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

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

02:53

### 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 𝑎 sub 𝑛, is the sum from 𝑖 equals one to 𝑛 of 𝑎 sub 𝑖. And that’s 𝑎 sub one plus 𝑎 sub two plus 𝑎 sub three all the way up to 𝑎 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 𝑎 sub one plus 𝑎 sub two plus 𝑎 sub three all the way up to 𝑎 sub 𝑛. But here, 𝑎 sub 𝑛 is 𝑒 to the power of one over 𝑛 minus 𝑒 to the power of one over 𝑛 plus one. 𝑎 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, 𝑎 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 𝑎 sub one. Plus 𝑒 to the power of one-half minus 𝑒 to the power of one-third. Remember, that’s 𝑎 sub two. Plus 𝑒 to the power of one-third minus 𝑒 to the power of a quarter. That was 𝑎 sub three. And we continue adding these terms until we get to 𝑎 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 𝑎 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 𝑎 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 — in fact, anything to the power of zero — 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.