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I've tried to calculate the convergence radius of the following power series: $$\sum_{n=1}^{\infty}\frac{3^n+4^n}{5^n+6^n}x^n$$

The Cauchy–Hadamard theorem doesn't help in this situation (I think). So what I did is I tried to apply the d'Alembert ratio test to it and got the following limit: $$\lim_{n\to\infty}\frac{\frac{3^n+4^n}{5^n+6^n}}{\frac{3^{n+1}+4^{n+1}}{5^{n+1}+6^{n+1}}}=\lim_{n\to\infty}\frac{(3^n+4^n)(5^{n+1}+6^{n+1})}{(5^n+6^n)(3^{n+1}+4^{n+1})}$$ but I haven't mannaged to solve in any way. I tried to calculate the limit of the function $$\lim_{x\to\infty}\frac{(3^x+4^x)(5^{x+1}+6^{x+1})}{(5^x+6^x)(3^{x+1}+4^{x+1})}$$ but of course that Lhospital's rule doesn't help (because it's in the power of n) so I was wondering:

  • Is there a different way to find the convergence radius by using something other than the ration test?
  • Might there be a identity regarding $$a^n+b^n=?$$ or $$\frac{a^n+b^n}{a^{n+1}+b^{n+1}}=?$$
nono
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    If $\lvert a\rvert > \lvert b\rvert$, it's quite helpful here to write $$a^n + b^n = a^n\cdot \Biggl(1 + \biggl(\frac{a}{b}\biggr)^n\Biggr).$$ – Daniel Fischer Dec 07 '16 at 15:36
  • @nono Just curious, is this a textbook problem? If so, what is the answer the book is providing? – imranfat Dec 07 '16 at 16:46
  • @imranfat Sorry about the delay... It's not a textbook problem, but my guess is that they would've given the same answer. – nono Dec 14 '16 at 19:57

2 Answers2

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$$\begin{align} \frac{(3^n+4^n)(5^{n+1}+6^{n+1})}{(5^n+6^n)(3^{n+1}+4^{n+1})} &=\frac{6\cdot24^n+6\cdot18^n+5\cdot20^n+5\cdot15^n}{4\cdot24^n+4\cdot20^n+3\cdot18^n+3\cdot15^n}\\ &=\frac{6+6(3/4)^n+5(5/6)^n+5(5/8)^n}{4+4(5/6)^n+3(3/4)^n+3(5/8)^n}\to\frac32 \end{align}$$

ajotatxe
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I'd go about like this:

$$\lim_{n\to\infty}\frac{\frac{3^n+4^n}{5^n+6^n}}{\frac{3^{n+1}+4^{n+1}}{5^{n+1}+6^{n+1}}}=\lim_{n\to\infty}\frac{(3^n+4^n)(5^{n+1}+6^{n+1})}{(5^n+6^n)(3^{n+1}+4^{n+1})}$$ $$=\lim_{n\to\infty}\frac{((\frac{3}{4})^n+1)(5\cdot (\frac{5}{6})^n+6)}{((\frac{5}{6})^n+1)((\frac{3}{4})^n\cdot 3+4)}$$ $$=\frac{(0+1)(5\cdot 0 + 6)}{(0 + 1)(3 \cdot 0 + 4)}$$ $$=\frac32 > 1$$

The conclusion is left for the interested reader.

SchrodingersCat
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    Ok, a silly question here. Why are the $n+1$ terms in the denominator, and the $n$ terms in the numerator? For the ratiotest, shouldn't that be the other way around? That is, aren't we looking at $\frac{a_{n+1}}{a_n}$ for $n$ going to infinity? That is how the ratiotest is shown in my book and...https://en.wikipedia.org/wiki/Ratio_test – imranfat Dec 07 '16 at 16:36
  • @imranfat Yes you are right. The expression will be inverted in that case and so will be the limit. – SchrodingersCat Dec 07 '16 at 18:58