Proof of geometric series formula

Question

So for, the above formula, how did they get $(n+1)$ a for the geometric progression when $r = 1$. I also am confused where the negative a comes from in the following sequence of steps.

Answer 1

When $r=1$ we have

$$S_{n}=\sum_{j=0}^{n}ar^{j}=\sum_{j=0}^{n}a=\underbrace{a+a+a+...+a+a}_{(n+1)-\text{times}}=...$$

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For $r\neq 1$, let

$$S_{n}=\sum_{k=0}^{n}ar^{k}=a+ar+ar^{2}+...+ar^{n-1}+ar^{n}$$

Multiplying by $r$ we have

$$rS_{n}=\sum_{k=0}^{n}ar^{k+1}=ar+ar^{2}+ar^{3}+...+ar^{n}+ar^{n+1}$$

Then which terms cancel when we compute $rS_{n}-S_{n}$?

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Or adding and subtracting the term $a$ to the RHS of $rS_{n}$ we have

$$rS_{n}=\color{red}{a}+ar+ar^{2}+ar^{3}+...+ar^{n}+ar^{n+1}-\color{red}{a}$$ $$=\left(a+ar+ar^2+...+ar^{n}\right)+ar^{n+1}-a$$

and what is the term in the brackets equal to?

Answer 2

Consider the sum $$\sum_{j=0}^{\infty}ar^j$$.

Now for find the sum we need show that the sequence of partial sum of the series converges.

Let us consider the partial sum of the serie $$S_n=a+ar+ar^2+ar^3+\cdots ar^n$$

Consider $$rS_n=ar+ar^2+ar^3+\cdots ar^{n+1}$$

Now $$S_n-rS_n=a-ar^{n+1}\iff S_n(1-r)=a-ar^{n+1}$$

For $r\neq 1$ $$S_n=\frac{a-ar^{n+1}}{1-r}$$

Now $S_n$ is the $n$-th partial sum of your serie, for find the sum is sufficient take $\lim_{n\to \infty }S_n$ and if it exists to a number $s$ we say that the sum of the serie is $s$. But what can you say about

$$\lim_{n\to \infty} \frac{a-ar^{n+1}}{1-r}$$

what can you infer from $r$ for the limit exists?

In the case of the above limit exists then $$\sum_{j=0}^{\infty}ar^{j}=s$$