Video: Checking the Existence of the Limit of the Partial Sums to Decide If a Series Is Convergent or Divergent

True or false: if π‘Ž_(𝑛) ⟢ 0 as 𝑛 ⟢ ∞, then βˆ‘_(𝑛 = 0)^(∞) π‘Ž_(𝑛) is convergent.

03:07

Video Transcript

True or false, if π‘Ž 𝑛 approaches zero as 𝑛 approaches ∞, then the sum from 𝑛 equals zero to ∞ of our sequence π‘Ž 𝑛 is convergent.

If we believe this to be true, we would have to show that every sequence π‘Ž 𝑛 which has that π‘Ž 𝑛 approaches zero when 𝑛 approaches ∞ has that the sum from 𝑛 equals zero to ∞ of π‘Ž 𝑛 is convergent. However, if we believe this to be false, then we only have to find one sequence π‘Ž 𝑛 where our sequence π‘Ž 𝑛 approaches zero as 𝑛 approaches ∞, and the sum from 𝑛 equals zero to ∞ of π‘Ž 𝑛 is divergent. Now, we recall that to check the convergence or divergence of a series, we check the limit of the partial sums. We say that our series is convergent if the limit as the number of terms tends to ∞ of our partial sum exists. And we say that our series is divergent if the limit of the partial sum as the number of terms tends to ∞ does not exist.

Let’s take a look at some divergent series, for example, the harmonic series, which can be represented as the sum from 𝑛 equals zero to ∞ of one divided by 𝑛 plus one. We can then define the partial sum 𝑆 π‘˜ to be equal to one plus a half plus a third all the way up to adding one divided by π‘˜. Let’s now consider our partial sum 𝑆 sub eight. And that’s one plus a half all the way up to adding one-eighth. We can turn this into an inequality by reducing some of the values in our terms. We reduce the value of one-third to a quarter. Then, we reduce the value of one-fifth, one-sixth, and one-seventh all to the value of one-eighth. Evaluating this inequality gives us that 𝑆 eight is greater than or equal to one plus a half plus a half plus a half.

Now we ask the question, what would’ve happened if instead of calculating the partial sum 𝑆 sub eight, we had calculated the partial sum 𝑆 sub 16? First, we need to add the extra terms onto our partial sum. So, we add one-ninth then one-tenth all the way up to one sixteenth. We could generate a very similar inequality for our partial sum 𝑆 sub 16. We’ll just change the last eight terms in our partial sum to one sixteenth. We have eight terms of one sixteenth. So, we add all of these together to get a half. In particular, what we’ve shown is that our partial sum 𝑆 sub 16 is greater than or equal to our partial sum 𝑆 sub eight plus a half. But we could’ve done this again. We could’ve done this with our partial sum 𝑆 sub 32. And we would’ve got that it’s greater than or equal to our sixteenth partial sum plus a half.

Another way of looking at this is as we take more and more terms of our partial sum, we keep adding a half. So, by taking more and more terms, we can make our partial sum as large as we want. This is the same as saying that our partial sums are unbounded. Since the partial sum is unbounded, its limit does not exist. And therefore, our series is divergent. Therefore, we can conclude the statement given to us in the question is false because we know the limit as 𝑛 approaches ∞ of one divided by 𝑛 plus one is equal to zero. But the sum from 𝑛 equals zero to ∞ of one divided by 𝑛 plus one is divergent.

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