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I am trying to prove this $2( \sqrt{n+1}-\sqrt n )< \frac{1}{\sqrt n}<2( \sqrt{n}-\sqrt {n-1})$ if $n \ge 1$ and using this to prove $2\sqrt{m}-2<\sum^m_{n=1} \frac{1}{\sqrt n}<2( 2\sqrt{m}-1)$ if $m\ge 2$ and in particular i want to show that $m=10^6$ the last inequality is between 1998 and 1999

i am using prove by induction ,first ,for the first statement but i having trouble establish for n=1 and i am confuse how to prove if n is true by hypothesis then n+1.for the second part i didn't get to that part

Edit

reading the answers and comments bellow now i find myself trying to prove the second part that is prove $2\sqrt{m}-2<\sum^m_{n=1} \frac{1}{\sqrt n}<2( 2\sqrt{m}-1)$.

My attempt

$2( \sqrt{n+1}-\sqrt n )< \frac{1}{\sqrt n}<2( \sqrt{n}-\sqrt {n-1})\Leftrightarrow \sum^m_{n=1}2( \sqrt{n+1}-\sqrt n )< \sum^m_{n=1}\frac{1}{\sqrt n}<\sum^m_{n=1}2( \sqrt{n}-\sqrt {n-1})\Leftrightarrow $

i assume this to be true

$2\sum^m_{n=1}( \sqrt{n+1}-\sqrt n )< \sum^m_{n=1}\frac{1}{\sqrt n}<2\sum^m_{n=1}( \sqrt{n}-\sqrt {n-1})\Leftrightarrow $

now using telescoping that is $\sum^m_{n=1} (a_n - a_{n-1})=(a_m - a_0)$ i can not make sense of it

Community
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Eiston Fergon
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    Duplicate. – Workaholic May 26 '15 at 11:23
  • You can simply sum by telescoping. –  May 26 '15 at 11:24
  • @Workaholic in the other post is not answer how to established the trivial case n=1 that i am having trouble with can you help me? – Eiston Fergon May 26 '15 at 11:42
  • @EistonFergon To show that $2(\sqrt2-1)\lt1$ all you have to do is to simplify it first: $2\sqrt{2}-2\lt1$, then add to both sides $2$ to get $2\sqrt{2}\lt3$, dividing by $3$ yields: $\sqrt{2}\lt\tfrac32=1.5$ which is true as you can check. – Workaholic May 26 '15 at 12:02
  • @Workaholic can you see my changes on the question of this post – Eiston Fergon May 26 '15 at 18:47
  • @math can you see my changes on the question of this post – Eiston Fergon May 26 '15 at 18:47
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    The approach by a telescopic sum you describe yourself yields $$\sum_{n=1}^{m}\frac1{\sqrt{n}}\geqslant 2\sum_{n=1}^{m}(\sqrt{n+1}-\sqrt{n})=2\sqrt{m+1}-2>2\sqrt{m}-2.$$ On the other hand, keeping the term $n=1$ and applying the upper bound to all the others yields $$\sum_{n=1}^{m}\frac1{\sqrt{n}}\leqslant1+2\sum_{n=2}^{m}\sqrt{n}-\sqrt{n-1}=1+2(\sqrt{m}-1)=2\sqrt{m}-1,$$ which is probably, modulo some typos in your question, what you were asked to prove. – Did May 26 '15 at 19:55
  • thanks for your help it help me @Did – Eiston Fergon May 26 '15 at 20:04

2 Answers2

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Hint: Observe that $$\sqrt{n+1}-\sqrt{n}=\dfrac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\sqrt{n+1}+\sqrt{n}}.$$

Bebop
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  • in the other post on the comments above is not answer how to established the trivial case n=1 that i am having trouble with can you help me? the link is math.stackexchange.com/questions/1211973/… – Eiston Fergon – Eiston Fergon May 26 '15 at 11:53
  • can you see my changes on the question of this post – Eiston Fergon May 26 '15 at 18:47
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For a positive decreasing sequence $(a_n)$, we can write the inequality $$\int\limits_{k+1}^{n+1}a(x)dx\leq\sum\limits_{v=k+1}^{n}a_v\leq\int\limits_{k}^na(x)dx$$ Then, in special case, taking $a_n=\frac{1}{\sqrt{n}}$, $a(x)=\frac{1}{\sqrt{x}}$ and $k=n-1$ you get the desired result.

For the other inequality, you can take $k=0$ in the left side of the inequality above . Then take $k=1$ and add the first term $a_1=1$ in the right side.

guest
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  • in the other post on the comments above is not answer how to established the trivial case n=1 that i am having trouble with can you help me? the link is http://math.stackexchange.com/questions/1211973/estimating-partial-sums-sum-n-1m-frac1-sqrtn – Eiston Fergon May 26 '15 at 11:45
  • can you see my changes on the question of this post – Eiston Fergon May 26 '15 at 18:47