Practice: Sequences and Their Limits

(a) Three limits, three methods. $a_n=\frac{3n+1}{n+2}$: divide by the dominant power. Dividing numerator and denominator by $n$ gives $\frac{3+1/n}{1+2/n}$, and since $1/n\to 0$ the limit laws give $\frac{3+0}{1+0}=3$. The laws apply directly because both parts converge after the division; before it, both numerator and denominator diverge and the quotient law would not apply. Confirmation: $a_{100}=2.9509803922$, $a_{1000}=2.9950099800$, $a_{10000}=2.9995001000$, errors $4.9\times {10}^{-2}$, $5.0\times {10}^{-3}$, $5.0\times {10}^{-4}$ ✓. $b_n=\frac{n}{2^n}$: growth-rate comparison. Dividing by a power of $n$ does not help, since the denominator is not polynomial. The ratio of consecutive terms is

so eventually each term is close to half its predecessor and the sequence shrinks geometrically to $0$. Confirmation: $0.5$, $0.5$, $0.375$, $0.15625$, $0.009765625$, $1.907\times {10}^{-5}$, $2.794\times {10}^{-8}$, $4.441\times {10}^{-14}$ at $n=1,2,3,5,10,20,30,50$ ✓. Note $b_1=b_2$, so the sequence is not strictly decreasing at the start, which is irrelevant since convergence depends only on the tail. $c_n=\frac{\sin n}{n}$: squeeze. Neither method above applies: $\sin n$ has no limit, so no dominant term can be extracted and no ratio argument works. But $|\sin n|\le 1$ gives

and both bounds tend to $0$, so $c_n\to 0$ by the squeeze theorem. Confirmation: $+0.8414709848$, $-0.0544021111$, $-0.0050636564$, $+0.0008268795$ at $n=1,10,100,1000$, within $\pm 1/n$ throughout and changing sign repeatedly while converging. This refutes the idea that a convergent sequence must approach its limit from one side. (b) An explicit $N$. Simplify the distance first:

Then for any :

So $N=\lceil 5/\epsilon -2\rceil$ serves in general. For $\epsilon =0.001$: , so $\boxed{{\displaystyle N=4998}}$. Check: $a_{4999}=2.99900020$, error . At $n=4998$ the error is exactly $0.001$, not strictly less, so $N=4997$ would fail; the inequality is strict. The dependence. $N$ grows as $\epsilon$ shrinks: $\epsilon =0.1$ gives $N=48$, $\epsilon =0.01$ gives $N=498$, $\epsilon =0.001$ gives $N=4998$, roughly tenfold per decimal place since $N\approx 5/\epsilon$. This dependence is why the quantifiers cannot be exchanged: a single $N$ working for every $\epsilon$ would force $a_n=3$ exactly. (c) A recursive sequence. Bounded above by 2, by induction. Base: ✓. Step: if then ✓. Increasing, by induction. Base: ✓. Step: if then , and since $\sqrt{}$ is increasing, ✓. Existence. The sequence is increasing and bounded above, so by the monotone convergence theorem a limit $L$ exists. This is the step that establishes existence, and everything after it depends on it. Value. Now that both sides have limits, let $n\to \mathrm{\infty }$ in $a_{n+1}=\sqrt{2+a_n}$. Since $(a_{n+1})$ is the same sequence shifted, both sides tend to $L$:

Rejecting a root. $L=-1$ is impossible: and square roots are non-negative, so every term is positive. Hence $L=2$. Check: $\sqrt{2+2}=2$ ✓, and the computed terms $1$, $1.732051$, $1.931852$, $1.982890$, $1.995718$, $1.998929$, $1.999732$, $1.999933$, $1.999983$, $1.999996$, $1.999999$ climb toward 2, with $a_{11}$ within $1.05\times {10}^{-6}$ ✓. Why the order matters. Solving the fixed-point equation first proves nothing. For $a_1=1$, $a_{n+1}=2a_n$ the same manipulation gives $L=2L$, hence $L=0$, but the sequence is $1,2,4,8,\dots$, diverging to infinity. The equation says what the limit must be if one exists; existence is a separate argument. (d) Divergence of $(-1{)}^n$. Suppose $a_n\to L$. Apply the definition with $\epsilon =1$: there is $N$ with for all . Beyond any $N$ the sequence takes both values $1$ and $-1$, so both

would hold. The first gives and the second . No $L$ satisfies both, so no limit exists. Alternatively: the even-indexed subsequence is constantly $1$ and the odd-indexed constantly $-1$. A convergent sequence has every subsequence converging to the same value, so two subsequential limits refute convergence at once. What is wrong with the intuition. "A bounded sequence must settle somewhere" assumes the only alternative to converging is escaping to infinity. There is a third: oscillating forever within a bounded region. Boundedness forbids escape and says nothing about oscillation. The correct statements are that convergence implies boundedness, so an unbounded sequence can be dismissed immediately, and that the converse fails, which is why the monotone convergence theorem requires monotonicity as well: monotonicity removes the freedom to move back and forth. (e) Terms versus partial sums. The terms converge. Given , take ; then gives . So $1/n\to 0$. The partial sums diverge. Group the terms in blocks of doubling length:

The $k$th block has $2^{k-1}$ terms, each at least $2^{-k}$, so it contributes at least $2^{k-1}\cdot 2^{-k}=\frac{1}{2}$. Hence

and after $k$ doublings $H_{2^k}\ge 1+k/2$, which exceeds every bound. So $(H_n)$ is unbounded, and an unbounded sequence cannot converge. Verified: $H_{2n}-H_n$ equals $0.500000$, $0.583333$, $0.634524$, $0.662872$, $0.677766$ at $n=1,2,4,8,16$, all at least $\frac{1}{2}$. The sums themselves: $H_{10}=2.928968$, $H_{100}=5.187378$, $H_{1000}=7.485471$, $H_{{10}^4}=9.787606$, $H_{{10}^5}=12.090146$. Why no contradiction. These are two different sequences. The terms $(1/n)$ are decreasing and bounded, so they converge; the partial sums $(H_n)$ are increasing and unbounded, so they do not. Both are monotone, and boundedness is what separates them: the monotone convergence theorem applies to the first and not the second. That the increments shrink to zero does not stop the total growing: infinitely many contributions, each small, can still sum without limit, because there is no bound on how many of them there are. The numbers conceal this well. $H_n$ tracks $\ln n+\gamma$ with $\gamma =0.5772156649$ ($5.187378$ against $5.182386$ at $n=100$; $12.090146$ against $12.090141$ at $n={10}^5$), and $\ln n\to \mathrm{\infty }$ so slowly that reaching needs about ${10}^{43}$ terms. No computation could reveal this divergence, which is why the grouping argument is needed. What it says about series generally. A series is defined as the limit of its partial sums, so every claim about series convergence is a claim about a sequence. "The terms tend to zero" is necessary for convergence, since otherwise the partial sums could not settle, but it is nowhere near sufficient, as the harmonic series shows. That asymmetry is exactly the $n$th-term test: it can prove divergence and never convergence.