Show that $1/\sqrt{1} + 1/\sqrt{2} + ... + 1/\sqrt{n} \leq 2\sqrt{n}-1$ for $n\geq 1$
I attempted the problem but I get stuck trying to show that if the statment is true for some $k\geq1$ then $k+1$ is also true.
My attempt: Base case $n=1$ is true by inspection
Now assume the statement is true for some $k\geq1$. Then $1/\sqrt{1} + 1/\sqrt{2} + ... + 1/\sqrt{n} +1\sqrt{n+1} \leq 2\sqrt{n}-1 + 1/\sqrt{n+1}$ but I am not sure how to show the rest.
hint: $\dfrac{1}{\sqrt{k}}<2\sqrt{k}-2\sqrt{k-1}$. You let $k$ runs from $1$ to $n$, and add the inequalities up.
If the proposition holds true for $n=m$
$$\sum_{r=1}^m\dfrac1{\sqrt r}\le2\sqrt m-1$$
$$\implies\sum_{r=1}^{m+1}\dfrac1{\sqrt r}=\dfrac1{\sqrt{m+1}}+\sum_{r=1}^m\dfrac1{\sqrt r}\le\dfrac1{\sqrt{m+1}}+2\sqrt m-1$$
It if sufficient to show $$\dfrac1{\sqrt{m+1}}+2\sqrt m-1\le2\sqrt{m+1}-1$$
$$\iff\dfrac1{\sqrt{m+1}}\le2[\sqrt{m+1}-\sqrt m]=\dfrac2{\sqrt{m+1}+\sqrt m}$$
$$\iff \sqrt{m+1}\ge\sqrt m$$ which is true
Without induction you could use this inequality
$$\frac1{\sqrt k}\le2(\sqrt{k}-\sqrt{k-1})$$ and then you telescope. Notice also that the same inequality is useful for your attempt.