Is the series: $$\frac{\pi}{p_{1}!}+\frac{\pi}{p_{2}!}+...+\frac{\pi}{p_{n}!}$$ convergent or divergent, where $p_n$ is the $n$th odd prime?
And also why it is (the partial sums) transcendental?
Note: I've just started studying about a series being convergent or divergent. I know the definition of convergent/divergent series, but, I somewhat failed to understand the relation of "limit" of a series.
Regards
On
$$\sum_{n=0}^{\infty} \frac{\pi}{p_n!}= \pi \sum_{n=0}^\infty \frac{1}{p_n!}$$
It is clear, $$\sum_{n=0}^\infty \frac{1}{p_n!}<\sum_{n=0}^{\infty}\frac{1}{n!}$$
$$\sum_{n=0}^{\infty} \frac{1}{n!}$$ converges (to $e$) therefore $$\sum_{n=0}^\infty \frac{1}{p_n!}$$
converges. Multiplying it by $\pi$ will still yield convergence.
I don't know if it's transcendental or not, I have nothing to say on that matter.
On
Yes, the partial sums are trancendental. The partial sums can be written
$$S_n = \pi\left(\frac{1}{p_1!} + \ldots + \frac{1}{p_n!}\right) = \pi Q$$
where $Q$ is rational since a finite sum of rationals is rational. Now we know that $\pi$ is trancendental and a product of a rational and a trancendental number is trancendental. The proof of this is given below.
If $\alpha$ is transendental and $r$ rational then $\beta = r\alpha$ is trancendental.
Proof: Assume that $\beta = r\alpha$ is algebraic. This means that there exists a polynomial $f(x) = a_m x^m + \ldots +a_1 x + a_0$ with rational coefficients $a_i$ such that $f(\beta) = 0$. But then $g(x) = f(x\beta/\alpha) = [a_m r^m] x^m + \ldots +[a_1 r]x + a_0$ is a polynomial with rational coefficients that has $\alpha$ as a root. This contradics $\alpha$ being trancendental and $\beta$ must therefore also be trancendental.
The "$k$th odd prime" part is there to throw you off. Show ${\displaystyle \sum_{k=1}^{\infty} {\pi \over k!}}$ converges using one of your techniques (they all work for this), then the comparison test will show your series converges.
Transcendentalness is a lot harder... I'd try to just understand convergence first.