Convergence of (2^(n)+3^(n))/(4^(n)+5^(n)) using the Comparison Test

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Decide (with justification) if the following series converges or diverges;

Sum(1,infinty) (2^(n)+3^(n))/(4^(n)+5^(n))

I've tried using the ratio test but I couldn't see that it was helping in any way, should I be using a different type of test for this problem? I really can't see where to start with this one.
 
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Try and think of a clever comparison to which you can apply the ratio test. E.g. (2^n+3^n)/(4^n+5^n)<=(3^n+3^n)/(4^n+4^n). See, I substituted a larger numerator and a smaller denominator?
 
So if you apply the ratio test to the (3^(n)+3^(n))/(4^(n)+4^(n)) you find that this series converges as l<1 (l=3/4?). Is it then allowable to say that the original series converges as it is less than (3^(n)+3^(n))/(4^(n)+4^(n)) and therefore the limit must be lees than the limit of the above series and hence it must converge.
 
You tell me, ok? Look up the comparison test for series and make sure all the requirements are fulfilled. It's good practice.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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