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How can I prove (using the sequence limit definition )the following: $$\begin{equation*} \lim_{n \rightarrow \infty} \frac{n+1}{2n^2 - 1}=0 \end{equation*},$$

I am stucked because I could not get rid from $(n+1)$ in the numerator, any hint will be appreciated.

Thanks!

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For every $\epsilon>0$, choose some $N\in{\bf{N}}$ such that $1/N<\epsilon$, for all $n>N+2$ then \begin{align*} (n+1)/(2n^{2}-1)\leq(n+1)/(2n^{2}-8)&=\dfrac{1}{2}\dfrac{n+1}{(n+2)(n-2)}\\ &\leq\dfrac{1}{2}\dfrac{n+1}{(n+1)(n+2)}\\ &\leq\dfrac{1}{2}\dfrac{1}{N}\\ &<\epsilon. \end{align*}

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HINT : $\dfrac{n+1}{2n^2 - 1} = \dfrac{\frac{1}{n} + \frac{1}{n^2}}{2 - \frac 1{n^2}}$. Now, the limit of both numerator and denominator exist as $n \to \infty$.

But if you want the $n_0 / \epsilon$ argument, then pick $\epsilon > 0$, and let $n_0$ be such that $\frac{1}{n_0} < \frac{\epsilon}{2}$. Then, note that $\frac {1}{n^2} < \frac 1 n < \frac 1{n_0}$ for all $n > n_0$, and that $2 - \frac{1}{n^2} \geq 1$ for all $n \geq 1$. Therefore, for $n > n_0$: $$ \frac{\frac 1n + \frac 1{n^2}}{2 - \frac{1}{n^2}} \leq \frac{1}{n} + \frac{1}{n^2} < \frac{2\epsilon}{2} < \epsilon $$

Proving that the limit exists and equals $0$.

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  • $\begingroup$ no I want to prove by definition ...... which means that I have to find $n_{0}$ $\endgroup$ – Emptymind Nov 4 '17 at 2:56
  • $\begingroup$ Either way, this representation is useful. You can get an $n_0$ out of it easily. I can show you. $\endgroup$ – астон вілла олоф мэллбэрг Nov 4 '17 at 2:57
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$$2n^2-1 \geq n^2$$

$$n+1 \leq 2n$$

Hence, $$\frac{n+1}{2n^2-1}\leq \frac{2n}{n^2}=\frac{2}{n}$$

Finding $n_0$ to prove $\frac2n$ converges to $0$ is left as an exercise.

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