Root test vs. ratio test question

cue928
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I am doing the following practice problem in prep for an exam:
sum from n=0 to infinity: (3^n)/(n+1)^n
The book says to use the ratio test on it, which I did, but would the root test also apply to this?
 
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The root test applies, and in fact is probably the simplest test to use in this case. What was your conclusion based on the ratio test?
 
cue928 said:
I am doing the following practice problem in prep for an exam:
sum from n=0 to infinity: (3^n)/(n+1)^n
The book says to use the ratio test on it, which I did, but would the root test also apply to this?


Well...

\sqrt[n]{\frac{3^n}{(n+1)^n}} =

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.
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I had initially changed it to be:
(3/(1+n))^n. Applying the nth square root, I got lim n approaches infinity 3/1+n = 0.
 
cue928 said:
I got lim n approaches infinity 3/1+n = 0.

Ok, so what's the conclusion?
 
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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