Video Transcript
Determine whether the series, which
is the sum from 𝑛 equals zero to ∞ of 𝑒 to the negative 𝑛, converges or
diverges.
We can try to find the convergence
of this series using the integral test. The integral test tells us that if
we have a function, 𝑓 of 𝑥, which is continuous, positive, and decreasing on the
interval between 𝑘 and ∞ and that 𝑓 of 𝑛 is equal to 𝑎 𝑛. Then, if the integral from 𝑘 to ∞
of 𝑓 of 𝑥 with respect to 𝑥 is convergent, so is the sum from 𝑛 equals 𝑘 to ∞
of 𝑎 𝑛. And if the integral from 𝑘 to ∞ of
𝑓 of 𝑥 with respect to 𝑥 is divergent, so is the sum from 𝑛 equals 𝑘 to ∞ of 𝑎
𝑛. Now, the series we’re trying to
find the convergence of is this sum from 𝑛 equals zero to ∞ of 𝑒 to the negative
𝑛. Therefore, 𝑎 𝑛 is equal to 𝑒 to
the power of negative 𝑛. Since 𝑓 of 𝑛 is equal to 𝑎 𝑛,
we obtain that 𝑓 of 𝑥 is equal to 𝑒 to the negative 𝑥. Since we’re taking the sum from 𝑛
is equal to zero, this means that 𝑘 is equal to zero.
And we now need to check whether 𝑓
is a continuous, positive, and decreasing function on the interval between zero and
∞. 𝑒 to the negative 𝑥 can also be
written as one over 𝑒 to the 𝑥. Now, 𝑒 to the 𝑥 is positive for
any value of 𝑥. Therefore, one over 𝑒 to the 𝑥 is
also positive. So, we’ve satisfied the condition
that 𝑓 of 𝑥 is positive on our interval. Now, one over 𝑒 to the 𝑥 is
continuous since 𝑒 to the 𝑥 cannot be equal to zero at any value. Therefore, we’ve also satisfied
this condition. And finally, since 𝑒 to the 𝑥 is
an increasing function for all 𝑥, this means that one over 𝑒 to the 𝑥 is a
decreasing function for all 𝑥. And so, we’ve satisfied the last
condition on 𝑓 of 𝑥.
We’re therefore able to use the
integral test. We need to find the convergence of
the integral from zero to ∞ of 𝑒 to the negative 𝑥 with respect to 𝑥. We can use the fact that the
integral from 𝑒 to the 𝑎𝑥 with respect to 𝑥 is equal to one over 𝑎 𝑒 to the
𝑎𝑥 plus 𝑐. In our case, 𝑎 is equal to
negative one, and one over negative one is simply negative one. So, when we integrate 𝑒 to the
negative 𝑥, we get negative 𝑒 to the negative 𝑥. And this is, of course, between our
two bounds which is zero and ∞.
Since our upper bound is infinite,
we need to find the limit as 𝑥 tends to ∞ of negative 𝑒 to the negative 𝑥. And we mustn’t forget to subtract
negative 𝑒 to the negative zero. When trying to evaluate this limit,
we can consider what happens as 𝑥 gets larger and larger and larger. As 𝑥 gets bigger, negative 𝑥 gets
more and more negative. Therefore, 𝑒 to the power of
negative 𝑥 gets closer and closer to zero.
And so, we can say that this limit
is equal to zero. Then, we have 𝑒 to the power of
negative zero, which is simply 𝑒 to the power of zero. Anything to the power of zero is
one. And the two negative signs in front
of the term cancel out with one another to give us a positive sign. From here, we can evaluate that the
integral from zero to ∞ of 𝑒 to the negative 𝑥 with respect to 𝑥 is equal to
one. Therefore, we found that our
integral is convergent. And so, by the integral test, we
can say that the sum from 𝑛 equals zero to ∞ of 𝑒 to the negative 𝑛 is
convergent.
