Sum of the terms of a geometric progression

The objective is to add the first $n$ terms of a geometric progression.

We take the geometric progression of the first term $a_1=3$ and ratio $r=2$. We denote $S_n$ the sum of the first $n$ terms, and we are going to calculate the value of $S_n$ for $n=1,2,3,\ldots,10$.

The first ten terms are:

$$3,6,12,24,48,96,192,384,768,1.536$$

And the values of the sums:

$$S_1=3$$

$$S_2=3+6=9$$

$$S_3=3+6+12=21$$

$$S_4=3+6+12+24=45$$

$$S_5=3+6+12+24+48=93$$

$$S_6=3+6+12+24+48+96=189$$

$$S_7=3+6+12+24+48+96+192=381$$

$$S_8=3+6+12+24+48+96+192+284=765$$

$$S_9=3+6+12+24+48+96+192+284+768=1.533$$

$$S_{10}=3+6+12+24+48+96+192+284+768+1.536=3.069$$

As expected (we are adding positive terms), we obtain an increasing result. Then we ask ourselves: can it become infinitely big or will the numbers level off at some point?

Let's consider now the progression of the first term $a_1=7$, and ratio $r=\dfrac{1}{3}$.

We write its first ten terms:

$$7, \dfrac{7}{3}, \dfrac{7}{9}, \dfrac{7}{27}, \dfrac{7}{81}, \dfrac{7}{243}, \dfrac{7}{729}, \dfrac{7}{2.187}, \dfrac{7}{6.561}, \dfrac{7}{19.683}$$

And we calculate the sums:

$$S_1=7$$

$$S_2=7 + \dfrac{7}{3}=9,3$$

$$S_3=7 + \dfrac{7}{3}+ \dfrac{7}{9}=10,1$$

$$S_4=7+\dfrac{7}{3}+\dfrac{7}{9}+\dfrac{7}{27}=10,37037$$

$$S_5=7+\dfrac{7}{3}+\dfrac{7}{9}+\dfrac{7}{27}+\dfrac{7}{81}=10,45679012$$

$$S_6=7+\dfrac{7}{3}+\dfrac{7}{9}+\dfrac{7}{27}+\dfrac{7}{81}+\dfrac{7}{243}=10,4855967$$

$$S_7=7+\dfrac{7}{3}+\dfrac{7}{9}+\dfrac{7}{27}+\dfrac{7}{81}+\dfrac{7}{243}+\dfrac{7}{729}=10,4951989$$

$$S_8=7+\dfrac{7}{3}+\dfrac{7}{9}+\dfrac{7}{27}+\dfrac{7}{81}+\dfrac{7}{243}+\dfrac{7}{729}+\dfrac{7}{2.187}=10,498699639$$

$$S_9=7+\dfrac{7}{3}+\dfrac{7}{9}+\dfrac{7}{27}+\dfrac{7}{81}+\dfrac{7}{243}+\dfrac{7}{729}+\dfrac{7}{2.187}+\dfrac{7}{6.561}=10,49946654$$

$$S_{10}=7+\dfrac{7}{3}+\dfrac{7}{9}+\dfrac{7}{27}+\dfrac{7}{81}+\dfrac{7}{243}+\dfrac{7}{729}+\dfrac{7}{2.187}+$$

$$+\dfrac{7}{6.561}+\dfrac{7}{19.683}=10,49982218$$

Note that the sums of the second progression are also increasing, but not rising as fast as in the previous example. In fact, it seems to be a manageable growth: for the obtained results, the amounts are getting closer to $10,5$. Will it exceed this value at any point or, if not, will it become an upper bound of the values $S_n$? And, in this case, will we obtain an approximation to $10,5$ as good as we want it if we add enough terms?

Let's consider now a theoretical case:

If $a_1, a_2, \ldots ,a_n$ are the first $n$ terms of a geometric progression of ratio $r$. Then,

$$S_n=a_1+a_2+\ldots +a_n= a_1+a_1\cdot r + \ldots + a_1 \cdot r^{n-1}$$

Multiplying both members of the equality by $r$, the following is obtained:

$$r\cdot S_n=a_1\cdot r+a_1\cdot r^2 + \ldots + a_1 \cdot r^{n}$$

By reducing, member by member, these two equalities, we obtain:

Namely, we have:

$$S_n - r\cdot S_n = a_1 - a_1 \cdot r^n$$

So:

$$S_n(1-r)=a_1(1-r^n) \Rightarrow S_n=\dfrac{a_1(1-r^n)}{1-r}$$

Remembering the previous examples,

If $a_n=3\cdot 2^{n-1}$, then, $$S_n=\dfrac{a_1(1-r^n)}{1-r}=\dfrac{3(1-2^2)}{1-2}=3(2^n-1)$$

In such a way that, if we allow $n$ to grow indefinitely, $S_n$ will not stop growing, since $2^n$ can grow indefinitely if we choose an $n$ big enough.

If in this expression we substitute $n$ by any value, for example $10$, we will obtain the result of adding the first $10$ terms.

On the other hand, if $b_n=\dfrac{7}{3^{n-1}}$, then, $$S_n=\dfrac{7\Big(1-\dfrac{1}{3^n}\Big)}{1-\dfrac{1}{3}}=\dfrac{21}{2}\Big[1-\Big(\dfrac{1}{3}\Big)^n\Big]$$

In this case, the base of the nth power is less than the unit, which means that as the value of $n$ increases, $\Big(\dfrac{1}{3}\Big)^n$ decreases. Because of this, the value of $S_n$ stabilizes if we choose $n$ to be big enough.