Help needed solving an inequality involving the natural log

I'm trying to solve a simple alternative series question, but I cannot get past the following inequality to calculate my nth term for the required error:

e^(-n-1).(n+1) < 0.01

I solved it all the way to:

ln n < (n + 1) + ln 0.01

I can't figure out how to solve for n after this.

Any help will be greatly appreciated,

Giest

Re: Help needed solving an inequality involving the natural log

There is no elementary way to solve that. What we can do is let u= n+ 1 so the function can be written $\displaystyle e^{-u}u$. We can further let v= -u so that $\displaystyle e^{-u}u= e^v(-v)= -e^vv$. Now, the best way to solve such an **in**equality is to first solve the associate **equation**- here, $\displaystyle e^{-(n+1)}(n+1)= -ve^v= 0.01$ which is, of course, the same as $\displaystyle ve^v= -0.01$. And the solution to that is x= W(-0.01) where W is the "Lambert W function" (also known as the "omega function" or "product log" function) which is **defined** as the inverse function to $\displaystyle f(x)= xe^x$. Looking at a graph of $\displaystyle y= xe^x$ we can see there are two roots for $\displaystyle xe^x= -0.01$, one at about -0.01 and the other at about -6.47. The inequality is satisfied for x below the root around -6 and above the root around -.01. (I used the "Newton-Raphson" iteration method to solve for those two values.)

Re: Help needed solving an inequality involving the natural log

Thanks for the help on this one.