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# Use the Integral Test to determine whether the series is convergent or divergent.$\displaystyle \sum_{n = 1}^{\infty} n^2 e^{-n^3}$

## The given series converges

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in this problem, we are given a Siri's and we have to determine if it's divergent or convergent. And we're told that we have to use the integral test. So let's first review what were given were given a Siri's where n equals one to infinity of n square times e race negative n cubed. So the first thing that we need to dio is determined. Can we even use the integral test? Well, there are conditions that we have to satisfy in order to use it. And I would recommend looking over those conditions before watching this video just to get it, you know, just to understand it more to understand where I'm getting this information. So what we determine is we can use the integral test because our function ffx equal to X squared times erase negative X cubed is positive and decreasing for expert er than one. And so we have satisfied the conditions for the integral test so we can get the integral from one to infinity of X squared times erase night of X cubed in D X. But again, we have this improper integral. So we're going to take the limit as T approaches infinity of her integral from one to t of x square times e race and negative X cubed indie X What we can do, you substitution. We're going to say let u equal Negative X cubed. So do you. The derivative would be negative three x squared in D. X so d X would be equal to negative d'you over three x squared so we can rearrange our limit a little bit. But first we need to find the limits of integration. So we start with the limits of integration from one to t. Well, we can use our use substitution to change our limits of integration as negative one to negative t cute. So now we can rewrite our limit. We'll get the limit as T approaches infinity of the integral negative one to negative T cube of X square times e to the u times. Negative do you over three x squared and then we can pull out that negative 1/3. So we'll get the limit as t approaches infinity of negative 1/3 times the integral of negative one to negative t cubed of eu into you. Then we could take the anti derivative and weaken. Factor out that negative one third outside of the limit. You get negative 1/3 times the limit as t approaches infinity of you evaluated from negative one negative T cube. And when you do that, we'll find that our limit equals 1/3 ee. So what does that mean? Well are limit, and also our Siri's approached a specific number that wasn't infinity. So we can say that our Siri's converges by the integral test. So I hope this problem helped you understand how we can use the integral test to evaluate the convergence of a Siri's.

University of Denver

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