Does xn Converge? A Comparison Test

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Homework Statement



Does xn converge (Sum from n=1 to infinity) of xn = 1/(n + SQRTn)

Homework Equations



Using comparision test

The Attempt at a Solution



I separted into fractions of 1/SQRTn - 1/(1 + SQRTn) and i know that both of these diverge since the power of n is less than one but am stuck as to whether is converges or diverges and how to prove it...
 
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\frac{1}{n+n}\leq \frac{1}{n+\sqrt{n}}
 
therefore xn divereges...?
 
yeah that is right, since the harmonic series diverges, it also diverges when we multiply it by a constant.
 
sutupidmath said:
yeah that is right, since the harmonic series diverges, it also diverges when we multiply it by a constant.
Non-zero constant.

Not to offend you but that is what I like to call a "physics-type mistake".
 
Kummer said:
Non-zero constant.

Not to offend you but that is what I like to call a "physics-type mistake".

yeah that is what i actually meant, but thnx for pointing it out. and not am not offended in any way.
 
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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