Does the series converge or diverge?

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Homework Statement



Determine whether the series \sum_{n=1}^{\infty} \frac{1}{n+\sqrt{n}}
converges or diverges.

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The Attempt at a Solution



This is driving me nuts. I get nowhere with the root test or the ratio test. I can think of some larger series that diverge (no good) and some smaller series that converge (also no good). What test would you recommend?
 
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The integral test.

n+√n < n+n = 2n

\frac{1}{n + \sqrt{n}} &gt; \frac{1}{2n}
 
I'm confused. Doesn't the integral test have us say that if \int_{1}^{\infty}{\frac{1}{n+\sqrt{n}} dn} diverges, then our series diverges?

Where do you get the 2n from using the integral test?

Also, the relation n + \sqrt{n} \leq n+n[\tex] definitely holds, but isn&#039;t the series of 1/2n from n=1 to infinity a divergent harmonic series?
 
Ah, well, I suppose in either case, it's a divergent series. This makes me very happy. Thanks.
 
I should have said the the comparison test with the integral test . Sorry about any confusion.

1/2n diverges, and since it is greater than 1/(n+√n), the latter also diverges.
 
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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