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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 for every $k \in \Bbb N, k\gt 0,$ we have $a_n \ge a_k$, for every $n \ge k^2$.

The only method I know is computing the difference $a_{n+1}-a_n$, but I didn't get to any result.

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    own efforts $\quad$? – 2017-02-14
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    The only method I know is computing the difference $a_{n+1}-a_n$, but it doesn;t work. – 2017-02-14
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    can you pin down why? – 2017-02-14
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    Because we have some fractions with + and other with - and I can't compare them. – 2017-02-14
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    why not adding this to the question? – 2017-02-14

2 Answers 2

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HINT

$a_n \ge \frac {n}{\sqrt {n^2+n}}$ and $a_k \le \frac {k}{\sqrt {k^2+1}}$

Can you take it from here?

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    All too easy (+1) – 2017-02-14
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    Yes, thank yo so much. – 2017-02-15
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Notice that for $k=1$ and $n=1$, $n\geq k^2$ but $a_n=a_k$ and thus $a_n>a_k$ does not hold for $k \in \Bbb N, k\gt 0$.

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    Just edited thanks – 2017-02-14