Let us consider the sequence $(a_n)_{n \ge 1}$ such that $$a_n=\frac {1}{\sqrt {n^2+1}}+ \frac {1}{\sqrt {n^2+2}} + \dots +\frac {1}{\sqrt {n^2+n}}.$$ Show that the sequence is monotone.
This question is already posted, but the answers are wrong.
Monotonity and sequence problem
2
$\begingroup$
sequences-and-series
monotone-functions
-
0Do you mean [here](http://math.stackexchange.com/questions/2088985/how-can-i-solve-the-following-sequence)? – 2017-02-24
-
0Yes, it is a mistake in a fraction. – 2017-02-24
-
0It looks like each $a_n$ is clearly greater than $a_{n-1}$ since it gets a positive term added. Am I missing something? – 2017-02-24
-
0You also substract some of the first terms. – 2017-02-24
-
0https://en.wikipedia.org/wiki/Karamata's_inequality gives a way to go. – 2017-02-24
-
0How can I use it? @JackD'Aurizio – 2017-02-24
-
0Another, probably simpler way, is to exploit the [Hermite-Hadamard inequality][1], given that $\frac{1}{\sqrt{1+x}}$ is a convex function on $\mathbb{R}^+$. In particular $$-2+\sqrt{1+\frac{1}{n}}=\int_{0}^{1/n}\frac{dx}{\sqrt{1+x}}\geq \frac{1}{n^2}\left[\frac{1}{2}+\sum_{k=1}^{n-1}\frac{1}{\sqrt{1+\frac{k}{n^2}}}+\frac{1}{2\sqrt{1+\frac{1}{n}}}\right]$$ and a similar inequality in the opposite direction holds. Once we prove $l_n\leq a_n\leq r_n$ and $r_n\leq l_{n+1}$ we are done. [1]: https://en.wikipedia.org/wiki/Hermite%E2%80%93Hadamard_inequality – 2017-02-24
-
0What did you try? – 2017-03-05
-
0While the Accepted Answer on the duplicate Question does contain errors, this scarcely justifies posting a known duplicate. You have the reputation needed to Comment on the posts of others, so you should use that privilege rather than repeating the same problem. – 2017-03-06
-
0Possible duplicate of [How can I solve the following sequence](http://math.stackexchange.com/questions/2088985/how-can-i-solve-the-following-sequence) – 2017-03-06