Approximate a summation

Question

Approximate $3+ \displaystyle \sum_{x = 2}^{999}\dfrac{3(1000-x)}{1000+x}$. It may help to know that $\ln 2 = 0.69$.

I was thinking of doing the integral test to approximate this but I am unsure if this would work.

Answer 1

Since the problem says "approximate" and does not give an acceptable error, I would just say $0$ and be done. You are probably expected to pull out the $3$ in the sum and say $$\sum_{i=2}^{999}\frac{1000-i}{1000+i}\approx \int_2^{1000}\frac{1000-i}{1000+i}di$$ Can you see why this is a reasonable approximation? Can you see why the upper limit is $1000$? Can you do the integral?

Answer 2

This answer if more than likely off-topic. In any manner, it is too long for a comment.

Ross Millikan gave a simple way to approximate the summation. Making the problem more general $$S=\sum_{i=2}^{a-1}\frac{a-i}{a+i}\approx \int_2^{a}\frac{a-i}{a+i}\,di=2+a (2 \log (a)+\log (4)-1)-2 a \log (a+2)$$ which, for $a=1000$, would give $S\approx 384.298$. This comes from the fact that $$\int\frac{a-i}{a+i}\,di=\int\frac{2a-(a+i)}{a+i}\,di=2 a \log (a+i)-i$$

Just for you curiosity, the sum can be computed exactly since $$S=\sum_{i=2}^{a-1}\frac{a-i}{a+i}=2-a+2 a\, \psi (2 a)-2 a\, \psi (a+2)$$ where appears the digamma function.

For large argument $a$, series expansion would give $$S\approx a (2 \log (2)-1)-\frac{3}{2}+\frac{17}{8 a}-\frac{2}{a^2}+O\left(\frac{1}{a^3}\right)$$ which, for $a=1000$, would give $S\approx 1000 (2 \log (2)-1)-\frac{1497877}{1000000} \approx384.796$ while the exact value would be $\approx384.794$.

I suppose that this shows how good is Ross Millikan's answer based on so simple considerations.

Edit

If you really need to use the hint $\log(2)\approx 0.69$, you have, for $a=1000$, $$S=2+1000 (2 \log (1000)+\log (4)-1)-2000 \log (1002)$$ that is to say $$S=-998+1000\log(2^2)+2000\log\big(\frac{1000}{1002}\big)$$ The last logarithm is almost $0$ so $$S\approx-998+2000\log(2)\approx -998+2000\times 0.69=382$$