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1. What does it mean for a sequence {a} to converge to a € R? State the definition. (-1)n+1 2. Prove that lim = 0 n 2n 3. Pro
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Answer #1

1) The sequence \{ a_n\} is said to be convergent to a\in R if given \epsilon >0 we can find a M such that

|a_n-a|<\epsilon ~~~~~~~~~~\text{ for all }n>M

2) Since

\left |{ (-1)^{n+1} \over n} -0\right |= \left |{ (-1)^{n+1} \over n} \right |={1\over n}< \epsilon ~~~~~\text{ for all } n>M

Say if we take \epsilon =0.001 then M=1001

Hence \lim_{n \rightarrow \infty} {(-1)^{n+1}\over n}=0

3) Since

\left |{ 2n \over n+1} -2\right |= \left |{ -2 \over n+1} \right |={2\over n+1}< \epsilon ~~~~~\text{ for all } n>M

Say if we take \epsilon =0.001 then M=1000

Hence, \lim_{n\rightarrow \infty}{ 2n \over n+1} =2

4) Since

\left |{ 80 \over \sqrt{5n}} -0\right |= \left |{ 80 \over \sqrt{n}} \right |={ 80 \over \sqrt{n}} < \epsilon ~~~~~\text{ for all } n>M={6400\over \epsilon^2}

Hence, \lim_{n\rightarrow \infty}{ 80 \over \sqrt{n}} =0

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