Sum of the terms of a geometric progression

Example

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 $n=1,2,3,\dots ,10$.

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 $r={\displaystyle \frac{1}{3}}$.

We write its first ten terms:

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

And we calculate the sums:

$S_1=7$

$S_2=7+{\displaystyle \frac{7}{3}}=9,3$

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

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

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

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

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

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

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

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

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

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?

Example

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

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 $b_n={\displaystyle \frac{7}{3^{n-1}}}$, then, $$S_n={\displaystyle \frac{7(1-{\displaystyle \frac{1}{3^n}})}{1-{\displaystyle \frac{1}{3}}}}={\displaystyle \frac{21}{2}}[1-({\displaystyle \frac{1}{3}}{)}^n]$$

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