# Question Video: Finding the Partial Sum of a Series and Deciding Whether the Series Is Convergent or Divergent Mathematics • Higher Education

Find the partial sum from the series β_(π = 1) ^(β) π^(1/π) β π^(1/(π + 1)). Is the series convergent or divergent?

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### Video Transcript

Find the partial sum from the series the sum from π equals one to β of π to the power of one over π minus π to the power of one over π plus one. Is the series convergent or divergent?

Remember, the partial sum of our series is the sum of the first π terms. In general, the πth partial sum of the series, the sum from π equals one to β of π sub π, is the sum from π equals one to π of π sub π. And thatβs π sub one plus π sub two plus π sub three all the way up to π sub π. We generally define this as π sub π. So, letβs find the πth partial sum for our series.

We know itβs going to be π sub one plus π sub two plus π sub three all the way up to π sub π. But here, π sub π is π to the power of one over π minus π to the power of one over π plus one. π sub one is found by replacing π with the number one. So, we get π to the power of one over one minus π to the power of one plus one, which is simply π minus π to the power of one-half. Then, π sub two is π to the power of one-half minus π to the power of one over two plus one, or π to the power of one-third.

In the same way, we can define the remaining terms as shown. This means π sub π, the πth partial sum, is given by π minus π to the power of one-half. Thatβs π sub one. Plus π to the power of one-half minus π to the power of one-third. Remember, thatβs π sub two. Plus π to the power of one-third minus π to the power of a quarter. That was π sub three. And we continue adding these terms until we get to π sub π, which is π to the power of one over π minus π to the power of one over π plus one.

But if we look really carefully at our πth partial sum, we should see that some of the terms will cancel. We have negative π to the power of one-half plus π to the power of one-half. Well, thatβs zero. We then have negative π to the power of one-third plus π to the power of one-third, which is also zero. But letβs add in π sub π minus one. And when we do, we see that this process repeats all the way up to negative π to the power of one over π plus π to the power of one over π. And so, what this means is all weβre left with to describe our πth partial sum is π minus π to the power of one over π plus one. And so, the partial sum for our series is π minus π to the power of one over π 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 π approaches β of the πth partial sum is equal to some real number π, then that means the series, the sum of π π, is convergent. So, letβs evaluate the limit as π approaches β of π sub π.

In that case, thatβs the limit as π approaches β of π minus π to the power of one over π plus one. Now, actually, π itself is completely independent of π. And we should notice that as π grows larger, one over π plus one grows smaller. As π approaches β, one over π plus one approaches zero. So, the limit as π approaches β of the πth partial sum is π minus π to the power of zero. But of course, π 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 π minus one. This is indeed a real number. And so, we can say that our series is convergent.