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I still do not have the answers I want for this thread I made mathematical induction methods, or rather it made me more confused in what I potentially am doing wrong. So I will try to ask again and see what is wrong with the method (I do know how to do it the standard way, just wondering if this method also works):

Question: Prove by mathematical induction that for $n\ge2, n\in \mathbb{Z^+}$: $$\sum_{r=1}^{n}\frac{1}{\sqrt{r}}>\sqrt{n}$$

Method: If I go straight to the induction step to proving $P(k+1)$, then we have: $$\sum_{r=1}^{k+1}\frac{1}{\sqrt{r}}>\sqrt{k+1}$$ $$1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}-\sqrt{k+1}$$ $$1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}-\sqrt{k+1}>\sqrt{k}+\frac{1}{\sqrt{k+1}}-\sqrt{k+1}=\frac{1}{\sqrt{k+1}+\sqrt{k}}+\frac{1}{\sqrt{k+1}}$$ However $k\ge 2$, and therefore the expression: $$\frac{1}{\sqrt{k+1}+\sqrt{k}}+\frac{1}{\sqrt{k+1}}>0$$ Hence: $$1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}-\sqrt{k+1}>\frac{1}{\sqrt{k+1}+\sqrt{k}}+\frac{1}{\sqrt{k+1}}>0$$ $$1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}-\sqrt{k+1}>0$$ $$1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}>\sqrt{k+1}$$

Hence by mathematical induction, we proved the result.

Is this a valid method?

Aurora Borealis
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    If your current thread on the problem doesn't have the answers you desire, then don't open another thread. Continue the discussions in the original thread, and wait for someone to come along - or put a bounty on the question to get further attention. – PrincessEev Mar 21 '19 at 03:28
  • I have and I am not getting any, and why am I not allowed to open another thread? This thread is my confusion I had with an example of a new problem. And brushed up quite abit with respect to what I wanted to ask. – Aurora Borealis Mar 21 '19 at 03:30
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    Your original thread is concerned with the validity of the method. Changing the example used doesn't really change anything in the matter. If I might further critique, your lack of explanation as to the method in question being used - assuming we can just follow along - is likely the very reason you're not getting answers in the first place. It would be appropriate to edit your original question accordingly and hope someone can answer. – PrincessEev Mar 21 '19 at 03:33
  • I do not see what is wrong with making two posts on similar topics? I am simply asking what I do not get, and this kind of feedback is really unnecessary. – Aurora Borealis Mar 21 '19 at 03:35
  • They're not "similar" topics, they're the same topic. You're asking what is - for all intents and purposes - the exact same question. Sure, you used a different example. But your question is fundamentally the same, and that's the key point in all this, at least to me. Your new post also suffers from the same flaws as your original, too, so I'd be surprised - happily, and for your sake, of course - if someone could indeed decipher what you're trying to elucidate in these posts and give a proper answer that you're looking for. – PrincessEev Mar 21 '19 at 03:37
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    You seem very frustrated. I can understand that but it is not clear what the problem is. What is it about induction that you want to know? Are you merely looking for a proof check or is there a concept that you wish to have explained? – John Douma Mar 21 '19 at 03:38
  • It doesn't make sense for the upper limit of the sum to be $r-n$. Should it just be $n$? Or maybe you meant "$r=n$". – Minus One-Twelfth Mar 21 '19 at 03:42
  • What I am trying to ask is just the method I am going about these problems is a correct way to do it. If you see my previous thread, the comments I found quite confusing, so this thread I tried to make my solutions more clear on one particular aspect of the method I want to approach it as, and see if it works or holds to be true. – Aurora Borealis Mar 21 '19 at 03:42
  • Minus One-Twelfth Maybe there must have been a typo let me edit that. – Aurora Borealis Mar 21 '19 at 03:43
  • @AuroraBorealis This is still essentially unreadable. Your new example doesn't help much. I'll have to vote to close this as a duplicate of your first post. – Rushabh Mehta Mar 21 '19 at 03:53
  • Possible duplicate of mathematical induction methods – Rushabh Mehta Mar 21 '19 at 03:54
  • What kind of criticism is this, saying its unreadable? Like seriously do people have manners in this forum. I mean how is it not readable like at least share, if not you are just being very hostile and foolish. – Aurora Borealis Mar 21 '19 at 04:15

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