Show that for $n\geq 2$, $$1^{-1/2}+2^{-1/2}+3^{-1/2}+\dots+n^{-1/2}>n^{1/2}.$$
I tried to prove this one by many ways, but couldn't make it.
Thanks.
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Robert Z
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Ofek Pintok
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1have you solved the case $n=2$? – Dr. Sonnhard Graubner Oct 28 '17 at 12:02
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It's a very known problem. Search please this forum. – Michael Rozenberg Oct 28 '17 at 12:05
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1Yes, 1^(-1/2) + 2^(-1/2) > 2^(1/2) – Ofek Pintok Oct 28 '17 at 12:09
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From $\sum_{k=1}^{n}k^{-1/2}>n^{1/2}$ we get that $\sum_{k=1}^{n+1}k^{-1/2}>n^{1/2}+(n+1)^{-1/2}$. If you can prove that $n^{1/2}+(n+1)^{-1/2}>(n+1)^{1/2}$ you are done. Try writing the latter as $\frac{n^{1/2}((n+1)^{1/2}-n^{1/2})}{(n+1)^{1/2}}>0$, which is true. – effd Oct 28 '17 at 12:13
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the beginning is $n=2$ and we have $1+\frac{1}{\sqrt{2}}>\sqrt{2}$ this is true, since $\sqrt{2}>1$ and now we assume that for $n=k$ is hold that: $$1^{-1/2}+2^{-1/2}+3^{-1/2}+...+k^{-1/2}>k^{1/2}$$ and now we have to Show that $$1^{-1/2}+2^{-1/2}+3^{-1/2}+...+k^{-1/2}+(k+1)^{-1/2}>(k+1)^{1/2}$$ and now we have to Show that $$k^{1/2}+(k+1)^{-1/2}>(k+1)^{1/2}$$ multiplying the last inequality by $(k+1)^{1/2}$ we get $$k^{1/2}(k+1)^{1/2}+1>k+1$$ the last inequality is true since we have $k>0$
Dr. Sonnhard Graubner
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