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I am trying to estimate the value of the Sum below without using the calculator:

$$ \sum_{n=1}^{10000} \sqrt{i} $$

I have looked for other ways around and tried to turn it into $n\sqrt{n}\int_{0}^{1} \sqrt{x} dx$, but It doesn't work quite well ( the estimation isn't really close to the actual answer computed by the calculator ). Is there a comprehensible method for a calculus starter like me?

Thanks!

Thomas Andrews
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vietnamsteven
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2 Answers2

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Since $x\to \sqrt{x}$ is strictly increasing, then for $n\geq 1$, $$\sum_{i=1}^{n-1} \sqrt{i}<\underbrace{\int_0^{n}\sqrt{x}\,dx}_{=\frac{2}{3}n\sqrt{n}}< \sum_{i=1}^{n} \sqrt{i}.$$ Hence, by letting $n=10000$, we find that $\sqrt{n}=100$ and $$666666.666<\frac{2000000}{3}<\sum_{i=1}^{10000} \sqrt{i}< \frac{2000000}{3}+100<666766.667.$$

Since $x\to \sqrt{x}$ is also strictly concave, we can find a much better upper bound by using trapezoids instead of rectangles: $$\sum_{i=0}^{n-1} \frac{\sqrt{i}+\sqrt{i+1}}{2}< \int_0^{n}\sqrt{x}\,dx=\frac{2}{3}n\sqrt{n}$$ which leads to $$\sum_{i=1}^{10000} \sqrt{i}<\frac{2000000}{3}+50<666716.667.$$

Note that $\sum_{i=1}^{10000} \sqrt{i}\approx 666716.459$, so the above approximation is not too bad (by the above bounds we know that the absolute error is less than $50$).

Robert Z
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  • @ThomasAndrews I gave a precise upper and lower bound showing that the error is less than $100$ – Robert Z Aug 05 '21 at 14:54
  • Thanks for the answer! Actually I hoped to find a result correct to the decimals... But if the estimation I've made above is widely accepted, I think it is good enough! – vietnamsteven Aug 05 '21 at 15:07
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if we have $f(x) > 0$ and $f'(x) > 0,$ then $$ \int_{a-1}^{b} \; f(x) \; dx \; < \; \sum_{j=a}^b \; f(j) \; < \; \int_{a}^{b+1} \; f(x) \; dx $$

These are about bounding $f(n)$ above and below by an integral. We do better by integrating the function from $n - \frac{1}{2} $ to $n + \frac{1}{2} $

Either way, we do better with $$ \; \sum_{j=a}^b \; g(j) \; \approx \; \int_{a-\frac{1}{2}}^{b+\frac{1}{2} } \; g(x) \; dx $$

As the size of the error depends on the second derivative of $f,$ this comes out better. The second derivative gets small as the argument gets large. A good deal of error occurs ate the beginning, so we may take $a=10,$ for example, and carefully add on the square roots of the numbers from $1$ to $9$

First, $$ \; \sum_{j=1}^{10000} \; \sqrt j \; \approx \; \int_{1-\frac{1}{2}}^{10000+\frac{1}{2} } \; \sqrt x \; dx $$

Second try,

$$ \; \sum_{j=10}^{10000} \; \sqrt j \; \approx \; \int_{10- \frac{1}{2}}^{10000+\frac{1}{2} } \; \sqrt x \; dx $$ so

$$ \; \sum_{j=1}^{10000} \; \sqrt j \; \approx \; \; \sum_{j=1}^{9} \; \sqrt j \; \; + \; \; \int_{10- \frac{1}{2}}^{10000+\frac{1}{2} } \; \sqrt x \; dx $$

Will Jagy
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