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How to approximate this series, non-numerically?

$ S_n = \sum_{n=1}^{50} \sqrt{n}$

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  • $\begingroup$ What about "sum-of-like-powers" with exponent $1/2$ thus $\to$ generalized bernoulli-polynomials ? Or Hurwitz-zeta $\zeta(-1/2,1)-\zeta(-1/2,51) $ ? $\endgroup$ Oct 18, 2013 at 0:35
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    $\begingroup$ There are a handful of different ways you might go about it, with widely varying levels of sophistication. What level math courses are you in? Gottfried's hint would be good if this is for a 2nd or 3rd year number theory class, but this could also be an intro calculus question, in which case the above comment likely went straight over your head. :) $\endgroup$
    – David H
    Oct 18, 2013 at 1:02
  • $\begingroup$ Thanks David. I haven't taken any # theory courses yet. It is however on my to do list. $\endgroup$
    – Quester
    Oct 22, 2013 at 0:47

2 Answers 2

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For a very simple approximation, which is the first step to the Euler-McLaurin formula, use this $f'(n) \sim f(n)-f(n-1) $ so $f(n) \sim \int_{n-1}^{n} f(x) dx $ so $\sum_{n=1}^N f(n) \sim \int_0^{N} f(x) dx $.

(Actually, $f(n) \sim \int_{n-1/2}^{n+1/2} f(x) dx $ is more accurate, but this is an approximation, and definitely not the best.)

Letting $f(n) =\sqrt{n}$, $\sum_{n=1}^N \sqrt{n} \sim \int_0^{N} \sqrt{x} dx = \frac{x^{3/2}}{3/2}\big |_0^N =\frac23 N^{3/2} $.

Then, let $N = 50$.

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Consider that: $$2\left((n+1)\sqrt{n+1}-n\sqrt{n}\right)=3\sqrt{n+1}+\frac{n-1-\sqrt{n(n+1)}}{\sqrt{n}+\sqrt{n+1}}.$$ The last term is negative, but greater than $$-\frac{3}{2(\sqrt{n}+\sqrt{n+1})}$$ by the AM-GM inequality. This gives: $$\sum_{n=1}^{50}\sqrt{n}\geq\frac{2}{3}\sum_{n=1}^{50}\left(n\sqrt{n}-(n-1)\sqrt{n-1}\right)=\frac{2}{3}50\sqrt{50},$$ $$-\frac{2}{3}50\sqrt{50}+\sum_{n=1}^{50}\sqrt{n}\leq\frac{1}{2}\sum_{n=1}^{50}\frac{1}{\sqrt{n-1}+\sqrt{n}}=\frac{1}{2}\sum_{n=1}^{50}\left(\sqrt{n}-\sqrt{n-1}\right)=\frac{1}{2}\sqrt{50}.$$ The same argument proves that the sum $$\sum_{n=1}^{N}\sqrt{n}$$ is always between $$\frac{2}{3}N\sqrt{N}$$ and $$\frac{4N+3}{6}\sqrt{N}.$$ As pointed out in the previous comments, we can produce a tighter bound by considering that: $$0\leq 2\left((n+1/2)\sqrt{n+1/2}-(n-1/2)\sqrt{n-1/2}\right)-3\sqrt{n}\leq \frac{1}{16}\left(\frac{1}{\sqrt{n+1/2}}-\frac{1}{\sqrt{n-1/2}}\right).$$

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  • $\begingroup$ your demonstration is beautiful. But, are you sure that there is not a typo somewhere ? Computing with N=50 apparently does not agree with your upper and lower bounds. For sure, I can be wrong. $\endgroup$ Oct 18, 2013 at 8:38
  • $\begingroup$ @ClaudeLeibovici: the inequalities were right, but there was an off-by-one in the computation of the first telescopic sum, thank you for pointing that out. $\endgroup$ Oct 18, 2013 at 9:03
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    $\begingroup$ It's answers like this that remind that as talented as I am at math, there's yet a level of talent out there that leaves me feeling like a toddler. +1 $\endgroup$
    – David H
    Oct 18, 2013 at 10:40
  • $\begingroup$ Thank you Jack. I haven't even heard of AM-GM ineq. before. $\endgroup$
    – Quester
    Oct 22, 2013 at 1:58

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