Problem 26 Easy Difficulty

Determine whether the series converges or diverges.
$ \displaystyle \sum_{n = 1}^{\infty} \frac {1}{n \sqrt {n^2 -1}} $

Answer

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Video Transcript

Let's determine whether the Siri's converges laboratories here. I'LL use the limit comparison test. So let's call this given term here, not the whole Siri's. Just the term. Just a fraction Let's call that a M, and then I'll call being one over and radical and squared, but that could just simplified to be won over and square. So the choice when you're using woman comparison, says the choice for Bien could often be determined just by looking at Anne here. I just see one up top, so I just leave it in there and be and then in the denominator is and gets really, really big. This minus one doesn't really play a significant role compared to end. So I just dropped the minus one, and that's how I have my formula for being. Also, we know that the sum of the B end this will converge here. You can use the pee test by the pee test with P equals two. That's the two right there. That's the value of P. So hopefully we can use this test. But before we use limit comparison tests, we need to find this limit and let's just enough this limit by the value, see? So we need to look at a N over bian in the limit. So lim and goes to infinity. So am I. And then dividing by being well, that's dividing by one over and square. That's the same thing, is multiplying by and squared over one. And then in the denominator work, let me just go ahead and pull out and end from the radical. So I'm looking inside the radical out, pull out and square, and then I have one minus one over and square, and then I could pull the end square outside of the square roots, and it just becomes an end after I evaluate the square root. So we have this and that's already here. But then I have this whole term appear after I rewrite the radical. So this term it's coming from the radical, and that's radical was the term that was originally in the denominator. So now the reason for doing this is that we could cancel those and squares rooms and then evaluate the limit. So there's and square and then another and square we have won over radical one, minus one over and square. Now take that limit you have won over the square root of one minute zero, which is just one. And this is good, because any time the value of C satisfies this inequality in our case, C equals one. So this is true. This means that we can actually use the Lim comparison test. And so since we know the B end, the sum of one over and square converges so sense Deon converges and the limit as n goes to infinity of a n over bien equals one by the limit comparison test our Siri's, You also must converge. Okay. Yeah, and that's our final answer.