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

## Diverges

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##### Top Calculus 2 / BC Educators   ##### Kristen K.

University of Michigan - Ann Arbor Lectures

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### Video Transcript

let's determine whether the Siri's converges are diverges. Well, here we can go ahead and maybe compare this with another Siri's. So here, let's call this a M. And then let me compare this with BND equals one over the square root of and square, which happens to simplify it and just be won over it. Then let's use limit comparison test. So let's look at the limit as n goes to infinity and then we have a n over bien. So we take the limit as n goes to infinity. So if we do a n over bian, that should end up being and over the square root of n squared plus one. Now let me bring this limit over here. Since I'm running out of room, this is the limit and goes to infinity. Now what I'LL do here is I'LL go inside that radical Claudia and square So let's factor out and squared That's one plus one over and square. Then I pull the and square outside of the radical and then the square of and squared as we have seen already. That's just end. So then I could rewrite the denominator is n times one plus one over and square. Then we could cancel these ends. That's the whole point of factoring out the end of the radical. Now we just have the limit angles to infinity, one over a square room, one plus one over and square. The one over and squared will go to zero, and you just get one. So let me computer since says, if you take the limited and over being and you have a number, so this is positive. So we have a number one that satisfies this inequality. It's a positive number. It has to be bigger than zero, and it cannot be insanity, and that's what we have. So by living comparison test, the answer to our problem will be the same as the answer for the sum of Bien. Now we know the Siri's one over end this diverges. This is You can either use the pee test or just say it's a harmonic series. But if you use P equals one, then it's divergent by Peters, therefore, by Lim comparison. Since we got the answer and one in the limit, this tells us that our Siri's also diverges. So our Siri's diverges by Lim comparison says, and that's the final answer

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##### Top Calculus 2 / BC Educators   ##### Kristen K.

University of Michigan - Ann Arbor Lectures

Join Bootcamp