2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ looks good to me $\endgroup$ – gt6989b Dec 5 '18 at 20:19
3
$\begingroup$

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

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

$\endgroup$
1
$\begingroup$

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.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.