wan't know how to start this problem, tried to multiply by $\sqrt{n}$, but it does not work. Can someone help me with that? Thanks. Here is the inequality. $$2(\sqrt{n+1}-\sqrt{n}) < \frac{1}{\sqrt{n}} < 2(\sqrt{n}-\sqrt{n-1})$$
Asked
Active
Viewed 170 times
2
-
See also https://math.stackexchange.com/questions/263115/proof-of-inequality-2-sqrtn1-sqrtn-frac1-sqrtn-2-sqrtn, https://math.stackexchange.com/questions/1139563/proof-by-induction-2-sqrtn1-sqrtn-frac1-sqrtn-2-sqrtn – Martin R Jun 22 '17 at 05:01
-
1hint: use induction – MAN-MADE Jun 22 '17 at 06:25
-
hint: use integration on $1/\sqrt{x}$ – MAN-MADE Jun 22 '17 at 06:27
2 Answers
5
Hint $$\sqrt{n+1}-\sqrt{n}=(\sqrt{n+1}-\sqrt{n})\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}\\ \sqrt{n}-\sqrt{n-1}=(\sqrt{n}-\sqrt{n-1})\frac{\sqrt{n}+\sqrt{n-1}}{\sqrt{n}+\sqrt{n-1}}$$
N. S.
- 132,525
5
By MVT,
$$\sqrt {n}-\sqrt {n-1}=\frac {1}{2\sqrt {c}} $$
$$\sqrt {n+1}-\sqrt {n}=\frac {1}{2\sqrt {d}} $$ with $$n-1 <c <n<d <n+1 . $$
hamam_Abdallah
- 62,951