I am just starting out learning mathematical induction and I got this homework question to prove with induction but I am not managing.
$$\sum\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}\ge\sqrt{n}}$$
Perhaps someone can help me out I don't understand how to move forward from here:
$$\sum\limits_{k=1}^{n+1}{\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{n+1}}\ge \sqrt{n+1}}$$
proof and explanation would greatly be appreciated 🙂
Thanks 🙂
EDIT sorry meant GE not = fixed 🙂
Best Answer
If you wanted to prove that $$ \sum_{k=1}^n \frac 1{\sqrt k} \ge \sqrt n, $$ that I can do. It is clear for $n=1$ (since we have equality then), so that it suffices to verify that $$ \sum_{k=1}^{n+1} \frac 1{\sqrt k} \ge \sqrt{n+1} $$ but this is equivalent to $$ \sum_{k=1}^{n} \frac 1{\sqrt k} + \frac 1{\sqrt{n+1}} \ge \sqrt{n+1} \ $$ and again equivalent to $$ \sum_{k=1}^n \frac{\sqrt{n+1}}{\sqrt k} + 1 \ge n+1 $$ so we only need to prove the last statement now, using induction hypothesis. Since $$ \sum_{k=1}^n \frac 1{\sqrt k} \ge \sqrt n, $$ we have $$ \sum_{k=1}^n \frac{\sqrt{n+1}}{\sqrt k} \ge \sqrt{n+1}\sqrt{n} \ge \sqrt{n} \sqrt{n} = n. $$ Adding the $1$'s on both sides we get what we wanted.
Hope that helps,