Proof that $w_n=e^{\sqrt{n}} - e^{\sqrt{n-1}} \approx \dfrac{e^{\sqrt{n}}}{2\sqrt{n}}$

Question

According to a article, author wrote: $w_n=e^{\sqrt{n}} - e^{\sqrt{n-1}} \approx \dfrac{e^{\sqrt{n}}}{2\sqrt{n}}$ for $n>1$. Can you Explain for me this?

Answer 1

For large $n$, $e^{\sqrt n} - e^{\sqrt{n-1}} \approx \left.\frac{d}{dx}[e^\sqrt x]\right|_{x=n} = \frac{e^\sqrt n}{2\sqrt n}$

Answer 2

We have \begin{align} e^{\sqrt{n}}-e^{\sqrt{n-1}} & = e^{\sqrt{n}}\left(1-e^{\sqrt{n-1}-\sqrt{n}}\right) \approx e^{\sqrt{n}}\left(1-\left(1+\sqrt{n-1}-\sqrt{n}\right)\right)\\ & = e^{\sqrt{n}} \left(\sqrt{n}-\sqrt{n-1}\right) = \dfrac{e^{\sqrt{n}}}{\sqrt{n}+\sqrt{n-1}} \approx \dfrac{e^{\sqrt{n}}}{2\sqrt{n}} \end{align}

Answer 3

You can even go further. Writing, as Leg did in his/her answer, $$w_n=e^{\sqrt{n}}-e^{\sqrt{n-1}} = e^{\sqrt{n}}\left(1-e^{\sqrt{n-1}-\sqrt{n}}\right)$$ consider $$\sqrt{n-1}-\sqrt{n}=\sqrt{n}\left(\sqrt{1-\frac 1n}-1 \right)$$ and use Taylor (or generalized binomial expansion) for small values of $x$ $$\sqrt{1-x}=1-\frac{x}{2}-\frac{x^2}{8}+O\left(x^3\right)$$ Replace $x$ by $\frac 1n$ to get $$\sqrt{1-\frac 1n}=1-\frac{1}{2 n}-\frac{1}{8 n^2}+O\left(\frac{1}{n^3}\right)$$ $$\sqrt{n-1}-\sqrt{n}=-\frac{1}{2 \sqrt{n}}-\frac{1}{8 n^{3/2}}+O\left(\frac{1}{n^{3/2}}\right)$$ Using Taylor again $$e^{\sqrt{n-1}-\sqrt{n}}=1-\frac{1}{2 \sqrt{n}}-\frac{1}{8 n}+O\left(\frac{1}{n^{3/2}}\right)$$ $$1-e^{\sqrt{n-1}-\sqrt{n}}=\frac{1}{2 \sqrt{n}}+\frac{1}{8 n}+O\left(\frac{1}{n^{3/2}}\right)$$ This finally makes $$w_n=e^{\sqrt{n}}-e^{\sqrt{n-1}}=\frac{e^{\sqrt{n}}}{2 \sqrt{n}}+\frac{e^{\sqrt{n}}}{8 n}+\cdots$$