determine the values of positive k for which the series

from n=2 to infinity of ((n)^(1/2)) / (n-n^k)

Question

determine the values of positive k for which the series

from n=2 to infinity of ((n)^(1/2)) / (n-n^k)

sirm3d · Answer

is covergent?
$$\large \frac{\sqrt n}{n-n^k}\frac{\sqrt n}{\sqrt n(\sqrt n-n^{k-1/2})}=\frac{1}{n^{1/2}-n^{k-1/2}}$$
the series $$\large \sum \frac{1}{n^{1/2}-n^{n^k-1/2}}$$ can be compared to a p-series that is convergent when $p>1$.

Here, $$\large p=k-1/2$$

anonymous · Answer

thank you so much. i was just really stuck on how to approach this. i'm confused on how you got to series 1/ (n^(1/2)- n^(n^k-1/2). i'm not sure where the from

anonymous · Answer

*where the n came from

sirm3d · Answer

factor n^(1/2), so that you can cancel it with the numerator.

anonymous · Answer

i saw that. it's just for your final summation you put an n to the n power. i think that was just a little mistake.

anonymous · Answer

how would you do a p series test if there are two ps under 1 fraction.  you have n^(1/2) and n^(k-1/2)?