Determine for which values for $x\in \mathbb R$ series $\sum _{n=1}^\infty \displaystyle \frac{\sqrt{n+1}-\sqrt{n}}{n^x}$ convergent and divergent
My Attempts: I multiplied numerator and denumerator by conjugate so I obtained $\sum _{n=1}^\infty \displaystyle \frac{1}{n^x .(\sqrt{n+1}+\sqrt{n})}$. Apparently divergence test or n-th term test is useless as limit goes to $0$.
So I was looking for a test that can be useful for it. Actually, I want to compare series with $\sum _{n=1}^\infty \displaystyle \frac{1}{\sqrt{n+1}+\sqrt{n}}$. This series divergent by limit comparison test by series $\frac {1}{n}$. Now using limit comparison test we get $$\lim_{n\to \infty} \displaystyle \frac {1} {n^x}$$
If that limit exists and not equal to $0$ then our original series is divergent. But I believe that it is not possible for any $x$ because limit goes to $0$ for every time.
Moreover, I think ratio test is also useless. Any suggestion is preciated.
Best Answer
$$\frac{1}{n^x(\sqrt{n+1}+\sqrt{n})}\sim \frac{1}{2n^{x+1/2}}$$ hence the series converges iff $$x>\frac12.$$