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ProveSample claim

Let aa and bb be arbitrary odd integers. Prove that a+ba+b is even.

  1. Shuffled line 1

    Adding these two representations gives a+b=(2m+1)+(2n+1)a+b=(2m+1)+(2n+1).

  2. Shuffled line 2

    Since bb is odd, choose an integer nn such that b=2n+1b=2n+1.

  3. Shuffled line 3

    Since aa is odd, choose an integer mm such that a=2m+1a=2m+1.

  4. Shuffled line 4

    Adding and regrouping gives (2m+1)+(2n+1)=2(m+n)+1(2m+1)+(2n+1)=2(m+n)+1.

+3 more lines. Not all of them belong.
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Proof: make a limit precise problem

For each positive integer nn, let an=n/(n+2)a_n=n/(n+2). Prove from the definition of convergence that an→1a_n\to1.

Show a hintProof: make a limit precise hint

The definition asks: given any ε>0\varepsilon>0, can you make ∣an−1∣<ε|a_n-1|<\varepsilon for every sufficiently large nn? First simplify ∣an−1∣|a_n-1|.

Check against the worked answer

Let \(\varepsilon>0\). For every positive integer \(n\), \[ \left|\frac{n}{n+2}-1\right|=\left|\frac{-2}{n+2}\right|=\frac{2}{n+2}. \] Choose a positive integer \(N>2/\varepsilon\). Such an integer exists because the positive integers are unbounded. If \(n\ge N\), all denominators are positive, so \[ \frac{2}{n+2}\le\frac{2}{N+2}<\frac{2}{N}<\varepsilon. \] Thus, for every positive tolerance \(\varepsilon\), this \(N\), chosen for the given \(\varepsilon\), works for every \(n\ge N\). This is exactly the definition of \(a_n\to1\).

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