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I found a lot of ways to prove that $\sum\limits_{n=1}^\infty \frac 1 {n^2}$ converges.

I wondered if you could also prove it using the the fraction criteria ($\lim\limits_{n\to \infty} |\frac {a_n+1} {a_n}|<1)$ and that $\frac 1 {n^2} < \frac 1 {(n-1)\cdot n}$

Which results in: $\lim\limits_{n\to\infty} \frac {a_n+1} {a_n} = \lim\limits_{n\to\infty} \frac {n-1} {n} = 1 $

However my textbook says it you could still show that it converges with $=1$ but I do not understand what it does. Can someone of you help me out?

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    You are asking if one can apply a criterion requiring that $\lim\limits_{n\to \infty} |\frac {a_n+1} {a_n}|<1$ to a series such that $\lim\limits_{n\to \infty} |\frac {a_n+1} {a_n}|=1$? The answer is clearly "no", if only for reasons of *pure logic*, don't you think? – 2017-01-11
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    Did you want to write $a_n+1$ or $a_{n+1}$ in the numerator? You can typeset the latter as `a_{n+1}`. – 2017-01-11
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    Perhaps it would be also good if you were more specific about what you mean by *my textbook*. At least name of the book (and perhaps edition), chapter and page number where the claim you mention can be find. And you might also add a link (if it is online or in Google Books). Or retype relevant passage if it is not too long (perhaps even screenshot would be better than nothing). – 2017-01-11
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    I think what your textbook means is that if you get 1, you can still show it converges by other means; I don't think it's meant that you can still use the ratio test in that case. – 2017-01-11
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    You can't drive a screw to the wall with a hammer (at least not well). You need another tool. – 2017-01-11

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Not gonna work. You can show it converges in various ways, but this isn't one of them; once you get an indeterminate answer from the ratio test, it's time to move on to a different test. (The integral test is the easiest one here, I think, but from the timbre of your questions, I don't think you're there yet...)

So let's do it this way.

$$\frac{1}{n} - \frac{1}{n+1} = \frac{1}{n^2+n} \therefore$$ $$\sum_{n=1}^m\frac{1}{n^2+n} = 1 - \frac{1}{m+1} \therefore\sum_{n=1}^\infty\frac{1}{n^2+n} = 1$$

So you use the limit comparison test:

$$\lim_{n\to\infty}\frac{\frac{1}{n^2}}{\frac{1}{n^2+n}} = 1$$

Therefore they both converge or both diverge; therefore they both converge.

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Note that when $ \lim\limits_{n\to\infty} \frac {a_{n+1}} {a_n} = 1$ you can't conclude that the series diverges or converges.

For example:

$\sum\limits_{n=1}^\infty \frac{1}{n}$ diverges and $\lim\limits_{n\to\infty} \frac {a_{n+1}} {a_n} = 1$

$\sum\limits_{n=1}^\infty \frac 1 {n^2}$ converges and $\lim\limits_{n\to\infty} \frac {a_{n+1}} {a_n} = 1$