Corrected exercises: sequences, 201-SN3-RE (201-NYB)

Dominant term $n^{2}$ at the top and at the bottom: $u_{n}=\frac{\frac{5}{n^{2}}-2}{3+\frac{1}{n}}\to\frac{0-2}{3+0}=-\frac{2}{3}$. Pitfall: taking the terms in the order they are written and answering $\frac{5}{3}$; the dominant term is the one of highest degree, sign included.

Factor out each dominant term: $u_{n}=\frac{n^{3}\left(\frac{3}{n^{3}}-1\right)}{n^{2}\left(1+\frac{4}{n}\right)}=n\cdot\frac{\frac{3}{n^{3}}-1}{1+\frac{4}{n}}$. It is the product of $n\to+\infty$ by a factor that tends to $-1$: $u_{n}\to-\infty$. Pitfall: the sign, carried by the $-n^{3}$.

Form $\infty-\infty$: multiply by the conjugate. $u_{n}=\frac{(n^{2}+4n)-(n^{2}-2n)}{\sqrt{n^{2}+4n}+\sqrt{n^{2}-2n}}=\frac{6n}{n\left(\sqrt{1+\frac{4}{n}}+\sqrt{1-\frac{2}{n}}\right)}\to\frac{6}{1+1}=3$. Pitfall: concluding $\infty-\infty=0$.

Divide the top and the bottom by $5^{n}$, the largest base: $u_{n}=\frac{2\left(\frac{2}{5}\right)^{n}+1}{5-\left(\frac{3}{5}\right)^{n}}\to\frac{0+1}{5-0}=\frac{1}{5}$, since $r^{n}\to 0$ for $|r|<1$. Pitfall: $5^{n+1}=5\cdot 5^{n}$, so the $5$ stays in the denominator.

Form $\infty-\infty$: combine, $u_{n}=\ln\frac{3n^{2}+1}{(n+5)^{2}}$. The quotient tends to $3$ by the dominant term, so, by continuity of the logarithm, $u_{n}\to\ln 3\approx 1.099$. Pitfall: $2\ln(n+5)=\ln(n+5)^{2}$, and not $\ln\left(2(n+5)\right)$.

Squeeze: $-1\leq\sin n\leq 1$ gives $-\frac{1}{n}\leq\frac{\sin n}{n}\leq\frac{1}{n}$, so $\frac{\sin n}{n}\to 0$; likewise $\frac{\cos n}{n}\to 0$. Dividing by $n$: $u_{n}=\frac{2+\frac{\sin n}{n}}{1+\frac{\cos n}{n}}\to\frac{2+0}{1+0}=2$. Pitfall: applying L'Hôpital's rule to the associated function, which gives $\frac{2+\cos x}{1-\sin x}$, with no limit; the rule fails, yet the sequence converges.

$\frac{n}{n+1}\to 1$, so the even-indexed terms tend to $1$ and the odd-indexed ones to $-1$: two subsequences, two different limits, the sequence has no limit. Pitfall: confusing it with $\frac{(-1)^{n}}{n}\to 0$; the squeeze only concludes if the factor multiplying $(-1)^{n}$ tends to $0$, and here it tends to $1$.

$n$ is in the exponent: $\ln u_{n}=\frac{2}{n}\ln\left(2n^{3}\right)=\frac{2\ln 2+6\ln n}{n}\to 0$, since $\frac{\ln n}{n}\to 0$. So $u_{n}\to e^{0}=1$. Pitfall: the form is $\infty^{0}$, which is indeterminate; you cannot conclude “zero exponent, so $1$” without the logarithm.

Form $1^{\infty}$: $\frac{n+1}{n-2}=1+\frac{3}{n-2}$. Set $m=n-2$: $u_{n}=\left[\left(1+\frac{3}{m}\right)^{m}\right]^{\frac{n}{n-2}}$. The bracket tends to $e^{3}$ by the reference limit $\left(1+\frac{a}{m}\right)^{m}\to e^{a}$, and the exponent $\frac{n}{n-2}$ to $1$. So $u_{n}\to e^{3}\approx 20.086$. Pitfall: answering $1$, because the base tends to $1$.

Write $\frac{(n+1)!}{n^{n}}=(n+1)\cdot\frac{1}{n}\cdot\frac{2}{n}\cdot\frac{3}{n}\cdots\frac{n}{n}$: $n$ factors $\frac{k}{n}$, all between $0$ and $1$. Keep the first TWO and bound the others by $1$: for $n\geq 2$, $0<u_{n}\leq(n+1)\cdot\frac{1}{n}\cdot\frac{2}{n}=\frac{2(n+1)}{n^{2}}\to 0$. By the squeeze theorem, $u_{n}\to 0$. Pitfall: keeping only one factor gives $u_{n}\leq\frac{n+1}{n}$, which tends to $1$ and proves nothing.

For $(a_{n})$: $a_{1}=\frac{3}{4}=0.75$, $a_{2}=\frac{5}{5}=1$, $a_{3}=\frac{7}{6}\approx 1.167$, $a_{4}=\frac{9}{7}\approx 1.286$. The terms increase while getting closer to 2, which suggests $\lim a_{n}=2$ (confirmed in part b), since the ratio of the leading coefficients is $\frac{2}{1}$). For $(b_{n})$: $b_{1}=1$, $b_{2}=4$, $b_{3}=7$, $b_{4}=10$. We add 3 at each step: it is an arithmetic sequence with common difference 3, so $b_{n}=1+3(n-1)=3n-2$. Check: $b_{4}=12-2=10$ ✓. This sequence DIVERGES to $+\infty$, a reminder that a perfectly regular sequence is not necessarily convergent.

Definition: $\lim_{n\to\infty}a_{n}=L$ if for every $\varepsilon>0$, there exists an index $N$ such that for all $n>N$, $|a_{n}-L|<\varepsilon$. In other words, however small the margin $\varepsilon$ we demand, the sequence eventually enters that margin and never leaves it. Application with $\varepsilon=0.01$: $\left|\frac{2n+1}{n+3}-2\right|=\left|\frac{2n+1-2n-6}{n+3}\right|=\frac{5}{n+3}$. We want $\frac{5}{n+3}<0.01$, that is $n+3>500$, that is $n>497$. So we can take $N=497$: from index 498 on, every term is within $0.01$ of 2. Check: $a_{498}=\frac{997}{501}\approx 1.99002$, and the gap is $\frac{5}{501}\approx 0.00998<0.01$ ✓.

An EXPLICIT formula computes $a_{n}$ from $n$ alone: $a_{1000}$ is obtained by one substitution. A RECURRENCE relation computes $a_{n}$ from $a_{n-1}$: to get $b_{1000}$ you must, in principle, go through the 999 previous terms. Only the explicit formula therefore allows a direct calculation, and this is why we often try to convert a recurrence into an explicit expression, as we did with $b_{n}=3n-2$. It matters because the limit techniques of the chapter (dominant term, L'Hôpital's rule, conjugate) apply to an EXPRESSION in $n$: without an explicit form, they cannot be used, and you must then rely on the Monotone Convergence Theorem, which is what exercise 6 does.

The sequence never reaches 2: $\frac{2n+1}{n+3}=2$ would require $2n+1=2n+6$, so $1=6$, which is impossible for every $n$. “Tending to 2” does not mean reaching 2, but getting as close to it as we wish from some index on. The gap $\frac{5}{n+3}$ stays strictly positive for every $n$, while becoming smaller than any margin fixed in advance. The confusion is common because everyday language uses “tend to” and “reach” interchangeably; in analysis, that difference is precisely what the definition with $\varepsilon$ is about. Note all the same that a sequence CAN reach its limit, for example a constant sequence, or a sequence equal to 0 at every even index, but nothing forces it to.

Factor out the dominant power in the numerator and in the denominator, here $n^{2}$: $\frac{3n^{2}-5n+2}{7n^{2}+n}=\frac{n^{2}\left(3-\frac{5}{n}+\frac{2}{n^{2}}\right)}{n^{2}\left(7+\frac{1}{n}\right)}=\frac{3-\frac{5}{n}+\frac{2}{n^{2}}}{7+\frac{1}{n}}$. All the terms in $\frac{1}{n}$ tend to zero, so the limit is $\frac{3}{7}$. General rule for a quotient of polynomials: if the degrees are equal, the limit is the ratio of the leading coefficients; if the degree of the numerator is smaller, the limit is 0; if it is larger, the sequence diverges to infinity.

Form $\infty-\infty$, so use the conjugate: $\sqrt{n^{2}+n}-n=\frac{\left(\sqrt{n^{2}+n}-n\right)\left(\sqrt{n^{2}+n}+n\right)}{\sqrt{n^{2}+n}+n}=\frac{n^{2}+n-n^{2}}{\sqrt{n^{2}+n}+n}=\frac{n}{\sqrt{n^{2}+n}+n}$. Factor $n$ out of the denominator: $\frac{n}{n\left(\sqrt{1+\frac{1}{n}}+1\right)}=\frac{1}{\sqrt{1+\frac{1}{n}}+1}\to\frac{1}{2}$. The limit is $\frac{1}{2}$. The result deserves a comment: $\sqrt{n^{2}+n}$ and $n$ both become huge, but their gap settles at $\frac{1}{2}$, which shows that $\sqrt{n^{2}+n}$ behaves like $n+\frac{1}{2}$ for large $n$.

Consider the function $f(x)=\frac{\ln x}{x}$, which agrees with the sequence at the integers, that is $a_{n}=f(n)$. By L'Hôpital's rule (form $\frac{\infty}{\infty}$): $\lim_{x\to\infty}\frac{1/x}{1}=0$. The transfer theorem then guarantees that $\frac{\ln n}{n}\to 0$. The step is legitimate in THIS direction because the integers are among the reals: if all the real values eventually lie within the margin $\varepsilon$, the integer values do a fortiori. The converse is false because the sequence only sees isolated points and can completely miss the behaviour of the function between those points: $\sin(2\pi n)=0$ for every integer $n$, so the SEQUENCE converges to 0, while the FUNCTION $\sin(2\pi x)$ oscillates with no limit. Practical consequence: if the limit of the function does not exist, you have learned nothing about the sequence and you need another argument.

Both limits are $0$, by the SQUEEZE theorem. For the first: $-\frac{1}{n}\leq\frac{\cos n}{n}\leq\frac{1}{n}$ since $|\cos n|\leq 1$, and both bounds tend to 0, so the sequence does too. For the second: $\left|\frac{(-1)^{n}}{\sqrt{n}}\right|=\frac{1}{\sqrt{n}}\to 0$, and a sequence whose absolute value tends to 0 tends to 0. L'Hôpital's rule would be unusable for two combined reasons. First, these expressions are not indeterminate forms of type $\frac{0}{0}$ or $\frac{\infty}{\infty}$: the numerator is bounded, not infinite. Second, $(-1)^{n}$ is not even defined for an arbitrary real $x$, so there is no differentiable function to apply the rule to. The squeeze theorem is the natural tool whenever a factor oscillates while staying bounded, and that is the most frequent situation with sequences.

The four regimes. If $|r|<1$, for example $r=0.5$: $r^{n}\to 0$, the sequence converges, and it does so very fast. If $r=1$: the sequence is constant, equal to 1, so it converges to 1. If $r>1$, for example $r=2$: $r^{n}\to+\infty$, divergence to infinity. If $r\leq -1$, for example $r=-1$ (oscillation between $-1$ and 1) or $r=-2$ (oscillation of growing amplitude): the sequence diverges by oscillation, with no limit, neither finite nor infinite. The case $r=-1$ is the exact boundary: it is the only $r$ for which the sequence stays bounded while diverging. This classification will be reused in full in the chapter on series, where the geometric series $\sum r^{n}$ converges exactly when $|r|<1$.

The sequence equals $-1$ for odd $n$ and $1$ for even $n$: it is bounded, since $|a_{n}|=1$ for every $n$. Suppose, for a contradiction, that it converges to a limit $L$. Taking $\varepsilon=\frac{1}{2}$ in the definition, there should be an index $N$ beyond which ALL the terms are within $\frac{1}{2}$ of $L$. But beyond any index, we find both terms equal to 1 and terms equal to $-1$, which are a distance 2 apart: they cannot both lie in an interval of width 1. Contradiction, so the sequence diverges. The idea that fails is that the definition requires ALL the terms of large enough index to get close to the SAME value; here the sequence has two subsequential limits, 1 and $-1$, and favours neither. Being bounded prevents blowing up, but does not force the sequence to settle.

For $n=10$: $(0.9)^{10}\approx 0.349$ and $(1.1)^{10}\approx 2.594$, the gap is still modest. For $n=100$: $(0.9)^{100}\approx 2.66\times 10^{-5}$ and $(1.1)^{100}\approx 13{,}781$, a ratio of more than five hundred million. In the limit: the first tends to 0, the second to $+\infty$. The comparison shows that the regime switches at a THRESHOLD, the value $|r|=1$, and that two ratios only $0.2$ apart lead to opposite fates. That is the hallmark of exponential growth: the multiplying factor applies at every step, so a tiny gap per round becomes an enormous gap at the end. It is also why, in finance as in biology, the exact value of the rate matters more than the duration.

First half, every convergent sequence is bounded: if $a_{n}\to L$, then taking $\varepsilon=1$ there is an index $N$ beyond which all the terms lie in the interval $\left(L-1,L+1\right)$, so they are bounded; and the terms of index below $N$ are FINITELY many, so they are bounded too (a finite set of numbers always has a largest and a smallest element). Combining both, the whole sequence is bounded. Second half, the converse is false: $(-1)^{n}$ from part b) is bounded and diverges, which is enough to refute it. The right way to use this implication is its CONTRAPOSITIVE: if a sequence is not bounded, it does not converge. It is a quick divergence test, and often the cheapest one.

All the terms are strictly positive, so the ratio is legitimate: $\frac{a_{n+1}}{a_{n}}=\frac{\frac{n+1}{2^{n+1}}}{\frac{n}{2^{n}}}=\frac{n+1}{2^{n+1}}\cdot\frac{2^{n}}{n}=\frac{n+1}{2n}$. For $n=1$, this ratio equals $\frac{2}{2}=1$ EXACTLY: the first two terms are equal, $a_{1}=\frac{1}{2}$ and $a_{2}=\frac{2}{4}=\frac{1}{2}$. For $n\geq 2$, $\frac{n+1}{2n}<1$ is equivalent to $n+1<2n$, that is $n>1$, which holds: the ratio is then strictly less than 1 and the sequence is strictly decreasing. Correct wording: the sequence is decreasing (non-increasing) from index 1, and STRICTLY decreasing from index 2. Writing “the sequence is strictly decreasing” without qualification would be false at index 1, and it is exactly the kind of detail a grader penalizes. Successive terms: $0.5$; $0.5$; $0.375$; $0.25$; $0.15625$.

The associated function is $f(x)=\frac{2x-1}{3x+4}$, defined and differentiable for $x\geq 1$. Its derivative is $f'(x)=\frac{2(3x+4)-3(2x-1)}{(3x+4)^{2}}=\frac{6x+8-6x+3}{(3x+4)^{2}}=\frac{11}{(3x+4)^{2}}>0$: the function is strictly increasing on $[1,\infty)$, so the sequence $b_{n}=f(n)$ is too. Upper bound: $\frac{2n-1}{3n+4}<\frac{2}{3}$ is equivalent to $3(2n-1)<2(3n+4)$, that is $6n-3<6n+8$, that is $-3<8$, true for every $n$. The sequence is therefore increasing and bounded above by $\frac{2}{3}$: by the Monotone Convergence Theorem, it CONVERGES. Here the limit is computed directly by the dominant term: $\frac{2n-1}{3n+4}=\frac{2-\frac{1}{n}}{3+\frac{4}{n}}\to\frac{2}{3}$. First terms: $0.1429$; $0.3$; $0.3846$; $0.4375$; $0.4737$, the climb toward $\frac{2}{3}$ is slow but steady.

The counterexample works because $\sin(2\pi n)=0$ for every integer $n$: the SEQUENCE therefore reduces to $a_{n}=\frac{1}{n}$, which is strictly decreasing. The associated FUNCTION $f(x)=\sin(2\pi x)+\frac{1}{x}$, on the other hand, oscillates endlessly: between two consecutive integers it rises to about $+1$ and then falls back to about $-1$, and it is monotone on NO interval of length 1. The claim is therefore false. The deeper reason is that the sequence only sees ISOLATED points and ignores everything that happens between them, exactly as in the question on limits handled in exercise 2c. The true implication goes the other way: if $f$ is decreasing on $[1,\infty)$, then $a_{n}=f(n)$ is decreasing, since $n<n+1$ implies $f(n)\geq f(n+1)$. Practical consequence: the associated function is a tool to PROVE that a sequence is monotone, never to disprove it. If $f'$ changes sign, you have learned nothing and must go back to the difference or the ratio.

The difference $c_{n+1}-c_{n}=\frac{2^{n+1}}{(n+1)!}-\frac{2^{n}}{n!}$ requires a common denominator with factorials, which is feasible but tedious; above all, the associated function is simply unavailable, because $n!$ is not defined for an arbitrary real $x$. There is no elementary differentiable function extending the factorial that we could differentiate at the level of the course, so neither $f'$ nor L'Hôpital's rule can be used. The ratio, on the other hand, fits perfectly: $\frac{c_{n+1}}{c_{n}}=\frac{2^{n+1}}{(n+1)!}\cdot\frac{n!}{2^{n}}=\frac{2}{n+1}$. For $n=1$ the ratio equals 1 (so $c_{1}=c_{2}=2$), and for $n\geq 2$ it equals $\frac{2}{n+1}<1$: the sequence is strictly decreasing from index 2, exactly as in part a). It is also bounded below by 0, since all its terms are positive. By the Monotone Convergence Theorem, it CONVERGES. The limit is 0, which can be guessed by noticing that the ratio $\frac{2}{n+1}$ itself tends to 0: from some index on we multiply by a smaller and smaller factor, so the crushing speeds up. Check: $c_{7}\approx 0.0254$ and $c_{30}\approx 4\times 10^{-24}$. The factorial beats the exponential, which extends by one notch the hierarchy of the L'Hôpital chapter, and it is exactly the mechanism of the ratio test that will be seen in the chapter on series.

First way: $\frac{1}{n}\sum_{k=1}^{n}\frac{k^{2}}{n^{2}}=\frac{1}{n^{3}}\sum_{k=1}^{n}k^{2}=\frac{1}{n^{3}}\cdot\frac{n(n+1)(2n+1)}{6}=\frac{(n+1)(2n+1)}{6n^{2}}$. Expanding, $\frac{2n^{2}+3n+1}{6n^{2}}=\frac{2+\frac{3}{n}+\frac{1}{n^{2}}}{6}\to\frac{2}{6}=\frac{1}{3}$. Second way: the sum is exactly $\frac{1}{n}\sum_{k=1}^{n}f\left(\frac{k}{n}\right)$ with $f(x)=x^{2}$, that is the right Riemann sum of $f$ on $[0,1]$, with $\Delta x=\frac{1}{n}$ and the points $x_{k}=\frac{k}{n}$. Its limit is therefore $\int_{0}^{1}x^{2}dx=\frac{1}{3}$. The two results agree, which is reassuring but above all revealing: the formula for $\sum k^{2}$ is only available for a few very special sums, whereas reading the sum as a Riemann sum works for any integrable $f$. Numerical check: the exact value $\frac{(n+1)(2n+1)}{6n^{2}}$ is $0.385$ for $n=10$ and $0.3333383$ for $n=10^{5}$.

The reasoning is wrong because the NUMBER of terms grows while each term shrinks. Each term $\frac{1}{n+k}$ is of order $\frac{1}{n}$, but there are $n$ of them: the total stays of order $n\cdot\frac{1}{n}=1$, so nothing tends to 0. It is the same illusion as for the harmonic series: infinitely many tiny contributions can add up to a nonzero amount, and “taking the limit term by term” is only allowed if the number of terms is FIXED. Correct calculation: $\sum_{k=1}^{n}\frac{1}{n+k}=\sum_{k=1}^{n}\frac{1}{n\left(1+\frac{k}{n}\right)}=\frac{1}{n}\sum_{k=1}^{n}\frac{1}{1+\frac{k}{n}}$, which is the Riemann sum of $f(x)=\frac{1}{1+x}$ on $[0,1]$. The limit is therefore $\int_{0}^{1}\frac{dx}{1+x}=\left[\ln(1+x)\right]_{0}^{1}=\ln 2\approx 0.693$. Numerical check: the sum is $0.6688$ for $n=10$ and $0.69314$ for $n=10^{5}$, the convergence is slow and from below.

$I_{n}=\int_{0}^{1}x^{n}dx=\left[\frac{x^{n+1}}{n+1}\right]_{0}^{1}=\frac{1}{n+1}$, so $\lim_{n\to\infty}I_{n}=0$. Pointwise behaviour of $x^{n}$: for every fixed $x$ in $[0,1)$, $x^{n}\to 0$; but at $x=1$, $x^{n}=1$ for every $n$, so the pointwise limit is 1 at that precise point. Is there a contradiction? No, none. The integral measures an AREA, and an isolated point has no area: the fact that the limit function equals 1 at $x=1$ and 0 everywhere else does not change the fact that the area under $x^{n}$ tends to 0. Geometrically, the graph of $x^{n}$ flattens against the axis over almost the whole interval and only rises in a narrower and narrower zone near $x=1$: that zone does have height 1, but its width tends to 0, and it is the product of the two that decides. The lesson is that integration is not governed by the behaviour at an isolated point, which is precisely what makes it robust.

Let $u_{n}=\frac{\left(n!\right)^{1/n}}{n}$. Since all the terms are positive, take the logarithm: $\ln u_{n}=\frac{1}{n}\ln(n!)-\ln n=\frac{1}{n}\sum_{k=1}^{n}\ln k-\ln n$. Moving $\ln n$ inside the sum, which has exactly $n$ terms: $\ln u_{n}=\frac{1}{n}\sum_{k=1}^{n}\left(\ln k-\ln n\right)=\frac{1}{n}\sum_{k=1}^{n}\ln\left(\frac{k}{n}\right)$. We recognize the Riemann sum of $f(x)=\ln x$ on $[0,1]$, so $\ln u_{n}\to\int_{0}^{1}\ln x\,dx=-1$, and consequently $u_{n}\to e^{-1}=\frac{1}{e}\approx 0.3679$. A remark on rigour is in order: the function $\ln$ is not bounded near 0, so the integral is improper at that endpoint; it converges nonetheless, as the chapter on improper integrals shows, and the result still holds. Numerical check: $u_{10}\approx 0.4529$, $u_{100}\approx 0.3799$, $u_{2000}\approx 0.3687$, the convergence toward $0.3679$ being very slow. This result is a weak form of Stirling's formula, $n!\approx\left(\frac{n}{e}\right)^{n}\sqrt{2\pi n}$, from which it extracts exactly the dominant factor: the factorial behaves like $\left(\frac{n}{e}\right)^{n}$, which explains after the fact why it crushed the exponential in the previous exercise.