# what is the limit of $\frac1{n+1}$ as $n\to\infty$? [closed]

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.

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

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

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