1

$\frac1{\sqrt{1}} + \frac1{\sqrt{2}}+\frac1{\sqrt{3}}+...+\frac1{\sqrt{x}}\ge {\sqrt{x}}$

I proved the basic case: and realize it is equal to 1, but I have absolutely no idea how to create prove the left and right side using the induction hypothesis.

Please help :) I've solved equality's before using induction, but this is the first time I've done an inequality, and although I'm sure the process is fairly similar, I think the square root is messing me up.

robjohn
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Jess
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    What is the inequality? I see an equality in your question. – M Turgeon Oct 21 '13 at 01:45
  • @MTurgeon, the OP edited out the entirety of the question, which indeed involved square roots. – Will Jagy Oct 21 '13 at 01:54
  • @WillJagy : whatever the OP did, the title does not match the question. – Stefan Smith Oct 21 '13 at 01:56
  • @StefanSmith, the fourth edit says "posted wrong question." I have flagged for moderator attention. – Will Jagy Oct 21 '13 at 02:13
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    Please do not edit your question to fundamentally change the content after I have already answered your question. I find it extremely rude and disrespectful and have flagged this post for moderator attention. –  Oct 21 '13 at 02:24
  • @WillJagy : it's not a big deal, but I didn't see "posted wrong question" when I read it the first time, I don't see it there now, the title (with "inequality") does not match the question (which contains no inequality), and there is no square root in the question. I don't really understand all the flagging business. Thanks for flagging the question. Somehow T. Bongers saw a version of the question that made sense. – Stefan Smith Oct 21 '13 at 02:47
  • @StefanSmith, click in the bottom middle of the question where it says, in light blue, 'edited 1 hour ago' It should work for you, you have over 2000 points. – Will Jagy Oct 21 '13 at 02:55
  • If you wish to ask another question, ask another question – robjohn Oct 21 '13 at 04:10
  • This is a duplicate of this question and this question. – robjohn Oct 21 '13 at 04:14

1 Answers1

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Let $s_n$ represent the sum up to $n$, and suppose that $s_n \ge \sqrt{n}$ for induction. We want to show that $s_{n + 1} \ge \sqrt{n + 1}$; to this end, note

\begin{align*} s_{n + 1} - s_n &= \frac{1}{\sqrt{n + 1}} \end{align*}

Therefore,

\begin{align*} s_{n + 1} &= \frac{1}{\sqrt{n + 1}} + s_n \\ &\ge \frac{1}{\sqrt{n + 1}} + \sqrt{n} \\ &= \frac{1 + \sqrt{n}\sqrt{n + 1}}{\sqrt{n + 1}} \\ &\ge \frac{1 + \sqrt{n} \sqrt{n}}{\sqrt{n + 1}} \\ &= \frac{n + 1}{\sqrt{n + 1}} = \sqrt{n + 1} \end{align*}

as desired. The first inequality is by the induction hypothesis, and the second is just by noting that

$$\sqrt{n + 1} \ge \sqrt{n}$$

  • I am a bit confused. I have never seen this way. The way i am used is the base case and then deriving a proof. Is this is only way to do it? – Jess Oct 21 '13 at 01:28
  • @Jess The base case needs to be done in addition to the inductive step. I've only done the latter of the two. –  Oct 21 '13 at 01:29
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    @T.Bongers, see the edit history of the question. I have flagged for moderator attention. – Will Jagy Oct 21 '13 at 02:15
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    @WillJagy Thank you for letting me know :) –  Oct 21 '13 at 02:23
  • Well, because I was playing with it when it got erased, $s_n \approx -1.4603545 + 2 \sqrt {n + 0.5 }$ – Will Jagy Oct 21 '13 at 03:39