### Video Transcript

Does the series the sum from π
equals one to infinity of three times one over 10 to the power of π minus one
converge or diverge? If it converges, find the value of
the series.

The first thing we should recognize
for this question is that we have been given an infinite geometric series. The general form of this type of
question is shown here using sigma notation as with this question. Note that this type of series is
characterized by a first term π and a common ratio π. A general rule that we use for
geometric series is that if the absolute value of the common ratio is less than one,
then the series is convergent. And the value of the sum is equal
to the first term π divided by one minus the common ratio π. If instead the absolute value of
the common ratio is greater than or equal to one, the series is divergent and we
cannot assign its sum a value. Okay, this is all interesting, but
letβs see how it applies to our question.

If we look at our question, we can
see that the series that weβve been given exactly matches the general form for a
geometric series that we should be familiar with. We have a first term π of
three. And we have a common ratio π of
one over 10. Note that the index of π equals
one matches on both of our sums, as does the exponents on our common ratio π minus
one. If the index number had not
matched, we might need to perform an index shift. And if the exponents had no
matched, we might need to perform some factorization. Lucky for us, this is not the case,
and so we continue, safe in the knowledge that our common ratio is one over 10.

To move forward, our first step is
quite simple. We observed that the absolute value
of our common ratio π is less than one. This allows us to conclude that the
series is convergent. It also tells us that we can find
the value of our sum using this formula. If we apply this to our series, its
value is π divided by one minus π. Of course, substituting π equals
three and π equals one over 10, we obtain three divided by one minus one over
10. This is the same as three divided
by nine over 10 which is three times 10 over nine. Or if we cancel the common factor
of three on the top and bottom of our fraction, weβre left with 10 over three. If we tidy up our working, we see
that we have now answered the question. Using our knowledge of infinite
geometric series, we concluded that the series given by the question is convergent
and has a value of 10 over three.

Now, before we finish, this
question has an interesting characteristic which is worth mentioning. Let us look back at our series and
write out the terms. Of course, our first term is
three. We can obtain our next term by
multiplying by the common ratio one over 10 to get three over 10. Continuing this pattern, we get
three over 100 three over 1000 and so on. Now, if we were to represent these
fractions in decimal form, we might begin to see a pattern emerging. Since we started with a three and
multiplied by one over 10 for each successive term, we will have a three in every
position following the decimal point. And when we say every position,
this is really true. Since weβre dealing with an
infinite series, we will never run out of terms. And so our threes really do repeat
forever.

In fact, what we have created here
is a recurring decimal, specifically 3.3 recurring. Now, this is very interesting to
us, since we have essentially represented our infinite geometric series as a
recurring decimal. We wonβt go into too much detail
for this video, but suffice to say, the reverse process could also be conducted. If we were given a recurring
decimal, we could represent it as an infinite geometric series. Okay, but why do this? Well, weβve just shown that our
series is convergent on that we can find its value. The value we found was the
convenient fraction of 10 over three. It, therefore, follows that our
infinite geometric series is equal to 3.3 recurring, which is also equal to 10 over
three. So to conclude, it might not seem
like it at first, but geometric series give us a way to represent recurring
decimals. If we extend this concept a little
bit further, they also give us a way to express recurring decimals as fractions,
which is sometimes much more convenient.