How do I find the value of the following summation:
$$\sum_{k=0}^{48} \frac{30^k}{k!}$$
I tried using the ratio test to find convergence but maybe that's not the way it should be.
And also I'm not allowed to use graphic/programmable calculator for this.
On
$\frac{15101357483145095543048608278289608881317}{1414382685380845484392545457}$ using Wolfram Alpha
On
By the Stirling approximation, the first missing term in the infinite summation would be
$$\frac{30^{49}}{49!}\approx\frac{(30e)^{49}}{\sqrt{98\pi}\,49^{49}}\approx3940716560.$$
This is to be compared to the infinite sum, $e^{30}\approx10686474581524$.
As the tail of the summation does not exceed $e$ times the first omitted term, the approximation as $e^{30}$ yields a relative error not exceeding $10^{-3}$.
The answer will have a massive denominator; it's basically just a complicated rational number. In my opinion, you're better off saying the answer is roughly $e^{30}$. The difference will be tiny. Indeed, the ratio test tells us that the partial sums of $\exp$ converge faster than geometrically beyond the $x$th partial sum.