I am learning about natural logarithm, and I have an inequality I can’t solve, so please help.
Prove that $$\sum_{i=1}^{n-1} \dfrac{1}{\sqrt{i^2+i}} < \ln n$$
where $n>1$ is a positive integer.
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3 Answers
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By the Cauchy-Schwarz inequality, for any $n\geq 1$ we have $$ \log(n+1)-\log(n) = \int_{0}^{1}\frac{dt}{n+t} \leq \sqrt{\int_{0}^{1}\frac{dt}{(n+t)^2}} = \frac{1}{\sqrt{n+n^2}}$$ hence by telescoping $$ \sum_{n=1}^{N-1}\frac{1}{\sqrt{n+n^2}}\;{\large\color{red} \geq }\;\log(N).$$ I prefer to avoid using $i$ as a summation index, since horrible things might happen if the imaginary unit is also involved in a sum.
answered Apr 12, 2018 at 13:58
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$\begingroup$ If I run out of the standard letters ($n, k, i$) to use as a summation index, I use $s$. I don't know why people don't really use $s$. $+1$ for the good answer though :) $\endgroup$– Mr PieApr 12, 2018 at 14:05
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Your inequation does not even hold for $n=2$ : $\frac{1}{\sqrt2}\approx0.71> \text{ln}(2)\approx 0.69$.
By any chance, is it not the other way around?
answered Apr 12, 2018 at 13:53
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Can't you approximate by:
$$\sum_{i=1}^{n-1}{\frac{1}{\sqrt{i^2 + i}}} < \sum_{i=1}^{n-1}{\frac{1}{\sqrt{i^2 + i^2}}} = \frac{1}{\sqrt{2}}\sum_{i=1}^{n-1}{\frac{1}{i}} \rightarrow O(ln(n))$$
?
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$\begingroup$ Is is pretty obvious that $\sum_{k=1}^{n-1}\frac{1}{\sqrt{k^2+k}}$ is $\Theta(\log n)$ but that is not the same as proving a pointwise inequality. Also because the actual inequality holds in the opposite direction. $\endgroup$ Apr 12, 2018 at 14:29