1
$\begingroup$

Does this series converge or diverge? $$\sum_{n=2}^\infty\frac{n^{1/2}}{n-1}$$

I think it is divergent. This is how I proceeded: Divide numerator and denominator by $n^{1/2}$ and it is greater than or equal to $\frac1{n^{1/2}}$ and since this is divergent, by comparison test, given series is divergent. Am I done?

$\endgroup$
  • $\begingroup$ What do you think ? What did you try already ? $\endgroup$ – Claude Leibovici Sep 25 '16 at 10:32
  • $\begingroup$ I think its divergent. This is how i proceeded divide numerator and denominator by n^1/2 and it is greater than or equal to 1/(n)^1/2 and since this is divergent..by comparison test, given series is divergent...am I done? $\endgroup$ – user371841 Sep 25 '16 at 10:33
  • $\begingroup$ You can use the limit comparison test, for instance. $\endgroup$ – Mark Sep 25 '16 at 10:34
  • 1
    $\begingroup$ Please post your work. Otherwise, there are big chances that the question will be closed for missing context. $\endgroup$ – Claude Leibovici Sep 25 '16 at 10:34
  • 1
    $\begingroup$ Better ! And, by the way, welcome to MSE which is a fantastic site. Cheers :-) $\endgroup$ – Claude Leibovici Sep 25 '16 at 10:39
1
$\begingroup$

Hint: You can use the comparison test and the fact that $$\frac{n^{1/2}}{n-1} > \frac{1}{n-1} > \frac1n$$

In your question, you used the comparison test with $\frac1{\sqrt{n}}$. This also works, but I would write out a little more specifically why the terms are larger than $\frac1{\sqrt{n}}$, like I did above with $\frac1n$.

$\endgroup$
0
$\begingroup$

Limit Comparison Test + Integral Test:

$$\sum_{n=2}^\infty\frac{n^{1/2}}{n-1}\implies\sum_{n=2}^\infty\frac{n^{1/2}}n=\sum_{n=2}^\infty\frac1{n^{1/2}}\implies\int_2^\infty x^{-1/2}dx$$

$\endgroup$
0
$\begingroup$

We can see that by D'Alembert's ratio test the limit becomes 1 and thus the test fails. Then we use Raabe's test and after some manipulation we get the limit as 0.5 when n tends to infinity. Thus the series is divergent. And also the method you have used is perfectly right as the terms of the series are less than n^0.5 for all n>1 .

$\endgroup$
0
$\begingroup$

Simplest (as often): Use equivalents, since it is a series with positive terms. The general term is $$\frac{n^{1/2}}{n-1}\sim_\infty\frac{n^{1/2}}{n}=\frac1{n^{1/2}},$$ which is a divergent Riemann series.

$\endgroup$

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.