Compute $e^5$ to 2 decimal digits precision 
How to compute $e^5$ without calculator with 2 decimal digits precision?

I was thinking about Taylor series:
$e^x=\Sigma_{k=0}^\infty \dfrac{x^k}{k!}$
, but it doesn't work.
(It's an interview question.)
 A: You should trust the Taylor Series expansion more, it is simpler than you probably think. If $e^x$ is approximately $1+x+{x^2}/{2}+{x^3}/{6}+...$, we can approximate your expression as $1+5+12.5+125/6+625/24=18.5+(20+5/6)+(20+145/24)$ which is approximately $59.3+6$. The next term will be $3125/120$, after which is $15625/720$, after which is about $78000/5000$. The terms now become smaller and smaller as the factorial is being multiplied faster than 5 each term. Some rough guessing gives $65.3+27+22+16$. Perhaps we can guess the next terms can continue to get smaller such as $10,5,2$ and $1$. Adding these up yields $142.3$. If you feel lucky and throw in an extra $5$ for the infinite terms, you will be only $1.03$ lower than the actual value of $148.3362...$.
No calculator used until the very last "actual" figure.
A: For starters, $e - s_n = \frac{1}{(n+1)!} + \frac{1}{(n+2)!} + ... < \frac{1}{(n+1)!} * \{1 + \frac{1}{n+1} + \frac{1}{(n+1)^2} + ...\} = \frac{1}{n!n}$, where $s_n$ is the nth partial sum of taylor series for $e$. See if you can use this to approximate $e^5$.
