
Problem 11.1. Given the sequence $a_{n}=n+a \sqrt{n^{2}+1}, n=1,2, \ldots$, where $a$ is a real number:

a) Find the values of $a$ such that the sequence $\left\{a_{n}\right\}_{n=1}^{\infty}$ is convergent.

b) Find the values of $a$ such that the sequence $\left\{a_{n}\right\}_{n=1}^{\infty}$ is monotone increasing.


Solution: a) If $a=-1$, the sequence $\left\{a_{n}\right\}_{n=1}^{\infty}$ is convergent because $a_{n}=n-\sqrt{n^{2}+1}=\frac{-1}{n+\sqrt{n^{2}+1}}=-\frac{1}{n} \frac{1}{1+\frac{1}{n^{2}}} \rightarrow 0$ when $n \rightarrow \infty$. Conversely, let the sequence $\left\{a_{n}\right\}_{n=1}^{\infty}$ be convergent. Since $a_{n}=n-\sqrt{n^{2}+1}+(a+1) \sqrt{n^{2}+1}$, we get that the sequence $(a+1) \sqrt{n^{2}+1}$ is also convergent. Since $\sqrt{n^{2}+1} \rightarrow \infty$ when $n \rightarrow \infty$, it follows that $a+1 \neq 0$, so $a=-1$.

b) Let $\left\{a_{n}\right\}_{n=1}^{\infty}$ be a monotone increasing sequence, i. e., $a_{n+1} \geq$
$a_{n}$ for any $n$. This inequality is equivalent to

$$
\frac{a(2 n+1)}{\sqrt{(n+1)^{2}+1}+\sqrt{n^{2}+1}} \geq-1
$$

Since

$\lim _{n \rightarrow \infty} \frac{2 n+1}{\sqrt{(n+1)^{2}+1}+\sqrt{n^{2}+1}}=\lim _{n \rightarrow \infty} \frac{2+\frac{1}{n}}{\sqrt{\left(1+\frac{1}{n}\right)^{2}+\frac{1}{n^{2}}}+\sqrt{1+\frac{1}{n^{2}}}}=1$

it follows from $(\star)$ that $a \geq-1$.

$$
\text { Conversely, let } a \geq-1 \text {. It follows from } \frac{2 n+1}{\sqrt{(n+1)^{2}+1}+\sqrt{n^{2}+1}}<
$$

$\frac{2 n+1}{n+1+n}=1$ that $(\star)$ holds true so the sequence $\left\{a_{n}\right\}_{n=1}^{\infty}$ is increasing. The required values of $a$ are $a \in[-1,+\infty)$.
