I've to resolve a inequality using the induction: for every natural number $\geq 1$
$$ \sum_{k=1}^n \frac{1}{\sqrt{k}} \geq \sqrt{n} $$
At the end, I arrive at this result, but I don't know how I can continue: $\sqrt{n}+1/\sqrt{n+1} \geq \sqrt{n+1}$
$\begingroup$
$\endgroup$
2
-
$\begingroup$ Also: math.stackexchange.com/q/2149448/42969, math.stackexchange.com/q/1371267/42969. $\endgroup$– Martin RCommented Sep 11, 2017 at 13:07
-
$\begingroup$ I have removed (proof-verification) tag, since the post actually contains no proof/solution which is supposed to be checked. $\endgroup$– Martin SleziakCommented Sep 22, 2017 at 16:38
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
7
it is $$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}>n\frac{1}{\sqrt{n}}=\sqrt{n}$$
answered Sep 11, 2017 at 12:51
-
$\begingroup$ sorry, i don't have understand... $\endgroup$– fenigo69Commented Sep 11, 2017 at 12:54
-
$\begingroup$ since $$\frac{1}{\sqrt{n}}$$ is the smallest summand $\endgroup$ Commented Sep 11, 2017 at 12:56
-
$\begingroup$ ok, but sorry, i don't understand how can i proove the inequation using that $\endgroup$– fenigo69Commented Sep 11, 2017 at 13:01
-
$\begingroup$ we get $n$ summands and $$\frac{1}{\sqrt{n}}$$ is the smallest so your sum is graeter than $$n\cdot \frac{1}{\sqrt{n}}$$ $\endgroup$ Commented Sep 11, 2017 at 13:05
-