Convergent of Divergent of sequence having Rational terms

Question

Finding whether the sequence

$$a_{n}=\sqrt{1+n}\sqrt{2-n}-\bigg(\sqrt{1+n}\bigg)^2+2n+1$$ is convergent or Divergent.

If convergent, Then $\lim_{n\rightarrow \infty}a_{n}$ equals

What i try:

$$\displaystyle \lim_{n\rightarrow \infty} a_{n}=\lim_{n\rightarrow \infty}\bigg[\sqrt{n+1}\bigg(\sqrt{n-2}-\sqrt{n+1}\bigg)\bigg]+2n+1$$

$$\lim_{n\rightarrow \infty}-\frac{3\sqrt{n+1}}{\sqrt{n-2}+\sqrt{n+1}}+2n+1$$

How i find that limits, please Help me Thanks

$\color{Blue}{\text{Edited::}}$

Instead of $\sqrt{n-2}$ actually it is $\sqrt{2-n}$

$$\displaystyle \lim_{n\rightarrow \infty} a_{n}=\lim_{n\rightarrow \infty}\bigg[\sqrt{n+1}\bigg(\sqrt{2-n}-\sqrt{n+1}\bigg)\bigg]+2n+1$$

$$\lim_{n\rightarrow \infty}\frac{\sqrt{n+1}(1-2n)}{\sqrt{2-n}+\sqrt{n+1}}+2n+1$$

With $\sqrt{2-n}$, what is $a_3$ beside a complex number ? Is the sequance supposed to be complex ? — Claude Leibovici, May 23 '20 at 01:34

Kavi Rama Murthy · Accepted Answer · 2020-05-23T01:39:58.100

2

$\sqrt {2-n}$ must be a typo since it is defined (as a real number) only when $n \leq 2$. So. I will take it as $\sqrt {n-2}$

This is of the form $b_n+2n+1$ where $b_n \to -\frac 3 2$ (as seen by dividing numerator and denominator by $\sqrt n$). Hence the limit is $\infty$.

edited May 23 '20 at 01:39

answered May 23 '20 at 00:31

Kavi Rama Murthy

311,013

Thanks Professor .So can we say that the sequence is Diverges as limit of sequence goes to infinity. – jacky May 23 '20 at 00:45
1

@jacky That is correct. – K.defaoite May 23 '20 at 01:01
Professor please have a look on that problem. I have done calculation mistake insteat of $\sqrt{n-2}$ it is $\sqrt{2-n}$. Thanks – jacky May 23 '20 at 01:14
1

$\sqrt {2-n}$ is defined only when $n=1$ or $2$. So I thought you had typed $2-n$ in stead of $n-2$. @jacky – Kavi Rama Murthy May 23 '20 at 01:38

Convergent of Divergent of sequence having Rational terms

1 Answers1