If it is, how would you go about doing it?
$\displaystyle \frac{\left(\frac{1}{2}\right)^{(2n+3)}+\left(\fra c{1}{3}\right)^{(2n+3)}}{2n+3}~<~10^{-1000001}$
If it is, how would you go about doing it?
$\displaystyle \frac{\left(\frac{1}{2}\right)^{(2n+3)}+\left(\fra c{1}{3}\right)^{(2n+3)}}{2n+3}~<~10^{-1000001}$
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.