Question

Prove whether the series converges or diverges.

Answer 1

I would probably use comparison test
hint: n+1>n

Answer 2

What would I compare it too?

Answer 3

sqrt(1/n) ?

Answer 4

Usually when I come to things and want to try a comparison test, I don't really "try" to look for something to compare it to that'll make it converge or diverge simply cause I don't really know yet. So I just ask myself, "What looks similar?" and then go from there.

So here we have:

\[\sqrt{\frac{n+1}{n}}\]

and it looks like if we just take that 1 out it would be pretty similar to:

\[\sqrt{\frac{n}{n}}\]

Is this greater than or less than the thing we're comparing it to? Since we took 1 out, I think we can see that it's less than:

\[\sqrt{\frac{n}{n} } < \sqrt{\frac{n+1}{n}}\]

So now of course that thing on the left simplifies, and we can sum from 1 to infinity now:

\[\sum_{n=1}^\infty 1 \le \sum_{n=1}^\infty \sqrt{\frac{n+1}{n}}\]

Since the left hand side is infinite, and less than our sum, then our sum is also infinite too!

Answer 5

so it diverges ?