Let $ \ f(x)=\sum_{n=0}^{\infty} \frac{x^n}{n!} \ $.
Then conclude whether $ \ f(0) , \ f(4) , \ f(50), \ f(-5) , \ f(-100) \ $ are summable.
Plot the partial sums $ \ f(x) \ $ for $ \ N=4,20,100,101 \ $ and explain interval of convergence from the graphs.
Answer:
Since $ \ f(0)=0 \ $, it converges and hence $ \ f(0) \ $ is summable.
Since $ \ f(4)=\sum_{0}^{\infty} \frac{4^n}{n!} \ $ diverges , $ \ f(4) \ $ is not summable.
Since $ \ f(-5) =\sum_{0}^{\infty} (-1)^n \frac{5^n}{n!} $ converges by Alternating series test , $ \ f(-5) \ $ is summable.
Am I Right so far?
Next I have drawn the graph of partial sums for $ N=5,20,100,101 \ $
as follows:
what does this imply for interval of convergence ?
There are a few issues. First of all, $f(4)$ is summable; I suggest using the ratio test. In fact, using the ratio test on all these values shows that these series are summable.
Second, I'd caution you in using the alternating series test. It's difficult to tell, since you haven't written much of your working, but the alternating series test requires not just that the series be alternating, but that the absolute value of the terms converge (as a sequence) monotonically to $0$. You need to establish that the sequence $$\left|(-1)^n\frac{5^n}{n!}\right| = \frac{5^n}{n!} \to 0$$ as $n \to \infty$. Moreover, you need to establish that the sequence is monotone decreasing, i.e. $$\frac{5^{n+1}}{(n+1)!} \le \frac{5^n}{n!}$$ for all $n$. Unfortunately, this isn't the case, as $$\frac{5^2}{2!} = 12.5 > 5 = \frac{5^1}{1!}.$$ Fortunately, it is possible to show that the sequence is eventually decreasing, which could save the argument, but there's some work that needs to be done first.
Third, while you're right about the convergence of $f(0)$, there is something you've overlooked: particularly the term where $n = 0$. The first term takes the form $\frac{0^0}{0!} = 0^0$, which is conventionally $1$ when looking at power series. So, the sum is, in fact, $1$.