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I need to prove $\frac{1}{2\sqrt1} + \frac{1}{3\sqrt2} + ... + \frac{1}{(n+1)\sqrt n} < 2 - \frac{2}{\sqrt{(n+1)}}$ by induction for every $n \in \mathbb{N} $. Please help, I am stuck on this from the past 2 days.. Thanks.

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2 Answers 2

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Great answer by trancelocation, but in case you still want it, here is how to do induction step for an inductive proof. First we note the following general rule of quadratics:

$$(n + \tfrac{a+b}{2})^2 - (n+a)(n+b) = (\tfrac{a-b}{2})^2.$$

Using this rule with $a=1$ and $b=2$, we have:

$$(n+\tfrac{3}{2})^2 > (n+1)(n+2).$$

Rearranging this inequality gives:

$$\frac{(n+\tfrac{3}{2})^2}{n+1} > n+2.$$

Our inductive step can now be accomplished as follows:

$$\begin{equation} \begin{aligned} \sum_{k=1}^{n+1} \frac{1}{(k+1)\sqrt{k}} &= \sum_{k=1}^{n} \frac{1}{(k+1)\sqrt{k}} + \frac{1}{(n+2)\sqrt{n+1}} \\[6pt] &< 2 - \frac{2}{\sqrt{n+1}} + \frac{1}{(n+2)\sqrt{n+1}} \\[6pt] &= 2 - \frac{2(n+2)-1}{(n+2)\sqrt{n+1}} \\[6pt] &= 2 - \frac{2n+3}{(n+2)\sqrt{n+1}} \\[6pt] &= 2 - \frac{2}{(n+2)} \cdot \frac{n+\tfrac{3}{2}}{\sqrt{n+1}} \\[6pt] &< 2 - \frac{2}{(n+2)} \cdot \sqrt{n+2} \\[6pt] &= 2 - \frac{2}{\sqrt{n+2}} . \\[6pt] \end{aligned} \end{equation}$$

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  • $\begingroup$ Thank you so much! $\endgroup$
    – Gamenoob
    Feb 14, 2019 at 11:54
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    $\begingroup$ I think it is easier to just refer to the AM mean inequality, that is: $$\begin{aligned}\frac{(n+1)+(n+2)}2>\sqrt{(n+1)(n+2)}&\iff n+\frac32>\sqrt{(n+1)(n+2)}\\&\iff\frac{n+\frac32}{\sqrt{n+1}}>\sqrt{n+2}\end{aligned}$$ $\endgroup$
    – PinkyWay
    Jul 28, 2020 at 17:20
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Without induction you can show it immediately as follows:

$$\sum_{k=1}^n\frac{1}{\sqrt{k}(k + 1)}\leq \sum_{k=1}^n\frac{1}{k \sqrt{k}} < \int_1^{n+1}\frac{1}{x^{\frac{3}{2}}}\,dx =\left[-2\frac{1}{\sqrt{x}} \right]_1^{n+1} = 2 -\frac{2}{\sqrt{n+1}}$$

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  • $\begingroup$ Thank you, but it is not with induction and I need it with.. For college.. $\endgroup$
    – Gamenoob
    Feb 14, 2019 at 12:15
  • $\begingroup$ @Gamenoob, popravni iz elementarne? 😄 Smatraš li Benov odgovor korisnim, molim te, prihvati ga. $\endgroup$
    – PinkyWay
    Jul 28, 2020 at 17:07
  • $\begingroup$ Sorry but from where did that integral came ? Can you please give a link to this technique ? I want to learn it. $\endgroup$ Jan 31, 2023 at 22:55
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    $\begingroup$ @An_Elephant Check this out: en.wikipedia.org/wiki/Integral_test_for_convergence $\endgroup$ Feb 1, 2023 at 4:52
  • $\begingroup$ Thanks very much sir :) $\endgroup$ Feb 1, 2023 at 9:36

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