$\lim_{n\to\infty} \frac{a^{2n}+1}{a^n+2}.$

Question

I would like to ask why is the sequence convergent to $\dfrac{1}{2}$ when the parameter $|a|<1$ ? At the same time, I would like to ask if there is a limit for $a<-1$, because the sequence would alternate with signs.

$$\lim_{n\to\infty} \frac{a^{2n}+1}{a^n+2}$$

Answer 1

If $\lvert a\rvert<1,\ $ then $\displaystyle\lim_{n\to\infty} (a^{2n}+1) = 1,\ $ and $\displaystyle\lim_{n\to\infty} (a^{n}+2) = 2,\ $ so by the product rule for limits,

$\displaystyle\lim_{n\to\infty} \frac{a^{2n}+1}{a^n+2} = \frac{1}{2}.$

If $ a=1,\ $ then $\ \displaystyle\lim_{n\to\infty} \frac{a^{2n}+1}{a^n+2} = \frac{2}{3}.$

If $\ a = -1,\ $ or if $\ \lvert a\rvert>1,\ $ then the limit does not converge: if $\ a=-1,\ $ the sequence alternates between $2$ and $\frac{2}{3},\ $ if $\ a > 1,\ $it diverges to $\ +\infty,\ $ and finally if $\ a<-1,\ $ the sequence diverges by alternating between increasingly large negative and large positive numbers.

Answer 2

I have tried to make the answer as simple as possible, so forgive me if it sounds a bit silly to the many experts among you.... :~)

First we simplify $\frac{a^{2n}+1}{2+a^n}=\frac{a^{2n}}{2+a^n}+\frac{1}{2+a^n}$

we know that if |a|<1, then, $\lim _{x\rightarrow \infty }a^n=0$ (as when numbers smaller than one are raised to higher powers, they get smaller, eg. (0.01)^2=0.001,etc ...)

Hence, $\lim _{n\rightarrow \infty }\ \ \frac{a^{2n}+1}{2+a^n}=\frac{a^{2n}}{2+a^n}+\frac{1}{2+a^n}=\frac{0}{2+0}+\frac{1}{2+0}=\frac{1}{2}$

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1 raised to any positive power is 1, hence at a=1, we have $\lim _{n\rightarrow \infty }\ \ \frac{a^{2n}+1}{2+a^n}=\frac{\left(1^{\infty }+1\right)}{2+1^{\infty }}=\frac{2}{3}$

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But if |a|>1, then $\lim _{x\rightarrow \infty }a^n=\infty$ (as when numbers bigger than one are raised to higher powers, they get bigger, eg. 10^2=100,etc ...)

Hence, $\lim _{n\rightarrow \infty }\ \ \frac{a^{2n}+1}{2+a^n}=\frac{a^{2n}}{2+a^n}+\frac{1}{2+a^n}=\frac{\left(a^n\cdot a^n\right)}{2+a^n}+\frac{1}{2+a^n}=\frac{\left(a^n\right)}{\frac{2}{a^n}+1}+\frac{1}{2+a^n}=\frac{\infty }{\frac{2}{\infty }+1}+\frac{1}{2+\infty }=\infty $