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# Determine whether the series converges or diverges.$\displaystyle \sum_{n = 1}^{\infty} \frac {1}{n^n}$

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don't let's determine whether the Siri's conversions or diverges first. Not that this is less than or equal to one over and square. Why? To see if this is true, Just multiply the denominators to the other side. And here, in order for this to be true, we should take and this ex for in here to be bigger than or equal to tool. So eventually if we take this sum here, if we're letting and be bigger than two, then this is smaller than this. And eventually Lee and his bigger than or equal to two because we started one and then we go all the way to infinity. So we passed two right away so I can go ahead and right this if you let's go and pull off that first term and then I have the sum from n equals two to infinity. The reason I'm doing this I'm pulling out the first term and rewriting the sum is because I would like to use comparison test. But if I want to compare our expression toe one over and square, I need end to be tour larger. And this is why I originally and started at one so I took that term out. Now I just have to to infinity so I can go ahead and replace this with one over and square and again this already using is this. This inequality is just using this fact here and then taking a sum on both sides. And then the plus one just came along for the ride here. Now I know that this Siri's here converges from section eleven point three. This is what we call P series, and P is bigger than one. In any time that happened. Opens. It's conversion. And if we just add one through a convergence Siri's, that's not going to change the fact that we still have a real number. So our Siri's is less than or equal to a conversion. Siri's Our Siri's is less than or equal to a conversion series. So by the comparison test, our Siri's also that Burgess Funny convergence, John, And that's our final answer

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