Video: Finding the Partial Sum of a Series and Deciding Whether the Series Is Convergent or Divergent

Find the partial sum from the series βˆ‘_(𝑛 = 1) ^(∞) 𝑒^(1/𝑛) βˆ’ 𝑒^(1/(𝑛 + 1)). Is the series convergent or divergent?

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Video Transcript

Find the partial sum from the series the sum from 𝑛 equals one to ∞ of 𝑒 to the power of one over 𝑛 minus 𝑒 to the power of one over 𝑛 plus one. Is the series convergent or divergent?

Remember, the partial sum of our series is the sum of the first 𝑛 terms. In general, the 𝑛th partial sum of the series, the sum from 𝑛 equals one to ∞ of π‘Ž sub 𝑛, is the sum from 𝑖 equals one to 𝑛 of π‘Ž sub 𝑖. And that’s π‘Ž sub one plus π‘Ž sub two plus π‘Ž sub three all the way up to π‘Ž sub 𝑛. We generally define this as 𝑆 sub 𝑛. So, let’s find the 𝑛th partial sum for our series.

We know it’s going to be π‘Ž sub one plus π‘Ž sub two plus π‘Ž sub three all the way up to π‘Ž sub 𝑛. But here, π‘Ž sub 𝑛 is 𝑒 to the power of one over 𝑛 minus 𝑒 to the power of one over 𝑛 plus one. π‘Ž sub one is found by replacing 𝑛 with the number one. So, we get 𝑒 to the power of one over one minus 𝑒 to the power of one plus one, which is simply 𝑒 minus 𝑒 to the power of one-half. Then, π‘Ž sub two is 𝑒 to the power of one-half minus 𝑒 to the power of one over two plus one, or 𝑒 to the power of one-third.

In the same way, we can define the remaining terms as shown. This means 𝑆 sub 𝑛, the 𝑛th partial sum, is given by 𝑒 minus 𝑒 to the power of one-half. That’s π‘Ž sub one. Plus 𝑒 to the power of one-half minus 𝑒 to the power of one-third. Remember, that’s π‘Ž sub two. Plus 𝑒 to the power of one-third minus 𝑒 to the power of a quarter. That was π‘Ž sub three. And we continue adding these terms until we get to π‘Ž sub 𝑛, which is 𝑒 to the power of one over 𝑛 minus 𝑒 to the power of one over 𝑛 plus one.

But if we look really carefully at our 𝑛th partial sum, we should see that some of the terms will cancel. We have negative 𝑒 to the power of one-half plus 𝑒 to the power of one-half. Well, that’s zero. We then have negative 𝑒 to the power of one-third plus 𝑒 to the power of one-third, which is also zero. But let’s add in π‘Ž sub 𝑛 minus one. And when we do, we see that this process repeats all the way up to negative 𝑒 to the power of one over 𝑛 plus 𝑒 to the power of one over 𝑛. And so, what this means is all we’re left with to describe our 𝑛th partial sum is 𝑒 minus 𝑒 to the power of one over 𝑛 plus one. And so, the partial sum for our series is 𝑒 minus 𝑒 to the power of one over 𝑛 plus one.

The next part of this question asks us whether the series is convergent or divergent. Well, remember, we say that if the limit as 𝑛 approaches ∞ of the 𝑛th partial sum is equal to some real number 𝑆, then that means the series, the sum of π‘Ž 𝑛, is convergent. So, let’s evaluate the limit as 𝑛 approaches ∞ of 𝑆 sub 𝑛.

In that case, that’s the limit as 𝑛 approaches ∞ of 𝑒 minus 𝑒 to the power of one over 𝑛 plus one. Now, actually, 𝑒 itself is completely independent of 𝑛. And we should notice that as 𝑛 grows larger, one over 𝑛 plus one grows smaller. As 𝑛 approaches ∞, one over 𝑛 plus one approaches zero. So, the limit as 𝑛 approaches ∞ of the 𝑛th partial sum is 𝑒 minus 𝑒 to the power of zero. But of course, 𝑒 to the power of zero β€” in fact, anything to the power of zero β€” is equal to one. So, our limit is equal to 𝑒 minus one. This is indeed a real number. And so, we can say that our series is convergent.

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