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.
