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$x_n= \sqrt{\frac{n+1}{n}}$where $x_n$ is the $n^{th}$ term in the sequence.

Now, $x_n \rightarrow 1$ as $n\rightarrow\infty$.

Then applying definition of limit: $\left|\sqrt{\frac{n+1}{n}}- 1 \right|= \frac{\frac{n+1}{n}-1}{\sqrt{\frac{n+1}{n}}+1}<\frac{1}{2n}<\epsilon$

whenever $n>N=1/(2\epsilon)$.

-The difficulty for me is inability to understand how $1/(2n)$ was deduced.

Thank you for assistance and apologies for poor notations

the equation in the code should read after absolute value equation = {[(n+1)/n]-1}/{([(n+1)/n]^0.5)+1}<1/2n< epsilon

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    ill rewrite what i wrote initially, not sure what happened to the edits!2012-09-26

2 Answers 2

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Multiply the expression $\left(1+\frac{1}{n}\right)^{1/2}-1$ by $1$, in the form $\displaystyle\frac{\left(1+\frac{1}{n}\right)^{1/2}+1}{\left(1+\frac{1}{n}\right)^{1/2}+1}$.

The numerator is then $\left(1+\dfrac{1}{n}\right)-1$, which simplifies to $\dfrac{1}{n}$.

The denominator $\left(1+\dfrac{1}{n}\right)^{1/2}+1$ is $\gt 2$, since $1+\dfrac{1}{n}\gt 1$, and therefore its square root is $\gt 1$.

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I see that since $n(n+1)>n$ then $\sqrt{n(n+1)}+n>2n$ and so $\bigg|\frac{1}{\sqrt{n(n+1)}+n}\bigg|<\frac{1}{2n}$ But, $\bigg|\frac{1}{\sqrt{n(n+1)}+n}\bigg|=\bigg|\frac{\frac{n+1}{n}-1}{\sqrt{\frac{n+1}{n}}+1}\bigg|$ Therefore $\bigg|\sqrt{\frac{n+1}{n}}-1\bigg|=\bigg|\frac{\frac{n+1}{n}-1}{\sqrt{\frac{n+1}{n}}+1}\bigg|=\bigg|\frac{1}{\sqrt{n(n+1)}+n}\bigg|<\frac{1}{2n}$

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    @student101: Mostly right. In fact you need a logical and proper relation, containing $n$, just to make a connection between $n$, $\epsilon$ from one side and then with $N$. So, you get $\epsilon=1/2n$ and then $n=1/2\epsilon$. This is what you took as $N$. Of course $1/2n$ might be something different for example $\epsilon=1/3n$.2012-09-26