Convergence testing involving factorial and square root

Question

I'm trying to find the convergence of this using the ratio test:

$$\displaystyle\sum_{i=1}^{\infty}\dfrac{1}{\sqrt{t!}}.$$

But I'm getting no luck! Can anyone help?

(sorry I've not quite mastered MathJax notation, I'm getting there!)

Answer 1

Observe:

$$\frac{a_{n+1}}{a_n}=\dfrac{\frac{1}{\sqrt{(t+1)!}}}{\frac{1}{\sqrt{t!}}}=\sqrt{\frac{t!}{(t+1)!}}.$$

Now apply ratio test (couple steps in between are left to you).

Answer 2

$$\sum_{i=1}^\infty \frac{1}{\sqrt t!}$$

Ratio Test

$$\lim_{x \to \infty} |\frac{a_{n+1}}{a_n}| = L$$

$$|\frac{1}{\sqrt {(t+1)!}} \cdot \frac{\sqrt t!}{1}| = \frac{\sqrt{t!}}{\sqrt{(t+1)!}}$$

From Wolfram, $$\lim_{t \to \infty} \frac{\sqrt{t!}}{\sqrt{(t+1)!}}= 0$$

Since $L= 0 <1$, $$\sum_{i=1}^\infty \frac{1}{\sqrt t!}$$ converges absolutely $\implies$ it converges in the usual sense.

Edit From Don Antonio's comment:

$$\lim_{t \to \infty} \frac{\sqrt{t!}}{\sqrt{(t+1)!}}= \lim_{x\to \infty} \sqrt{\frac{t!}{(t+1)!}}$$

$$\frac{t!}{(t+1)!} = \frac{t \cdot t-1 \cdot t-2 \cdots 2 \cdot 1}{t+1 \cdot t \cdot t-1\cdot t-2 \cdots 2 \cdot 1} = \frac{1}{t+1}$$

$$\lim_{x \to \infty} \sqrt{\frac{1}{t+1}} = \lim_{x \to \infty} \frac{1}{\sqrt{t+1}} = \frac{1}{\infty} = 0$$