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Show that $lim_{ntoinfty}left(frac 1n−frac 1{n+1}right)=0$ using epsilon-delta definition.

$$|frac{1}{n}-frac{1}{n+1}-0|=frac{1}{n(n+1)}$$

$$<frac{1}{n^2}$$

Given $epsilon>0$, we look for $Nin Bbb N$ such that

$$nge Nimplies frac{1}{n(n+1)}<epsilon$$ It is easier to look for $N$ s.t

$$nge Nimplies frac{1}{n^2}<epsilon$$

or
$$nge Nimplies n>sqrt{frac{1}{epsilon}}$$

We can take $$N=lfloor sqrt{frac{1}{epsilon}}rfloor +1.$$

Think about what it means to show that $lim_{ntoinfty}a_n = L$. By definition,

For any $epsilon > 0$, there exists a $Ninmathbb N$ such that for all $nge N$, we have $$|a_n - L| < epsilon.$$

From here, you need to show that there exists some natural number $N$ such that $$left|left({1over n} - {1over n+1}right) - 0right| < epsilon.$$ You can say that $$begin{align}left|left({1over n} - {1over n+1}right) - 0right| &= left|{(n+1) - nover n(n+1)}right| \&= {1over n^2 + n}\ &le cdots \ &<epsilonend{align}$$ From here, your goal is to complete the scratchwork by being able to create another valid inequality at the $cdots$ section. This is usually done by trying to isolate $n$ in terms of $epsilon$. How can you reduce the number of $n$'s appearing in the 3rd to last line that allows you to have just one $n$?