# Thread: Is it possible to solve this inequality for n?

1. ## Is it possible to solve this inequality for n?

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

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

2. Lack of a response tells me no. Can anyone give me advice on approximating n?

3. 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?

4. It is old thread , i know

but am interesting to know the answer *_*

5. Originally Posted by leverin4
If it is, how would you go about doing it?

$\frac{\left(\frac{1}{2}\right)^{(2n+3)}+ \left(\frac{1}{3}\right)^{(2n+3)}}{2n+3}~<~10^{-1000001}$
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, $\textstyle\frac{(\frac12)^{2n+3}+ (\frac13)^{2n+3}}{2n+3} < (\frac12)^{2n+3}+ (\frac13)^{2n+3} < 2(\frac12)^{2n+3} < (\frac12)^{2n}$. If you can find a value of N such that $\textstyle(\frac12)^{2n} < 10^{-K}$ 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 $\textstyle(\frac12)^{2n} < 10^{-K}\ \Rightarrow\ 2^{2n}>10^K\ \Rightarrow\ 2n\ln2>K\ln10\ \Rightarrow\ n>\frac{K\ln10}{2\ln2}$, so take N to be any integer greater than $\frac{1000001\ln10}{2\ln2}\approx 1.661\times 10^6$ and your inequality will hold for all n>N.

Just to emphasise, $1.661\times 10^6$ 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 $1.661\times 10^6$, and it will tell you which is the first one to make the inequality true.