Four limits

Calculus (Integrals, Series)
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Grigorios Kostakos
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Four limits

#1

Post by Grigorios Kostakos »

Find the limits:

\(\begin{aligned} &(a)\quad\displaystyle\mathop{\lim}\limits_{n\rightarrow{+\infty}}\frac{n!+5^{n}}{7^{n}+n!}\,,\\\\ &(b)\quad\displaystyle\mathop{\lim}\limits_{n\rightarrow{+\infty}}\sqrt[n]{2+\frac{1}{n}}\,,\\\\ &(c)\quad\displaystyle\mathop{\lim}\limits_{n\rightarrow{+\infty}}\frac{n\,\sin(n!)}{n^2+1}\,,\\\\ &(d)\quad\displaystyle\mathop{\lim}\limits_{n\rightarrow{+\infty}}\left({\frac{1}{n^2}+\frac{2}{n^2}+\ldots+\frac{n}{n^2}}\right)\,. \end{aligned}\)
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Papapetros Vaggelis
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Re: Four limits

#2

Post by Papapetros Vaggelis »

\(\displaystyle{(a)}\) If \(\displaystyle{n\in\mathbb{N}}\) , then :

\(\displaystyle{\dfrac{n!+5^{n}}{7^{n}+n!}=\dfrac{\displaystyle{1+\dfrac{5^{n}}{n!}}}{\displaystyle{\dfrac{7^{n}}{n!}+1}}}\) .

It's known that \(\displaystyle{\sum_{n=0}^{\infty}\dfrac{7^{n}}{n!}=e^7<\infty \,,\sum_{n=0}^{\infty}\dfrac{5^{n}}{n!}=e^5<\infty}\), so :

\(\displaystyle{\lim_{n\to \infty}\dfrac{7^{n}}{n!}=\lim_{n\to \infty}\dfrac{5^{n}}{n!}=0}\) .

Finally, \(\displaystyle{\lim_{n\to \infty}\dfrac{n!+5^{n}}{7^{n}+n!}=1}\) .

\(\displaystyle{(c)}\) For each \(\displaystyle{n\in\mathbb{N}}\) holds :

\(\displaystyle{\left|\dfrac{n\,\sin\,(n!)}{n^2+1}-0\right|=\dfrac{n\,\left|\sin\,(n!)\right|}{n^2+1}\leq \dfrac{n}{n^2+1}<\dfrac{1}{n}}\)

cause \(\displaystyle{n^2<n^2+1}\) . Due to the fact that \(\displaystyle{\lim_{n\to \infty}\dfrac{1}{n}=0}\), it follows that :

\(\displaystyle{\left(\forall\,\epsilon>0\right)\,\left(\exists\,n_0=n_0(\epsilon)\in\mathbb{N}\right): n\geq n_0\implies \left|\dfrac{n\,\sin\,(n!)}{n^2+1}-0\right|<\dfrac{1}{n}<\epsilon}\)

which means that \(\displaystyle{\lim_{n\to \infty}\dfrac{n\,\sin\,(n!)}{n^2+1}=0}\) .
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Re: Four limits

#3

Post by jacks »

For \((d)\) Given \(\displaystyle \lim_{n\rightarrow \infty}\left(\frac{1}{n^2}+\frac{2}{n^2}+\ldots+\frac{n}{n^2}\right)\)

Now Convert the Infinite Series into Rein man Sum

\(\displaystyle =\lim_{n\rightarrow \infty}\frac{1}{n}\cdot \left(\frac{1}{n}+\frac{2}{n}+\ldots+\frac{n}{n}\right) =\lim_{n\rightarrow \infty}\frac{1}{n}\sum{r=1}^{n}\left(\frac{r}{n}\right)\)

Now put \(\displaystyle \frac{r}{n} = x\) and \(\displaystyle \frac{1}{n}\) and Evaluate the Limit, We get

\(\displaystyle \int_{0}^{1}xdx = \frac{1}{2}\)

OR we can solve the above question using the formula \(\displaystyle 1+2+3+4+\ldots+n = \frac{n(n+1)}{2}\)

So \(\displaystyle \lim_{n\rightarrow \infty}\left(\frac{1}{n^2}+\frac{2}{n^2}+\ldots+\frac{n}{n^2}\right) = \lim_{n\rightarrow \infty}\frac{1+2+3+\ldots+n}{n^2}\)

So \(\displaystyle \lim_{n\rightarrow \infty}\frac{n(n+1)}{2n^2} = \frac{1}{2}\)
Tsakanikas Nickos
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Re: Four limits

#4

Post by Tsakanikas Nickos »

Let me answer (b).

Let \( \displaystyle a_{n} = \sqrt[n]{2 + \frac{1}{n}} \; , \; n \in \mathbb{N} \). It is obvious that
\[ \displaystyle \sqrt[n]{ \frac{1}{n} } \leq \sqrt[n]{ 2 + \frac{1}{n}} = a_{n} \; , \; \forall n \in \mathbb{N} \]
Hence, \[ \displaystyle \lim_{n} a_{n} \geq \lim_{n} \frac{1}{\sqrt[n]{n}} = 1 \] It is also obvious that \[ \displaystyle a_{n} = \sqrt[n]{2 + \frac{1}{n}} \leq \sqrt[n]{3} \; , \; \forall n \in \mathbb{N} \] Hence, \[ \displaystyle \lim_{n} a_{n} \leq \lim_{n} \sqrt[n]{3} = 1 \] It follows that \[ \displaystyle \lim_{n} a_{n} = \lim_{n} \sqrt[n]{2 + \frac{1}{n}} = 1 \]
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Grigorios Kostakos
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Re: Four limits

#5

Post by Grigorios Kostakos »

\((b)\) Or, for every \(n\in\mathbb{N}\): \(\sqrt[n]{2}<\sqrt[n]{2+\frac{1}{n}}\leq\sqrt[n]{3},\) and becauce \(\sqrt[n]{2}\to1,\,\sqrt[n]{3}\to1\,,\) we have that \(\sqrt[n]{2+\frac{1}{n}}\to1\,.\)
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