\le\frac1{2^n}. $$

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