Reading guide · Proof index

Facts about limits of sequences

Jiří Lebl, Basic Analysis I–II, version 6.3. Free author edition of this section. Selection and attribution · Notation.

L2.2.1: Squeeze lemma

Proof.

Let x≔lim⁡n→∞an=lim⁡n→∞bn.x \coloneqq \lim_{n\to\infty} a_n = \lim_{n\to\infty} b_n\text{.} Let ϵ>0\epsilon > 0 be given. Find an M1M_1 such that for all n≥M1,n \geq M_1\text{,} we have ∣an−x∣<ϵ,\sabs{a_n-x} < \epsilon\text{,} and an M2M_2 such that for all n≥M2,n \geq M_2\text{,} we have ∣bn−x∣<ϵ.\sabs{b_n-x} < \epsilon\text{.} Set M≔max⁡{M1,M2}.M \coloneqq \max \{M_1, M_2 \}\text{.} Suppose n≥M.n \geq M\text{.} In particular, x−an<ϵ,x - a_n < \epsilon\text{,} or x−ϵ<an.x - \epsilon < a_n\text{.} Similarly, bn<x+ϵ.b_n < x + \epsilon\text{.} Putting everything together, we find
x−ϵ<an≤xn≤bn<x+ϵ.\begin{equation*} x - \epsilon < a_n \leq x_n \leq b_n < x + \epsilon . \end{equation*}
In other words, −ϵ<xn−x<ϵ-\epsilon < x_n-x < \epsilon or ∣xn−x∣<ϵ.\sabs{x_n-x} < \epsilon\text{.} So {xn}n=1∞\{x_n\}_{n=1}^\infty converges to x.x\text{.} See Figure 2.3.

L2.2.3: Nonstrict inequalities pass to limits

Proof.

Let x≔lim⁡n→∞xnx \coloneqq \lim_{n\to\infty} x_n and y≔lim⁡n→∞yn.y \coloneqq \lim_{n\to\infty} y_n\text{.} Let ϵ>0\epsilon > 0 be given. Find an M1M_1 such that for all n≥M1,n \geq M_1\text{,} we have ∣xn−x∣<ϵ ⁣/ ⁣2.\sabs{x_n-x} < \nicefrac{\epsilon}{2}\text{.} Find an M2M_2 such that for all n≥M2,n \geq M_2\text{,} we have ∣yn−y∣<ϵ ⁣/ ⁣2.\sabs{y_n-y} < \nicefrac{\epsilon}{2}\text{.} In particular, for any n≥max⁡{M1,M2},n \geq \max\{ M_1, M_2 \}\text{,} we have x−xn<ϵ ⁣/ ⁣2x-x_n < \nicefrac{\epsilon}{2} and yn−y<ϵ ⁣/ ⁣2.y_n-y < \nicefrac{\epsilon}{2}\text{.} We add these inequalities to obtain
yn−xn+x−y<ϵ,oryn−xn<y−x+ϵ.\begin{equation*} y_n-x_n+x-y < \epsilon, \qquad \text{or} \qquad y_n-x_n < y-x+ \epsilon . \end{equation*}
Since xn≤yn,x_n \leq y_n\text{,} we have 0≤yn−xn0 \leq y_n-x_n and hence 0<y−x+ϵ.0 < y-x+ \epsilon\text{.} In other words,
x−y<ϵ.\begin{equation*} x-y < \epsilon . \end{equation*}
Because ϵ>0\epsilon > 0 was arbitrary, we obtain x−y≤0.x-y \leq 0\text{.} Therefore, x≤y.x \leq y\text{.}

L2.2.5: Full addition, multiplication and reciprocal proofs; subtraction completed locally

Proof.

We start with i. Suppose {xn}n=1∞\{ x_n \}_{n=1}^\infty and {yn}n=1∞\{ y_n \}_{n=1}^\infty are convergent sequences and write zn≔xn+yn.z_n \coloneqq x_n + y_n\text{.} Let x≔lim⁡n→∞xn,x \coloneqq \lim_{n\to\infty} x_n\text{,} y≔lim⁡n→∞yn,y \coloneqq \lim_{n\to\infty} y_n\text{,} and z≔x+y.z \coloneqq x+y\text{.} Let ϵ>0\epsilon > 0 be given. Find an M1M_1 such that for all n≥M1,n \geq M_1\text{,} we have ∣xn−x∣<ϵ ⁣/ ⁣2.\sabs{x_n - x} < \nicefrac{\epsilon}{2}\text{.} Find an M2M_2 such that for all n≥M2,n \geq M_2\text{,} we have ∣yn−y∣<ϵ ⁣/ ⁣2.\sabs{y_n - y} < \nicefrac{\epsilon}{2}\text{.} Take M≔max⁡{M1,M2}.M \coloneqq \max \{ M_1, M_2 \}\text{.} For all n≥M,n \geq M\text{,} we have
∣zn−z∣=∣(xn+yn)−(x+y)∣=∣xn−x+yn−y∣≤∣xn−x∣+∣yn−y∣<ϵ2+ϵ2=ϵ.\begin{equation*} \begin{split} \sabs{z_n - z} &= \babs{(x_n+y_n) - (x+y)} \\ & = \sabs{x_n-x + y_n-y} \\ & \leq \sabs{x_n-x} + \sabs{y_n-y} \\ & < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon. \end{split} \end{equation*}
Therefore, i is proved. The proof of ii is almost identical and is left as an exercise.
Let us tackle iii. Suppose again that {xn}n=1∞\{ x_n \}_{n=1}^\infty and {yn}n=1∞\{ y_n \}_{n=1}^\infty are convergent sequences and write zn≔xnyn.z_n \coloneqq x_n y_n\text{.} Let x≔lim⁡n→∞xn,x \coloneqq \lim_{n\to\infty} x_n\text{,} y≔lim⁡n→∞yn,y \coloneqq \lim_{n\to\infty} y_n\text{,} and z≔xy.z \coloneqq xy\text{.} Let ϵ>0\epsilon > 0 be given. Let K≔max⁡{∣x∣,∣y∣,ϵ ⁣/ ⁣3,1}.K \coloneqq \max\{ \sabs{x}, \sabs{y}, \nicefrac{\epsilon}{3} , 1 \}\text{.} Find an M1M_1 such that for all n≥M1,n \geq M_1\text{,} we have ∣xn−x∣<ϵ3K.\sabs{x_n - x} < \frac{\epsilon}{3K}\text{.} Find an M2M_2 such that for all n≥M2,n \geq M_2\text{,} we have ∣yn−y∣<ϵ3K.\sabs{y_n - y} < \frac{\epsilon}{3K}\text{.} Take M≔max⁡{M1,M2}.M \coloneqq \max \{ M_1, M_2 \}\text{.} For all n≥M,n \geq M\text{,} we have
∣zn−z∣=∣(xnyn)−(xy)∣=∣(xn−x+x)(yn−y+y)−xy∣=∣(xn−x)y+x(yn−y)+(xn−x)(yn−y)∣≤∣(xn−x)y∣+∣x(yn−y)∣+∣(xn−x)(yn−y)∣=∣xn−x∣∣y∣+∣x∣∣yn−y∣+∣xn−x∣∣yn−y∣<ϵ3KK+Kϵ3K+ϵ3Kϵ3K(now notice that ϵ3K≤1 and K≥1)≤ϵ3+ϵ3+ϵ3=ϵ.\begin{equation*} \begin{split} \sabs{z_n - z} &= \babs{(x_ny_n) - (xy)} \\ & = \babs{(x_n-x+x)(y_n-y+y) - xy} \\ & = \babs{(x_n-x)y + x(y_n-y) +(x_n-x)(y_n-y)} \\ & \leq \babs{(x_n-x)y} + \babs{x(y_n - y)} + \babs{(x_n-x)(y_n-y)} \\ & = \sabs{x_n -x}\sabs{y} + \sabs{x}\sabs{y_n -y} + \sabs{x_n -x}\sabs{y_n -y} \\ & < \frac{\epsilon}{3K} K + K \frac{\epsilon}{3K} + \frac{\epsilon}{3K} \frac{\epsilon}{3K} \qquad \qquad \text{(now notice that } \tfrac{\epsilon}{3K} \leq 1 \text{ and } K \geq 1\text{)} \\ & \leq \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon . \end{split} \end{equation*}
Finally, we examine iv. We prove the following simpler claim:
Claim: If {yn}n=1∞\{ y_n \}_{n=1}^\infty is a convergent sequence such that lim⁡n→∞yn≠0\lim_{n\to\infty} y_n \neq 0 and yn≠0y_n \neq 0 for all n∈N,n \in \N\text{,} then {1 ⁣/ ⁣yn}n=1∞\{ \nicefrac{1}{y_n} \}_{n=1}^\infty converges and
lim⁡n→∞1yn=1lim⁡n→∞yn.\begin{equation*} \lim_{n\to\infty} \frac{1}{y_n} = \frac{1}{\lim_{n\to\infty} y_n} . \end{equation*}
Once the claim is proved, we take the sequence {1 ⁣/ ⁣yn}n=1∞,\{ \nicefrac{1}{y_n} \}_{n=1}^\infty\text{,} multiply it by the sequence {xn}n=1∞,\{ x_n \}_{n=1}^\infty\text{,} and apply item iii.
Proof of claim: Let ϵ>0\epsilon > 0 be given. Let y≔lim⁡n→∞yn.y \coloneqq \lim_{n\to\infty} y_n\text{.} As ∣y∣≠0,\sabs{y} \neq 0\text{,} we have min⁡{∣y∣2ϵ2, ∣y∣2}>0.\min \left\{ \sabs{y}^2\frac{\epsilon}{2}, \, \frac{\sabs{y}}{2} \right\} > 0\text{.} Find an MM such that for all n≥M,n \geq M\text{,} we have
∣yn−y∣<min⁡{∣y∣2ϵ2, ∣y∣2}.\begin{equation*} \sabs{y_n - y} < \min \left\{ \sabs{y}^2\frac{\epsilon}{2}, \, \frac{\sabs{y}}{2} \right\} . \end{equation*}
For all n≥M,n \geq M\text{,} we have ∣y−yn∣<∣y∣ ⁣/ ⁣2,\sabs{y - y_n} < \nicefrac{\sabs{y}}{2}\text{,} and so
∣y∣=∣y−yn+yn∣≤∣y−yn∣+∣yn∣<∣y∣2+∣yn∣.\begin{equation*} \sabs{y} = \sabs{y - y_n + y_n } \leq \sabs{y - y_n} + \sabs{ y_n } < \frac{\sabs{y}}{2} + \sabs{y_n}. \end{equation*}
Subtracting ∣y∣ ⁣/ ⁣2\nicefrac{\sabs{y}}{2} from both sides we obtain ∣y∣ ⁣/ ⁣2<∣yn∣,\nicefrac{\sabs{y}}{2} < \sabs{y_n}\text{,} or in other words,
1∣yn∣<2∣y∣.\begin{equation*} \frac{1}{\sabs{y_n}} < \frac{2}{\sabs{y}} . \end{equation*}
We finish the proof of the claim:
∣1yn−1y∣=∣y−ynyyn∣=∣y−yn∣∣y∣∣yn∣≤∣y−yn∣∣y∣ 2∣y∣<∣y∣2ϵ2∣y∣ 2∣y∣=ϵ.\begin{equation*} \begin{split} \abs{\frac{1}{y_n} - \frac{1}{y}} &= \abs{\frac{y - y_n}{y y_n}} \\ & = \frac{\sabs{y - y_n}}{\sabs{y} \sabs{y_n}} \\ & \leq \frac{\sabs{y - y_n}}{\sabs{y}} \, \frac{2}{\sabs{y}} \\ & < \frac{\sabs{y}^2 \frac{\epsilon}{2}}{\sabs{y}} \, \frac{2}{\sabs{y}} = \epsilon . \end{split} \end{equation*}
And we are done.

L2.2.11: For 0<c<1, c^n tends to zero; the c>1 part is not needed

Proof.

First consider c<1.c < 1\text{.} As c>0,c > 0\text{,} we get cn>0c^n > 0 for all n∈Nn \in \N by induction. Then c<1c < 1 implies that cn+1<cnc^{n+1} < c^n for all n.n\text{.} The sequence {cn}n=1∞\{ c^n \}_{n=1}^\infty is thus bounded below and decreasing. Hence, it is convergent. Let x≔lim⁡n→∞cn.x \coloneqq \lim_{n\to\infty} c^n\text{.} The 1-tail {cn+1}n=1∞\{ c^{n+1} \}_{n=1}^\infty also converges to x.x\text{.} Taking the limit of both sides of cn+1=c⋅cn,c^{n+1} = c \cdot c^n\text{,} we obtain x=cx,x = cx\text{,} or (1−c)x=0.(1-c)x=0\text{.} It follows that x=0x=0 as c≠1.c \neq 1\text{.}
Now consider c>1.c > 1\text{.} Let B>0B > 0 be arbitrary. As 1 ⁣/ ⁣c<1,\nicefrac{1}{c} < 1\text{,} the sequence {(1 ⁣/ ⁣c)n}n=1∞\bigl\{ {(\nicefrac{1}{c})}^n \bigr\}_{n=1}^\infty converges to 0.0\text{.} Hence, for some large enough n,n\text{,} we get
1cn=(1c)n<1B.\begin{equation*} \frac{1}{c^n} = {\left(\frac{1}{c}\right)}^n < \frac{1}{B} . \end{equation*}
In other words, cn>B,c^n > B\text{,} and BB is not an upper bound for {cn}n=1∞.\{ c^n \}_{n=1}^\infty\text{.} As BB was arbitrary, {cn}n=1∞\{ c^n \}_{n=1}^\infty is unbounded.
A diagram of the real line. Points are marked on the line in order from left to right: x minus epsilon, a sub n, x, x sub n, b sub n, and x plus epsilon. Arrows emphasize that the distance from x minus epsilon to x is epsilon and that the distance from x to x plus epsilon is also epsilon.
Figure 2.3. Squeeze lemma proof in picture.