Question Video: Finding the Intervals Where Functions Involving Logarithmic Functions Increase and Decrease | Nagwa Question Video: Finding the Intervals Where Functions Involving Logarithmic Functions Increase and Decrease | Nagwa

Question Video: Finding the Intervals Where Functions Involving Logarithmic Functions Increase and Decrease Mathematics • Higher Education

Determine whether the series ∑_(𝑛 = 0)^(∞) 𝑒^(−𝑛) converges or diverges.

03:05

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.

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