Prove that $\sum_{n=1}^{\infty}(-1)^n\sin\left(\sqrt{n+1}-\sqrt{n}\right)$ converges by Leibniz.
The answer says that $\sin$ is continuous, monotonic around $0$ and the limit there is $0$, therefore the series conditionally converges by the Leibniz test.
I don't understand, Leibniz says that we need an $a_n$ which is monotonic decreasing to $0$, How they took this sentence and say that just around $0$ it is ok and enough for the proof?
Best Answer
The point is you're looking at the magnitude $\sin\left(\sqrt{n+1}-\sqrt{n}\right)$ as $n\rightarrow\infty$.
Take a look at what happens to $\sqrt{n+1}-\sqrt{n}$ - it monotonically decreases to $0$. That means for large enough $n$, the argument to $\sin$ is close to $0$.
At this point, since $\sin$ is itself monotonic around $0$, applying a monotonic function to a monotonic sequence leaves it monotonic.