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So far, I'ave used the squeeze theorem with functions $\frac1n$ and $-\frac1n$, and so got the limit $0$, but the answer is supposedly infinity... which makes little sense to me.

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closed as unclear what you're asking by TheSimpliFire, Namaste, user223391, Sri-Amirthan Theivendran, Mohammad Riazi-Kermani Feb 10 '18 at 17:20

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    $\begingroup$ And it makes no sense to me. $\endgroup$ – José Carlos Santos Feb 10 '18 at 16:23
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    $\begingroup$ Are you talking about $$\sum_{n=0}^\infty \frac 1{n+1}$$? $\endgroup$ – Jaideep Khare Feb 10 '18 at 16:24
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    $\begingroup$ Or is it supposed to be $$\lim_{n\to-1}\frac1{n+1}$$? $\endgroup$ – TheSimpliFire Feb 10 '18 at 16:25
  • $\begingroup$ lim n-> infinity 1/(n+1) $\endgroup$ – Cookiedough Feb 10 '18 at 16:26
  • $\begingroup$ sorry i can't get the format right ._. $\endgroup$ – Cookiedough Feb 10 '18 at 16:26
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Your idea is correct

$$-\frac1n\le \frac1{n+1}\le \frac1n$$

then conclude by squeeze theorem noting that the LHS and RHS both tend to zero.

As an alternative you can also observe that

$$0\le \frac1{n+1}\le \frac1n$$

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If you're talking about $lim_{n\rightarrow \infty}{\frac{1}{n+1}}$ try to think about how $\frac{1}{n+1}$ acts for really big values of $n$, whereas if you have $lim_{n\rightarrow 0}{\frac{1}{n+1}}$ try to compute $\frac{1}{n+1}$ when $n$ is a number really close to $0$.

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