Solved Exercises — Recursively Defined Sequences
Theoretical Recall
A sequence {aₙ} is a function:
\[a:\mathbb{N}\to\mathbb{R}.\]The only accumulation point of ℕ in the extended real line ℝ ∪ {±∞} is +∞, so limits of sequences are always taken as n → +∞.
Key Results
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If {aₙ} converges, then it is bounded. The converse does not hold.
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Monotone Convergence Theorem. A monotone and bounded sequence converges.
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Recursive sequences. For k ∈ ℕ and a function:
a recursive sequence can be defined by:
\[\begin{cases} a_{n+1}=f(n,a_n,\dots,a_{n-k}), & n\ge k,\\ a_0,\dots,a_k\in\mathbb{R}, & \text{initial data}. \end{cases}\]Unlike explicit formulas, here each term depends on the previous ones; behavior depends critically on the initial values.
Worked Exercises
Below are ten selected recursive sequences. Each solution shows:
- study of monotonicity and boundedness,
- induction arguments when needed,
- passage to the limit via the fixed-point equation,
- final result isolated.
Exercise 1
\[\begin{cases} a_1=-\frac{3}{2},\\ a_{n+1}=a_n^{2}+4a_n+2. \end{cases}\]Solution.
We first show that the interval [−2, −1] is invariant.
If:
\[-2\le a_n\le-1,\]then:
\[a_{n+1}+2 = a_n^2+4a_n+4 = (a_n+2)^2 \ge0.\]Therefore:
\[a_{n+1}\ge-2.\]Moreover:
\[a_{n+1}+1 = a_n^2+4a_n+3 = (a_n+1)(a_n+3).\]For −2 ≤ aₙ ≤ −1:
\[a_n+1\le0,\]while:
\[a_n+3>0.\]Hence:
\[a_{n+1}+1\le0,\]so:
\[a_{n+1}\le-1.\]Since a₁ = −3/2 belongs to [−2, −1], induction gives:
\[-2\le a_n\le-1\]for every n.
Now:
\[a_{n+1}-a_n = a_n^2+3a_n+2.\]Factorizing:
\[a_{n+1}-a_n = (a_n+1)(a_n+2).\]On [−2, −1]:
\[(a_n+1)(a_n+2)\le0.\]Therefore:
\[a_{n+1}\le a_n.\]The sequence is decreasing and bounded below by −2, so it converges.
Let:
\[a_n\to\ell.\]Passing to the limit:
\[\ell=\ell^2+4\ell+2.\]Therefore:
\[\ell^2+3\ell+2=0.\]Thus:
\[(\ell+1)(\ell+2)=0.\]Hence:
\[\ell\in\{-2,-1\}.\]Since the sequence is decreasing and starts from:
\[a_1=-\frac32,\]it cannot converge to −1.
Final Result
\[\lim_{n\to\infty}a_n=-2\]Exercise 2
\[\begin{cases} a_1=2,\\ a_{n+1}=2\sqrt{a_n}. \end{cases}\]Solution.
We show that:
\[0\le a_n\le4.\]The initial value satisfies:
\[0\le a_1=2\le4.\]If 0 ≤ aₙ ≤ 4, then:
\[0\le2\sqrt{a_n}\le4.\]Hence:
\[0\le a_{n+1}\le4.\]Therefore [0,4] is invariant.
For 0 ≤ aₙ ≤ 4:
\[2\sqrt{a_n}\ge a_n.\]Thus:
\[a_{n+1}\ge a_n.\]The sequence is increasing and bounded above by 4, so it converges.
Let:
\[a_n\to\ell.\]Then:
\[\ell=2\sqrt{\ell}.\]Squaring:
\[\ell^2=4\ell.\]Therefore:
\[\ell(\ell-4)=0.\]Since aₙ ≥ 2:
\[\ell=4.\]Final Result
\[\lim_{n\to\infty}a_n=4\]Exercise 3
\[\begin{cases} a_1=5,\\ a_{n+1}=\dfrac{a_n}{\frac12+a_n}. \end{cases}\]Solution.
For aₙ > 0:
\[\frac{a_n}{\frac12+a_n}\le a_n\]is equivalent to:
\[\frac{1}{\frac12+a_n}\le1.\]Hence:
\[a_n\ge\frac12.\]We now show that [1/2, +∞) is invariant.
If:
\[a_n\ge\frac12,\]then:
\[\frac{a_n}{\frac12+a_n}\ge\frac12.\]Indeed, this is equivalent to:
\[2a_n\ge\frac12+a_n,\]that is:
\[a_n\ge\frac12.\]Thus:
\[a_{n+1}\ge\frac12.\]Since a₁ = 5, induction gives:
\[a_n\ge\frac12.\]Therefore:
\[a_{n+1}\le a_n.\]The sequence is decreasing and bounded below, so it converges.
Let:
\[a_n\to\ell.\]Then:
\[\ell = \frac{\ell}{\frac12+\ell}.\]Multiplying:
\[\ell\left(\frac12+\ell\right)=\ell.\]Hence:
\[\ell\left(\ell-\frac12\right)=0.\]Since:
\[\ell\ge\frac12,\]we obtain:
\[\ell=\frac12.\]Final Result
\[\lim_{n\to\infty}a_n=\frac12\]Exercise 4
\[\begin{cases} a_1=\frac12,\\ a_{n+1}=a_n^3. \end{cases}\]Solution.
By induction:
\[0\le a_n\le1.\]For every aₙ in [0,1]:
\[a_n^3\le a_n.\]Therefore:
\[a_{n+1}\le a_n.\]The sequence is decreasing and bounded below by zero, so it converges.
Let:
\[a_n\to\ell.\]Passing to the limit:
\[\ell=\ell^3.\]Thus:
\[\ell(\ell-1)(\ell+1)=0.\]Hence:
\[\ell\in\{-1,0,1\}.\]Since:
\[0\le\ell\le\frac12,\]only zero is possible.
Final Result
\[\lim_{n\to\infty}a_n=0\]Exercise 5 — Parameter a > 0
\[\begin{cases} a_1=a,\\ a_{n+1}=\frac12(a+a_n^2). \end{cases}\]Solution.
A finite limit ℓ must satisfy:
\[\ell=\frac12(a+\ell^2).\]Therefore:
\[\ell^2-2\ell+a=0.\]For 0 < a < 1, the two fixed points are:
\[\ell_-=1-\sqrt{1-a},\]and:
\[\ell_+=1+\sqrt{1-a}.\]We compare the initial value a with the smaller fixed point.
Since:
\[a = (1-\sqrt{1-a})(1+\sqrt{1-a}),\]and:
\[1+\sqrt{1-a}>1,\]we obtain:
\[a>\ell_-.\]Moreover:
\[a<1<\ell_+.\]Thus:
\[\ell_-<a_1<\ell_+.\]Now:
\[a_{n+1}-a_n = \frac12(a+a_n^2-2a_n).\]Factorizing:
\[a_{n+1}-a_n = \frac12(a_n-\ell_-)(a_n-\ell_+).\]For:
\[\ell_-<a_n<\ell_+,\]we therefore have:
\[a_{n+1}-a_n<0.\]Thus the sequence is decreasing.
We also show that it remains above ℓ₋.
Since:
\[a=2\ell_- -\ell_-^2,\]we obtain:
\[a_{n+1}-\ell_- = \frac12(a_n^2-\ell_-^2).\]Hence:
\[a_{n+1}-\ell_- = \frac12(a_n-\ell_-)(a_n+\ell_-).\]If aₙ ≥ ℓ₋, then:
\[a_{n+1}\ge\ell_-.\]The sequence is therefore decreasing and bounded below by ℓ₋.
Hence:
\[\ell=1-\sqrt{1-a}.\]If a = 1, then:
\[a_1=1,\]and the sequence is constant:
\[a_n=1.\]Now suppose a > 1.
Then:
\[a_{n+1}-a_n = \frac12\left[(a_n-1)^2+a-1\right].\]Since a > 1:
\[a_{n+1}-a_n>0.\]Thus the sequence is strictly increasing.
If it had a finite limit, that limit would satisfy:
\[\ell^2-2\ell+a=0.\]But the discriminant is:
\[4-4a<0.\]There is no real fixed point. Therefore the increasing sequence cannot have a finite limit and is unbounded above.
Final Result
\[\lim_{n\to\infty}a_n= \begin{cases} 1-\sqrt{1-a}, & 0<a<1,\\ 1, & a=1,\\ +\infty, & a>1. \end{cases}\]Exercise 6
\[\begin{cases} a_1=\frac32,\\ a_{n+1}=\frac{a_n}{2}+\frac{1}{a_n}. \end{cases}\]Solution.
For aₙ > 0:
\[a_{n+1}-\sqrt2 = \frac{a_n}{2} + \frac{1}{a_n} - \sqrt2.\]Combining the terms:
\[a_{n+1}-\sqrt2 = \frac{(a_n-\sqrt2)^2}{2a_n}.\]Therefore:
\[a_{n+1}\ge\sqrt2.\]Since:
\[a_1=\frac32>\sqrt2,\]we have:
\[a_n\ge\sqrt2\]for every n.
Now:
\[a_{n+1}-a_n = -\frac{a_n}{2}+\frac{1}{a_n}.\]Hence:
\[a_{n+1}-a_n = \frac{2-a_n^2}{2a_n}.\]Since aₙ ≥ √2:
\[a_{n+1}-a_n\le0.\]The sequence is decreasing and bounded below by √2, so it converges.
Let:
\[a_n\to\ell.\]Then:
\[\ell = \frac{\ell}{2} + \frac{1}{\ell}.\]Multiplying by 2ℓ:
\[2\ell^2=\ell^2+2.\]Therefore:
\[\ell^2=2.\]Since all terms are positive:
\[\ell=\sqrt2.\]Final Result
\[\lim_{n\to\infty}a_n=\sqrt2\]Exercise 7 — Parameter α > 0
\[\begin{cases} a_0=\alpha,\\ a_{n+1}=a_n^2-a_n+1. \end{cases}\]Solution.
Compute:
\[a_{n+1}-a_n = a_n^2-2a_n+1.\]Therefore:
\[a_{n+1}-a_n = (a_n-1)^2\ge0.\]The sequence is increasing.
If:
\[0<\alpha\le1,\]we show that the sequence remains in [0,1].
If:
\[0<a_n\le1,\]then:
\[a_{n+1} = 1-a_n(1-a_n).\]Therefore:
\[0<a_{n+1}\le1.\]Thus [0,1] is invariant.
The sequence is increasing and bounded above by 1, so it converges.
Let:
\[a_n\to\ell.\]Then:
\[\ell=\ell^2-\ell+1.\]Hence:
\[(\ell-1)^2=0.\]Therefore:
\[\ell=1.\]If α > 1, the sequence is strictly increasing and:
\[a_n\ge\alpha>1.\]A finite limit would again have to equal 1, which is impossible.
Therefore the sequence is unbounded above.
Final Result
\[\lim_{n\to\infty}a_n= \begin{cases} 1, & 0<\alpha\le1,\\ +\infty, & \alpha>1. \end{cases}\]Exercise 8
\[\begin{cases} a_1=\frac12,\\ a_{n+1}=\frac{1}{4-a_n}. \end{cases}\]Solution.
The fixed points satisfy:
\[\ell=\frac{1}{4-\ell}.\]Therefore:
\[\ell^2-4\ell+1=0.\]Hence:
\[\ell=2\pm\sqrt3.\]Let:
\[\alpha=2-\sqrt3.\]Since:
\[\alpha<\frac12,\]the initial value satisfies:
\[\alpha\le a_1\le\frac12.\]Consider:
\[f(x)=\frac{1}{4-x}.\]This function is increasing on the interval under consideration.
Since α is a fixed point:
\[f(\alpha)=\alpha.\]If:
\[\alpha\le a_n\le\frac12,\]then:
\[a_{n+1}\ge\alpha.\]Moreover:
\[a_{n+1} \le f\left(\frac12\right) = \frac{2}{7} < \frac12.\]Thus [α, 1/2] is invariant.
Now:
\[a_{n+1}-a_n = \frac{1}{4-a_n}-a_n.\]Combining terms:
\[a_{n+1}-a_n = \frac{a_n^2-4a_n+1}{4-a_n}.\]Factorizing:
\[a_{n+1}-a_n = \frac{ (a_n-(2-\sqrt3))(a_n-(2+\sqrt3)) }{ 4-a_n }.\]For:
\[2-\sqrt3\le a_n\le\frac12,\]the numerator is non-positive and the denominator is positive.
Therefore:
\[a_{n+1}\le a_n.\]The sequence is decreasing and bounded below by 2 − √3, so it converges.
The only fixed point in the invariant interval is:
\[2-\sqrt3.\]Final Result
\[\lim_{n\to\infty}a_n=2-\sqrt3\]Exercise 9
\[\begin{cases} a_1=4,\\ a_{n+1}=2\sqrt{a_n^2-6}. \end{cases}\]Solution.
Since:
\[a_1=4,\]the recursion is well defined.
For positive terms:
\[a_{n+1}\ge a_n\]is equivalent to:
\[2\sqrt{a_n^2-6}\ge a_n.\]Squaring:
\[4(a_n^2-6)\ge a_n^2.\]Thus:
\[3a_n^2\ge24.\]Therefore:
\[a_n\ge2\sqrt2.\]Since:
\[a_1=4>2\sqrt2,\]and the sequence is increasing, this condition remains satisfied.
Hence:
\[a_{n+1}\ge a_n.\]Suppose the sequence had a finite limit ℓ. Then:
\[\ell = 2\sqrt{\ell^2-6}.\]Squaring:
\[\ell^2 = 4\ell^2-24.\]Therefore:
\[3\ell^2=24.\]Hence:
\[\ell=2\sqrt2.\]But:
\[a_n\ge a_1=4>2\sqrt2,\]which is incompatible with convergence to 2√2.
Thus the increasing sequence is not bounded above.
Final Result
\[\lim_{n\to\infty}a_n=+\infty\]Exercise 10
\[\begin{cases} a_1=\frac{\pi}{2},\\ a_{n+1}=\sin a_n. \end{cases}\]Solution.
On [0, π/2]:
\[0\le\sin t\le t.\]Since:
\[a_1=\frac{\pi}{2},\]induction gives:
\[0\le a_n\le\frac{\pi}{2}.\]Moreover:
\[a_{n+1}=\sin a_n\le a_n.\]Thus the sequence is decreasing and bounded below by zero.
Therefore it converges.
Let:
\[a_n\to\ell.\]By continuity of the sine function:
\[\ell=\sin\ell.\]On [0, π/2], the only solution is:
\[\ell=0.\]Final Result
\[\lim_{n\to\infty}a_n=0\]Continue Exploring Calculus
Recursively defined sequences show how local rules can generate global behavior. Monotonicity, invariant intervals, boundedness, and fixed points provide a systematic way to study convergence.
The initial value is often decisive: the same recurrence relation can converge or diverge depending on where the sequence begins.