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I am trying to solve the following inequality:

$$2+\left(\frac{3}{2}\right)^{\frac{1}{2}}+\left(\frac{4}{3}\right)^{\frac{1}{3}}+\cdots+\left(\frac{n+1}{n}\right)^{\frac{1}{n}}<n+2$$

but I do not know how to do that. Please help me solve this inequality.

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    $\begingroup$ Should your first term be $2$? $\endgroup$ – preferred_anon Nov 16 '12 at 18:46
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    $\begingroup$ @DanielLittlewood, It was correct. But I added 1 to both sides of inequality to be more clear. $\endgroup$ – Mohammad Javad Naderi Nov 16 '12 at 18:50
  • $\begingroup$ I've removed algebra tag, since we don't use algebra tag anymore, see meta for details. $\endgroup$ – Martin Sleziak Nov 18 '12 at 20:32
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Use the fact that $(1 + \frac1n)^{\frac1n} < 1 + \frac{1}{n^2}$ for $n > 1$. Thus your sum is $\sum_{k=1}^n (1 + \frac1k)^{\frac1k} < \sum_{k=1}^n (1 + \frac{1}{k^2}) < n + \frac{\pi^2}{6} < n+2$.

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  • $\begingroup$ $(1+1/n)^{1/n}>1+1/n^2$ not $(1+1/n)^{1/n}>1+1/n^2$. Now I think your proof will not work. $\endgroup$ – Amr Nov 16 '12 at 21:56
  • $\begingroup$ No, it's true that $(1+1/n)^{1/n} < 1 + 1/n^2$ for $n \ge 1$. To see this, just raise both sides to $n$th power, you get $1+1/n$ on the LHS and $(1+1/n^2)^n = 1+1/n + \cdots$ on the RHS. $\endgroup$ – Alan Guo Nov 19 '13 at 16:10
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Since $\log(1+x)\leq x$ and $e^x\leq\frac{1}{1-x}$ we have: $$(1+1/k)^{1/k}=\exp\left(\frac{1}{k}\,\log(1+1/k)\right)\leq e^{1/k^2}\leq 1+\frac{1}{k^2-1}=1+\frac{1}{2}\left(\frac{1}{k-1}-\frac{1}{k+1}\right),$$ so: $$\sum_{k=2}^{n}(1+1/k)^{1/k} \leq (n-1)+\sum_{k=2}^n \frac{1}{k^2-1}\leq (n-1)+\sum_{k=2}^{+\infty}\frac{1}{k^2-1}=(n-1)+\frac{3}{4}< n.$$

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