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Limits of Sequences

As part of studying sequences and series in Calculus 2, we often have to find the limits of different sequences.  That is, we would like to know what happens to terms of the form \(a_{n}\) when \(n\) goes to \(\infty\).  In trying to find these limits, we will most commonly use two techniques: compare a sequence to a function which agrees at the natural numbers, or compare dominating sequences.  Here, we will look at how these work to give us limits of most the sequences we will need to find.

Functions limits imply sequence limits

Suppose that you are given a sequence, say \(a_{k}=\sqrt[k]{k}\).  In this case, we can find a function \(f(x)=\sqrt[x]{x}\) such that \(f(x)\) is continuous for \(x \geq 1\) and \(f(k)=a_{k}\) for all natural numbers \(k\).  What we would like to know is, do these converge to the same point?

In order to answer this we will begin by looking at \(\lim\limits_{x \to \infty}\sqrt[x]{x}\).  We find this by rewriting it and finding,

\begin{align*}
\lim _{x \to \infty} \sqrt[x]{x}&=\lim_{x \to \infty}e^{\ln(x^{\frac{1}{x}})}.
\end{align*}

Here we have rewritten the problem noting that \(e^{\ln(y)}=y\) for all positive \(y\) and \(\sqrt[x]{x}=(x)^{\frac{1}{x}}\). Noting that \(e^{y}\) is continuous, we can find that
\begin{align*}
\lim _{x \to \infty} \sqrt[x]{x}&=\lim_{x \to \infty}e^{\ln(x^{\frac{1}{x}})} \\
&=e^{\lim\limits_{x \to \infty} \ln(x^{\frac{1}{x}})} \\
&=e^{\lim\limits_{x \to \infty}\frac{1}{x}\ln(x)}
\end{align*}

Since \(\lim_{x \to \infty}(\ln(x))=\infty\) and \(\lim_{x \to \infty}x=\infty\), we can use L’Hopital’s rule to find that
\begin{align*}
\lim _{x \to \infty} \sqrt[x]{x}&=\lim_{x \to \infty}e^{\ln(x^{\frac{1}{x}})} \\
&=e^{\lim_{x \to \infty} \ln(x^{\frac{1}{x}}) } \\
&=e^{\lim_{x \to \infty}\frac{1}{x}\ln(x)} \\
&=e^{\lim_{x \to \infty}\frac{(\frac{1}{x})}{1}} \\
&=e^{0}=1.
\end{align*}

This then means that all of them terms of \(\sqrt[x]{x}\) tend toward 1 as \(x\) tends toward \(\infty\). In particular, since all of the terms tend to \(1\), all of the function values at integer values also tend to \(1\). We, therefore, get that \(\lim_{k \to \infty}\sqrt[k]{k}=1\) as well. In general if the \(\lim_{x \to \infty}f(x)=L\) and \(f(x)=a_{k}\) for all \(k \in \mathbb{N}\), then \(\lim_{k \to \infty}a_{k}=L\).

Using this, we realize that if we are presented with a sequence, we can look at function limits instead of sequence limits as long as we can find a function that agrees at the integer values. This is extremely helpful in the fact that we can revert back to using L’Hopital’s rule in order to find this limits with the techniques we used in Calculus 1.

Dominating terms

Another techniques we will use to calculate limits is to compare the functions within the sequence. Certain functions will grow fast enough that other function become negligible in comparison when looking at large values of \(k\), that is it will dominate the second function. If we can determine which functions dominate which other functions, we can quickly find the limits by ignoring all but the dominating functions. Therefore, we will make a list of dominating functions.

To begin with, we know that \(a_{k}\) dominates \(b_{k}\), \(a_{k} >> b_{k}\), if \(\lim_{k \to \infty}\frac{b_{k}}{a_{k}}=0\). We can now compare functions.

\(k^{p} >> \ln(k)\).

Here we let \(p > 0\). Since \(x^{p}\) and \(\ln(x)\) are defined for \(x \geq 1\), we can use related functions to find
\begin{align*}
\lim_{x \to \infty}\frac{\ln(x)}{x^{p}}&=\lim_{x \to \infty}\frac{(\frac{1}{x})}{px^{p-1}} \\
&=\lim_{x \to \infty}\frac{1}{px^{p}} \\
&=0.
\end{align*}

Therefore, as \(k\) gets large, \(\ln(k)\) will become negligible compared to \(k^{p}\).

\(k^{m} >> k^{n}\)

Here, we let \(m > n \). First note that \(m-n >0\). We then see that \(x^{m}\) and \(x^{n}\) are continuous for \(x \geq 1\). We can therefore find that
\begin{align*}
\lim_{x \to \infty}\frac{x^{n}}{x^{m}}&=\lim_{x \to \infty}\frac{1}{x^{m-n}} \\
&=0.
\end{align*}

Therefore, the sequence \(k^{m} >> k^{n}\) whenever \(m > n\).

\(a^{k} >> k^{m}\)

We now let \(a > 1\). In this case, we know that \(a^{x}\) and \(x^{m}\) are continuous for \(x \geq 1\). We then have that, through multiple uses of L’Hopital’s rule, we find that
\begin{align*}
\lim_{x \to \infty}\frac{x^{m}}{a^{x}}&=\lim_{x \to \infty}\frac{m*(m-1)* \cdots (m-\lceil m \rceil)*x^{m-\lceil m \rceil}}{(\ln(a))^{\lceil m \rceil}a^{x}} \\
&=0.
\end{align*}

Note that we need to round up to the nearest integer in case \(m\) is not an integer. If \(m\) is an integer we end with \(\frac{k}{a^{x}}\) for some constant \(k\). Otherwise we end up \(\frac{k}{x^{p}a^{x}}\) for some constant \(k\) and some \(p >0\). In either case, the limit is still 0, so a power function will become negligible compared to an exponential.

\(b^{k} >> a^{k}\)

In this case, we assume that \(b > a\). We then have that
\begin{align*}
\lim_{k \to \infty}\frac{a^{k}}{b^{k}}&=\lim_{k \to \infty} \left(\frac{a}{b}\right)^{k} \\
&=0.
\end{align*}
since \(\frac{a}{b} <1 \).

\(k! >> b^{k}\)

In this case, we can’t compare these sequences to functions, because \(x!\) is not defined for numbers other than non-negative integers. In this case, we will instead look at the terms to determine what they will look like. Note that \(b^{k}=b*b*b \cdots b\) \(k\) times and \(k!=k(k-1) \cdots 2*1\). Therefore, if we will get
\begin{align*}
\frac{b^{k}}{k!}=\frac{b}{k}\frac{b}{k-1}\cdots \frac{b}{2}\frac{b}{1}.
\end{align*}

We can now see that, for small \(k\), we will get \(\frac{b}{k}\) is a finite number. However, as \(k\) gets large, this number will become arbitrarily small. Therefore, we will end up with terms that look like numbers that tend to 0 multiplied by constant numbers. Hence,
\begin{align*}
\lim_{k \to \infty}\frac{b^{k}}{k!}=0,
\end{align*}
so \(k!\) dominates \(b^{k}\) for any \(b\).

\(k^{k} >> k!\)

Again, like we had last time, we cannot compare these to continuous functions. Therefore, we will have to look at these terms in order to determine what is happening. We then get that
\begin{align*}
\frac{k!}{k^{k}}=\frac{k}{k}\frac{k-1}{k}\cdots \frac{2}{k}\frac{1}{k}.
\end{align*}
In this case, we note that each of the terms on top will be smaller than \(k\) by the definition of \(k!\). Therefore,
\begin{align*}
\frac{k!}{k^{k}}&=\frac{k}{k}\frac{k-1}{k}\cdots \frac{2}{k}\frac{1}{k} \\
&<1*1*1 \cdots *1*\frac{1}{k}. \end{align*} Therefore, as \(k\) gets large, these terms will become arbitrarily small. We then get that \begin{align*} \lim_{k \to \infty}\frac{k!}{k^{k}}=0, \end{align*} so \(k^{k} >> k!\).

Calculating limits

Now that we have an ordering of dominating sequences, we can use these to quickly calculate limits of sequences. For an example, let’s suppose that we want to find the limit of the sequence
\begin{align*}
\left\{a_{k}\right\}=\left\{\frac{k^{3}-4k+\ln(k)-(2k)!}{k-5k^{54}+k^{k}}\right\}.
\end{align*}
We can then see that the dominating term on top is the \(-(2k)!\) and the dominating term on bottom is \(k^{k}\). Furthermore, since \(k^{k}\) dominates \((2k)!\), we find that
\begin{align*}
\lim_{k \to \infty}\frac{k^{3}-4k+\ln(k)-(2k)!}{k-5k^{54}+k^{k}}&=\lim_{k \to \infty}\frac{-(2k)!}{k^{k}} \\
&=0.
\end{align*}

Conclusion

After going through the initial work of showing that if the function has a limit, the corresponding sequence will have a limit, and finding an ordering of dominating functions, we were able to find limits of sequences fairly quickly. Since this will be the first step in determining whether any series converges or diverges, we will have to find a lot of limits of sequences in Calculus 2, or any time we are dealing with series. Therefore, the initial set up is worth the time in the long run.

I hope you learned something and enjoyed the post along the way. If you did, make sure to like the post and share it with anyone else that may need some help with sequences.  Also, if you need more help with Calculus, check out the study help related posts.

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