### 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 πth 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 thatβs 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, letβs find the πth partial sum
for our series.

We know itβs 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 πth
partial sum, is given by π minus π to the power of one-half. Thatβs π sub one. Plus π to the power of one-half
minus π to the power of one-third. Remember, thatβs 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 πth 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, thatβs 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 letβs 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
weβre left with to describe our πth 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 πth partial sum is equal to some real number π,
then that means the series, the sum of π π, is convergent. So, letβs evaluate the limit as π
approaches β of π sub π.

In that case, thatβs 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 πth 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.