Question Video: Deciding if a Given Series Converges or Diverges | Nagwa Question Video: Deciding if a Given Series Converges or Diverges | Nagwa

Question Video: Deciding if a Given Series Converges or Diverges Mathematics • Higher Education

Does the series βˆ‘^(∞) _(𝑛 = 1) 3(1/10)^(𝑛 βˆ’ 1) converge or diverge? If it converges, find the value of the series.

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Video Transcript

Does the series the sum from 𝑛 equals one to infinity of three times one over 10 to the power of 𝑛 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‘Ž and a common ratio π‘Ÿ. 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‘Ž divided by one minus the common ratio π‘Ÿ. 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 let’s see how it applies to our question.

If we look at our question, we can see that the series that we’ve been given exactly matches the general form for a geometric series that we should be familiar with. We have a first term π‘Ž of three. And we have a common ratio π‘Ÿ of one over 10. Note that the index of 𝑛 equals one matches on both of our sums, as does the exponents on our common ratio 𝑛 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‘Ÿ 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‘Ž divided by one minus π‘Ÿ. Of course, substituting π‘Ž equals three and π‘Ÿ 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, we’re 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 we’re 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 won’t 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, we’ve 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.

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