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Prove $a_n=\sqrt[2n+1]{n^2+n}$ tends to $1$ as $n$ tends to infinity.

In the textbook, a hint was given: Let $a_n=1+h$, then $n^2+n=(1+h)^{2n+1}\gt \binom{2n+1}{3}(h^3)$

Then the consecutive steps are some algebraic manipulation, I managed to prove that $$\sqrt[3]{\frac{1}{n+1}+\frac{1}{n^2}}\gt \sqrt[2n+1]{n^2+n}-1\gt0$$ So $a_n$ tends to $1$ as $n$ gets sufficiently large.

But I don't understand why $a_n$ was set as $1+h$ at the first place, in retrospective, this does make calculations a lot convenient. And how did the author know when to use the binomial coefficient and compare the fourth term, it seems the author just plucked it out of thin air.

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  • $\begingroup$ Because if $a_n$ tends to $1$ then $h$ tends to $0$, and that's nice. $\endgroup$ – Vincenzo Oliva Aug 9 '15 at 10:28
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Well here is an intuitive (but non-rigorous) argument $$(n^{2} + n)^{1/(2n + 1)} \approx (n^{2})^{1/(2n + 1)} = n^{2/(2n + 1)} \approx n^{1/n} \to 1$$ where the last result $n^{1/n} \to 1$ is pretty standard. The idea to express $a_{n} = 1 + h$ and then show $h \to 0$ is by R. Courant (if I recall correctly) and he used it to show that $n^{1/n} \to 1$ as $n \to \infty$. He put $n^{1/n} = 1 + h$ so that $$n = (1 + h)^{n} > \frac{n(n - 1)}{2}h^{2}$$ so that $0 < h^{2} < \frac{2}{n - 1} \to 0$. Thus $h \to 0$ and $n^{1/n} = 1 + h \to 1$.

Note that using $n^{1/n} \to 1$ we can analyze the behavior of sequence $a_{n} = \{P(n)\}^{1/Q(n)}$ where $P(n), Q(n)$ are polynomials and this sequence also has limit $1$ (only restriction is that coefficient of highest power of $n$ in $P(n)$ must be positive otherwise the sequence is not defined for all $n$ and $Q(n)$ is of positive degree). The fact follows more easily if we use logs. Clearly $$\log a_{n} = \frac{\log P(n)}{Q(n)}$$ and if $P(n) = a_{0}n^{k} + a_{1}n^{k - 1} + \cdots$ then we can use $$\frac{a_{0}}{2}n^{k} < P(n) < 2a_{0}n^{k}$$ for all $n$ after a certain point and then on taking logs we get $$\frac{\log(a_{0}/2)}{Q(n)} + k\cdot\frac{\log n}{Q(n)} < \frac{\log(P(n))}{Q(n)} < \frac{\log(2a_{0})}{Q(n)} + k\cdot\frac{\log n}{Q(n)}\tag{1}$$ Since $n^{1/n}$ tends to $1$ it follows that $(\log n)/n \to 0$ and therefore $(\log n)/Q(n) \to 0$. Applying squeeze theorem on $(1)$ we get $(\log P(n))/Q(n) \to 0$ and hence $a_{n} \to 1$. Just for completeness note that if coefficient of highest power of $n$ in $Q(n)$ is negative then the inequalities in equation $(1)$ are reversed but result remains the same.

In your case $P(n) = n^{2} + n, Q(n) = 2n + 1$. The technique of Courant can also be used on $a_{n} = \{P(n)\}^{1/Q(n)}$ directly if $Q(n)$ is a positive integer for all large $n$. As in case of $n^{1/n}$ we put $a_{n} = 1 + h$ so that $$P(n) = (1 + h)^{Q(n)} = (1 + h)^{q} > \binom{q}{r}h^{r}$$ We just need to choose $r$ such that degree of $$\binom{q}{r} = \binom{Q(n)}{r} = F(n)$$ as a polynomial in $n$ is higher than that of $P(n)$ and then we get $0 < h^{r} < P(n)/F(n)$ where $F$ has degree greater than $P$ and hence $h^{r}$ tends to $0$ and so $h \to 0$.

If $Q(n) < 0$ for all large $n$ then we use $b_{n} = 1/a_{n} = \{P(n)\}^{1/R(n)}$ where $R(n) = -Q(n)$ so that $R(n) > 0$ for all large $n$ and $b_{n} \to 1$ so that $a_{n} \to 1$.

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  • $\begingroup$ Does your answer imply that, in the long run, for any arbitrary polynomial $P$ and $Q$, $P(n)$ will always be outrun by $Q(n)$th root, because the power function in $P(n)$ does not increase quick enough to be divided by $Q(n)$ into equal parts whose product will equal $P(n)$? $\endgroup$ – Dave Clifford Aug 9 '15 at 13:55
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    $\begingroup$ @DaveClifford: Yes. This is clearly seen in the log equation. $\log P(n) \to \infty$ much slower than $Q(n)$ and thus $Q(n)$ th root dominates $P(n)$. We have the following theorem $(\log x)^{b}/x^{a} \to 0$ when $x \to \infty$ where $a $ is any positive number however small it might be and $b$ is any positive number however large it may be. $\endgroup$ – Paramanand Singh Aug 9 '15 at 14:06
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    $\begingroup$ @DaveClifford: You may consider this simple argument: $$\sqrt[n]{n^{100}} = \sqrt[n]{n}\cdot\sqrt[n]{n}\cdot\cdots \text{ 100 times } \to 1\cdot 1\cdots 1 = 1$$ so $n$th root dominates the 100th power of $n$. $\endgroup$ – Paramanand Singh Aug 9 '15 at 14:22
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One method of solving this is to use L'Hôpital's rule.

$A_n = \lim_{n\to \infty} {(n^2 + n)}^{1\over(2n+1)}$

$ln(A_n) = \lim_{n\to \infty} {ln ((n^2 + n)^{1\over(2n+1)}})$

$ln(A_n) = \lim_{n\to \infty}{ln(n^2 + n)\over (2n+1)}$

$ln(A_n) = \lim_{n\to \infty}{{(2n+1) \over (n^2 + n)} \over 2}$ L'Hôpital's Rule $ {"\infty" \over "\infty"}$

$ln(A_n) = \lim_{n\to \infty}{(2n+1) \over 2(n^2 + n)}$

$ln(A_n) = \lim_{n\to \infty}{2 \over 4n+2}$ L'Hôpital's Rule $ {"\infty" \over "\infty"}$

$ln(A_n) = 0$

Therefore, $A_n = 1$

Apologies for the formatting, hopefully someone will be able to clean this up a bit to look better, not sure on how to do it myself (new at this).

Cheers, GaramMasala

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    $\begingroup$ put latex code within dollar signs $ $\endgroup$ – sav Aug 9 '15 at 10:35
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Easier: consider $e^{\log a_n} = e^{\frac{\log (n^2+n)}{2n +1}}$. Remember $exp(\cdot)$ is a continuous function. All you need to do is some algebra.

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Notice, we have $$a_n=\lim_{n\to \infty}\sqrt[2n+1]{n^2+n}=\lim_{n\to \infty}\left(n^2+n\right)^{\frac{1}{2n+1}}$$ Let, $n=\frac{1}{t}\implies t\to 0 \ as \ n\to \infty$, we get $$\lim_{t\to 0}\left(\frac{1}{t^2}+\frac{1}{t}\right)^{\frac{1}{\frac{2}{t}+1}}$$ $$=\lim_{t\to 0}\left(\frac{1+t}{t^2}\right)^{\frac{t}{2+t}}$$ $$=\lim_{t\to 0}\left(1+t\right)^{\frac{t}{2+t}}\times \lim_{t\to 0}\left(\frac{1}{t}\right)^{\frac{2t}{2+t}}$$ $$=1\times 1=1$$

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