Use the definition of limit to prove that both of the sequences { 1 √ n √ } and {(−1)n n } converges to 0.
Let an=1na_{n}=\frac{1}{\sqrt{n}}an=n1 . Let ϵ>0\epsilon>0ϵ>0 is given. Then there exist some N>0 satisfying N>1ϵ2N>\frac{1}{\epsilon^{2}}N>ϵ21.( Archimedean property of real numbers)
Therefore ∀\forall∀ n≥Nn\geq Nn≥N,
∣an−0∣=∣1n∣=1n≤1N<ϵ|a_{n}-0|=|\frac{1}{\sqrt{n}}|=\frac{1}{\sqrt{n}}\leq \frac{1}{\sqrt{N}}<\epsilon∣an−0∣=∣n1∣=n1≤N1<ϵ .
Hence from the definition of limits an→0.a_{n} \rightarrow 0.an→0.
Let bn=(−1)nn.b_{n}= \frac{(-1)^{n}}{n}.bn=n(−1)n. Then ∀n≥N\forall n\geq N∀n≥N,
∣bn−0∣=∣(−1)nn∣=1n≤1N<ϵ|b_{n}-0|=|\frac{(-1)^{n}}{n}|=\frac{1}{n}\leq \frac{1}{N}<\epsilon∣bn−0∣=∣n(−1)n∣=n1≤N1<ϵ .
Hence from the definition of limits bn→0.b_{n}\rightarrow 0.bn→0.
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