Question bank — Sequences — convergence, divergence, boundedness, monotonicity

True. If $a_n\to L$, take $\epsilon =1$: past some $N$ all terms sit in $(L-1,L+1)$, and the finitely many early terms have a max/min, so a global bound $K$ exists.

False. Boundedness stops terms from escaping to infinity but does not stop them wobbling — $a_n=(-1{)}^n$ is bounded by $1$ yet never settles. You need monotonicity too, or Bolzano–Weierstrass which only gives a convergent subsequence.

False. Monotone alone is not enough: $a_n=n$ increases forever and diverges to $\infty$. You must also add boundedness — that is exactly the Monotone Convergence Theorem.

False. This confuses the sequence of terms with the series of partial sums. $a_n=1/n\to 0$ yet $\sum 1/n$ diverges. Term $\to 0$ is necessary for series convergence, never sufficient.

False. Convergence is a statement about the tail ($n>N$ for arbitrarily large $N$). Altering any finite block of terms leaves the tail — hence the limit — untouched.

True. Since $||a_n|-0|=|a_n-0|$, the exact same $N$ works for both, so the $\epsilon$–$N$ condition transfers directly.

False. Only eventually: past some $N$ the terms lie within $\epsilon =L/2$ of $L$ so are positive, but early terms can be anything (e.g. $a_1=-100$).

True. In particular the whole sequence is a subsequence of itself. (The stronger fact: if $(a_n)$ diverges, some pair of subsequences head to different values.)

False. Limits are unique: if $a_n\to L$ and $a_n\to M$ with $L\ne M$, choose $\epsilon =|L-M|/2$ and the tail cannot sit inside two disjoint bands at once.

True in $\mathbb{R}$, precisely because of completeness. In $\mathbb{Q}$ it can fail — e.g. rational approximations to $\sqrt{2}$ are Cauchy but their limit is not rational.

==True only if $B\ne 0$==. The algebra-of-limits quotient rule needs a non-zero denominator limit; if $B=0$ the ratio may diverge or do anything.

The error is the second clause: bounded does not imply convergent. $(-1{)}^n$ is the standard counterexample — it oscillates between $1$ and $-1$ forever.

Convergence does not require the distance to shrink monotonically. Terms may overshoot back and forth (e.g. $a_n=(-1{)}^n/n\to 0$); all that's required is that eventually every term lies inside each band.

Wrong: apply the Squeeze Theorem. Because $-\frac{1}{n}\le \frac{\sin n}{n}\le \frac{1}{n}$ and both bounds $\to 0$, the sequence is squeezed to $0$ regardless of $\sin n$'s wobble.

Solving the fixed-point equation is only valid after you prove the sequence converges (via monotone + bounded here). The fixed-point step finds what the limit is, never that one exists.

"$\infty /\infty$" is not an answer — divide top and bottom by $n^2$ to get $\frac{3+1/n^2}{1-5/n^2}\to 3$. The dominant-power ratio is what governs the limit.

Divergence to infinity is a form of divergence, not convergence. Convergence requires a finite limit $L$; $\infty$ is not a real number and the $\epsilon$–$N$ definition cannot be satisfied.

Only if it is bounded above. An unbounded increasing sequence has $\sup =+\infty$ and diverges; the sup-is-the-limit fact is the bounded half of the Monotone Convergence Theorem.

Because "gets close" must hold for every tolerance, however tiny. Letting the challenger pick the hardest $\epsilon$ forces the terms to genuinely home in, not just get vaguely near.

The condition "$n>N\Rightarrow |a_n-L|<\epsilon$" only ever constrains terms beyond $N$, which can be pushed arbitrarily far out — so the first $N$ terms are never tested.

The theorem's proof needs the least upper bound $\sup \{a_n\}$ to exist for a bounded increasing sequence — and completeness is exactly the guarantee that every bounded-above set of reals has a supremum.

It's a cheap necessary filter (convergent $\Rightarrow$ bounded), so an unbounded sequence is instantly known to diverge; and combined with monotonicity it becomes sufficient via MCT.

Consecutive terms are $+1$ and $-1$, always a distance $2$ apart. Convergence would force the whole tail into a band of width $2\epsilon$; picking $\epsilon <1$ makes the band too narrow to hold both.

For products/factorials the terms multiply, so the ratio simplifies cleanly (to $(\frac{n}{n+1}{)}^n$) while the difference stays a messy subtraction. Comparing the ratio to $1$ reveals increase/decrease directly.

$L$ is an actual real number the terms cluster around; "$\to \infty$" is shorthand for terms eventually exceed every bound $M$ — there is no number they approach, so it is a description of unbounded growth.

If $a_n$ is trapped between two sequences both funneling to the same $L$, the outer bands close in and drag $a_n$ inside every $\epsilon$-band around $L$ too — it has no room to go anywhere else.

Yes, to $5$. For any $\epsilon$, $|5-5|=0<\epsilon$ holds for all $n$, so $N=1$ works. Constant sequences are the simplest convergent (and monotone-both-ways, bounded) case.

Only if it is constant. $a_{n+1}\ge a_n$ and $a_{n+1}\le a_n$ together force $a_{n+1}=a_n$ for all $n$, giving a constant sequence.

Yes, to $0$. Rationalise: $\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}\to 0$. A difference of two diverging pieces can still converge.

No. $(-1{)}^n$ takes just two values and diverges. It converges only if the values it repeats settle to a single one eventually.

It has no limit. $\cos (n\pi )=(-1{)}^n$, which oscillates between $1$ and $-1$ — the same divergence as the alternating sign sequence.

No. It may still converge outright (e.g. $\frac{(-1{)}^n}{n}\to 0$), and even if it diverges, Bolzano–Weierstrass guarantees a convergent subsequence.

Yes. $a_n=1-\frac{1}{n}$ increases to $1$ but no term equals $1$; the limit is the $\sup$, which may be un-attained.

No. Every term is strictly below $e$ (the sequence increases toward it); $e$ is the supremum approached but never hit.