To analyze the convergence or divergence of the series provided, let us break down the three series and apply appropriate convergence tests.

Problem Breakdown:

1. Series (i):
   $\sum_{n=1}^\infty \sqrt{(n^2 + 1)} - n$

   • We analyze the asymptotic behavior of $\sqrt{n^2 + 1} - n$ as $n \to \infty$.

2. Series (ii):
   $\sum_{n=1}^\infty \frac{n + 1}{n!}x^n$

   • The convergence depends on the factorial term in the denominator, and the term $x^n$ plays a role.

3. Series (iii):
   $\sum_{n=1}^\infty \frac{n^n}{n!} x^n$

   • Here, the factorial $n!$ and $n^n$ need comparison to determine the growth rate.

Solution:

Series (i):

The term $\sqrt{n^2 + 1} - n$ can be simplified for large $n$ by approximating: $\sqrt{n^2 + 1} - n = \frac{n^2 + 1 - n^2}{\sqrt{n^2 + 1} + n} = \frac{1}{\sqrt{n^2 + 1} + n}.$ For large $n$, this behaves like: $\frac{1}{2n}.$ Thus, the series becomes approximately: $\sum_{n=1}^\infty \frac{1}{2n},$ which is a multiple of the harmonic series $\sum \frac{1}{n}$. The harmonic series diverges, so this series diverges.

Series (ii):

The general term is: $a_n = \frac{n + 1}{n!}x^n.$ Using the Ratio Test: $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{(n+2)x^{n+1}}{(n+1)!} \cdot \frac{n!}{(n+1)x^n}.$ Simplify: $\lim_{n \to \infty} \frac{(n+2)x}{n+1} \cdot \frac{1}{n+1} = \lim_{n \to \infty} \frac{x}{(n+1)} = 0.$ Since the limit is $0$, the series converges for all $x$.

Series (iii):

The general term is: $a_n = \frac{n^n}{n!} x^n.$ Using the Ratio Test: $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{(n+1)^{n+1} x^{n+1}}{(n+1)!} \cdot \frac{n!}{n^n x^n}.$ Simplify the powers and factorials: $\lim_{n \to \infty} \frac{(n+1)^n (n+1)x}{n^n (n+1)} = \lim_{n \to \infty} \frac{(n+1)^n}{n^n} \cdot x.$ Rewriting $(n+1)^n/n^n$: $\left(\frac{n+1}{n}\right)^n = \left(1 + \frac{1}{n}\right)^n.$ As $n \to \infty$, this approaches $e$. Thus: $\lim_{n \to \infty} e \cdot x.$ For convergence, we need $e \cdot x < 1$, or $|x| < \frac{1}{e}$. The series converges for $|x| < \frac{1}{e}$ and diverges otherwise.

Final Answers:

1. (i): Diverges.
2. (ii): Converges for all $x$.
3. (iii): Converges for $|x| < \frac{1}{e}$, diverges otherwise.

Do you want detailed proofs or further analysis on these results? Here are some related questions for exploration:

1. What is the significance of the Ratio Test in determining convergence?
2. Can we use the Comparison Test for Series (i)?
3. How does factorial growth compare to exponential growth in Series (ii)?
4. What is the role of $e$ in Series (iii)?
5. How can you graph these series to better understand convergence and divergence?

Tip: Always approximate series terms for large $n$ to determine their asymptotic behavior.