limit ( {(a+1)^n - a^n) where n tends to inifnity assuming that a0 but also consider a=1.

Question

limit ( {(a+1)^n - a^n)  where  n tends to inifnity
assuming that a>0 but also consider a<0  and a>=1.

dls · Answer

$$\LARGE  \lim_{n \rightarrow \infty}(a+1)^{n}-a^n $$
you mean this?

anonymous · Answer

The two terms must be included in braces meaning that the limit of the
difference of the two must be determined.

dls · Answer

$$\LARGE  \lim_{n \rightarrow \infty}((a+1)^{n}-a^n)$$

anonymous · Answer

how about using this identity$$b^n-a^n=(b-a)(b^{n-1}+b^{n-2}a+...ba^{n-2}+a^{n-1})$$are u familiar with that?

anonymous · Answer

No not.Please give more information

anonymous · Answer

$\begin{align*}(a+1)^n&=a^n+a^{n-1}+a^{n-2}+\cdots+a^2+a+1\
(a+1)^n-a^n&=a^{n-1}+a^{n-2}+\cdots+a^2+a+1\end{align*}$

Note that the RHS can be written as
$$\large\sum_{k=0}^{n-1}a^k$$
Since you're taking the limit as n approaches infinity, you basically have
$$\large\sum_{k=0}^{\infty}a^k$$
Now let's use what we know about a geometric series like this one.

For $a\ge1,$ the series diverges.

For $0<a<1,$ the series converges.

For $a=0,$ the terms are all 0, so the sum is 0 (convergent).

For $-1<a<0,$ the series converges.

For $a\le-1,$ the series diverges.

anonymous · Answer

Thank you very much. This is well explained.

anonymous · Answer

oh oh, be careful$$(1+a)^n\neq a^n+a^{n-1}+...+a+1$$

anonymous · Answer

@Hennie, make note of what @mukushla pointed out. I forgot the coefficients on most of the terms, but you can still write it as a series. Thanks for the correction