I need the full proof/solution to this problem.
What I've done so far:
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Please show us some effort. Also, try to search this site for similar questions. I think this one has been asked before. – mickep Oct 12 '15 at 12:09
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OK, I will post what I've done so far. I think it's not completely correct. – Code Man Oct 12 '15 at 12:09
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Another way (not by induction) would be to observe $\frac{1}{\sqrt n}\geq\frac{2}{\sqrt n+\sqrt{n+1}}=2(\sqrt{n+1}-\sqrt n)$. – Element118 Oct 12 '15 at 12:11
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2sketchy move from second before last to before last .. those things arent equal – qwerty314 Oct 12 '15 at 12:24
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1See this post and other posts linked there. – Martin Sleziak Oct 12 '15 at 15:21
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Very classical looking handwriting. – Hypergeometricx Oct 12 '15 at 16:41
4 Answers
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suppose your formula is true for $n$, Then $$\sum^{n}_{i=1}{\frac{1}{\sqrt{i}}}>\sqrt{n}$$ we then try to prove the formula true for $n+1$ $$\sum^{n}_{i=1}{\frac{1}{\sqrt{i}}}+\frac{1}{\sqrt{n+1}}>\sqrt{n}+\frac{1}{\sqrt{n+1}}$$ $$=\frac{\sqrt{n}\sqrt{n+1}+1}{\sqrt{n+1}}>\frac{n+1}{\sqrt{n+1}}=\sqrt{n+1}$$
The base case is true so it's true for all n
Lessness Randomness
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qwerty314
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3how did you simplify sqrt(n) X sqrt(n+1) + 1 to n+1? Also, why is (n+1)/(sqrt(n+1)) = sqrt(n+1)? – Code Man Oct 12 '15 at 12:26
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- sqrt(n) is smaller than sqrt(n+1), so we can replace it and make the relation an inequality (meaning always bigger). Then the sqrt(n)^2=n
– qwerty314 Oct 12 '15 at 12:28 -
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1shouldn't your summation have an upper bound of n instead of n+1? – Mohammad Ali Dec 10 '19 at 21:27
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Multiply at both sides by $\sqrt{n}$. You get
$\sum_{k=1}^n\frac{\sqrt{n}}{\sqrt{k}}=\sum_{k=1}^n\frac{n}{\sqrt{kn}}\geq \sum_{k=1}^n\frac{n}{\sqrt{nn}}=\sum_{k=1}^n1=n$
NeitherNor
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9Or, more simply, with no multiplication by $\sqrt{n}$ nor any term $\sqrt{nk}$, $$\sum_{k=1}^n\frac1{\sqrt{k}}\geqslant\sum_{k=1}^n\frac1{\sqrt{n}}=\sqrt{n}.$$ – Did Oct 12 '15 at 12:15
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Your proof is basically OK except for near the end, where you write
$${\sqrt{k(k+1)}+1\over\sqrt{k+1}}={\sqrt{k+1}\times\sqrt{k+1}\over\sqrt{k+1}}$$
The "$=$" there should be a "$\ge$."
Barry Cipra
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