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 β_(π = 0)^(β) π^(βπ) converges or diverges.

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### Video Transcript

Determine whether the series, which is the sum from π equals zero to β of π to the negative π, 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, π of π₯, which is continuous, positive, and decreasing on the interval between π and β and that π of π is equal to π π. Then, if the integral from π to β of π of π₯ with respect to π₯ is convergent, so is the sum from π equals π to β of π π. And if the integral from π to β of π of π₯ with respect to π₯ is divergent, so is the sum from π equals π to β of π π. Now, the series weβre trying to find the convergence of is this sum from π equals zero to β of π to the negative π. Therefore, π π is equal to π to the power of negative π. Since π of π is equal to π π, we obtain that π of π₯ is equal to π to the negative π₯. Since weβre taking the sum from π is equal to zero, this means that π is equal to zero.

And we now need to check whether π is a continuous, positive, and decreasing function on the interval between zero and β. π to the negative π₯ can also be written as one over π to the π₯. Now, π to the π₯ is positive for any value of π₯. Therefore, one over π to the π₯ is also positive. So, weβve satisfied the condition that π of π₯ is positive on our interval. Now, one over π to the π₯ is continuous since π to the π₯ cannot be equal to zero at any value. Therefore, weβve also satisfied this condition. And finally, since π to the π₯ is an increasing function for all π₯, this means that one over π to the π₯ is a decreasing function for all π₯. And so, weβve satisfied the last condition on π of π₯.

Weβre therefore able to use the integral test. We need to find the convergence of the integral from zero to β of π to the negative π₯ with respect to π₯. We can use the fact that the integral from π to the ππ₯ with respect to π₯ is equal to one over π π to the ππ₯ plus π. In our case, π is equal to negative one, and one over negative one is simply negative one. So, when we integrate π to the negative π₯, we get negative π to the negative π₯. And this is, of course, between our two bounds which is zero and β.

Since our upper bound is infinite, we need to find the limit as π₯ tends to β of negative π to the negative π₯. And we mustnβt forget to subtract negative π to the negative zero. When trying to evaluate this limit, we can consider what happens as π₯ gets larger and larger and larger. As π₯ gets bigger, negative π₯ gets more and more negative. Therefore, π to the power of negative π₯ gets closer and closer to zero.

And so, we can say that this limit is equal to zero. Then, we have π to the power of negative zero, which is simply π 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 β of π to the negative π₯ with respect to π₯ 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 π equals zero to β of π to the negative π is convergent.

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