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Does this series converge or diverge?

$$\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}-\frac{2}3}$$

I tried using the limit comparison test with $\frac{1}{\sqrt{n}}$, which diverges.

$$\lim_{n\to\infty}{\frac{{\sqrt{n}}}{\sqrt{n}-\frac{2}3}}=1$$

Then the series diverges, is this right or I'm wrong?

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  • $\begingroup$ looks good to me $\endgroup$
    – gt6989b
    Commented Dec 5, 2018 at 20:19

2 Answers 2

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Yes it is absolutely right, indeed note that

$$\dfrac{1}{\sqrt{n}-\frac{2}3}\sim \dfrac{1}{\sqrt{n}}$$

and the latter diverges for p test.

As an alternative by direct comparison test

$$\dfrac{1}{\sqrt{n}-\frac{2}3}\ge \dfrac{1}{\sqrt{n}}$$

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Yeah, just as another answer shows, you can see the convergence of such a sum of sequence does not matter w.r.t first several terms.

Then, you can see

$\frac{1}{\sqrt{n} -\frac{2}{3}} \sim \frac{1}{\sqrt{n}} = \frac{1}{n^{1/2}}$

notice that for those power less than or equals 1,(here it's 1/2), it's a diverge sequence.

(You may refer to any analysis book for this result.)

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