If it is, how would you go about doing it?

$\displaystyle \frac{\left(\frac{1}{2}\right)^{(2n+3)}+\left(\fra c{1}{3}\right)^{(2n+3)}}{2n+3}~<~10^{-1000001}$

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- Apr 30th 2009, 04:08 PMleverin4Is it possible to solve this inequality for n?
If it is, how would you go about doing it?

$\displaystyle \frac{\left(\frac{1}{2}\right)^{(2n+3)}+\left(\fra c{1}{3}\right)^{(2n+3)}}{2n+3}~<~10^{-1000001}$ - May 1st 2009, 07:30 AMleverin4
Lack of a response tells me no. Can anyone give me advice on approximating n?

- May 3rd 2009, 08:47 PMleverin4
I tried using Mathematica, but it couldn't do it. I don't remember the exact error message, but I'll try it again tomorrow and post it. Meanwhile, isn't there anyone that can give me some advice on how to simplify it?

- Jun 24th 2009, 11:22 PMTWiX
It is old thread , i know

but am interesting to know the answer *_* - Aug 5th 2009, 11:22 AMOpalg
It would be hard to find a necessary and sufficient condition on n for this inequality to hold. However, problems like this usually occur in a first course on analysis, and the aim there is not to find the

*smallest*value of n that will satisfy the inequality, but to find*some*number N such that all values of n gretaer than N will satisfy the inequality. In that case, you can make the problem much easier by getting a rough upper bound for the left side of the inequality.

For example, $\displaystyle \textstyle\frac{(\frac12)^{2n+3}+ (\frac13)^{2n+3}}{2n+3} < (\frac12)^{2n+3}+ (\frac13)^{2n+3} < 2(\frac12)^{2n+3} < (\frac12)^{2n}$. If you can find a value of N such that $\displaystyle \textstyle(\frac12)^{2n} < 10^{-K}$ whenever n>N, it will follow that the original inequality will hold whenever n>N. (Here, K stands for some given large number, such as 1000001.)

But $\displaystyle \textstyle(\frac12)^{2n} < 10^{-K}\ \Rightarrow\ 2^{2n}>10^K\ \Rightarrow\ 2n\ln2>K\ln10\ \Rightarrow\ n>\frac{K\ln10}{2\ln2}$, so take N to be any integer greater than $\displaystyle \frac{1000001\ln10}{2\ln2}\approx 1.661\times 10^6$ and your inequality will hold for all n>N.

Just to emphasise, $\displaystyle 1.661\times 10^6$ is not the smallest number with that property. If you want to find that number, you can use a computer to check all the values of N from 1 to $\displaystyle 1.661\times 10^6$, and it will tell you which is the first one to make the inequality true.