Limits of Sequences

Let us set \(\varepsilon=1\). From the definition of the convergent sequence we see that there exists a real number \(L\) and an integer \(n_0\in\mathbb N\) such that \(|a_n-L| < 1\) for \(n\geq n_0\). Let \(R=\max\left\{|a_1-L|,|a_2-L|,\dots, |a_{n_0}-L|\right\}\). The number \(R\) is a real number (as opposed to \(+\infty\)) because it is a maximum of a finite set of numbers. Therefore, \(|a_n-L|\leq \max\{1,R\}\) for all \(n\in\mathbb N\) hence: \[|a_n|=|a_n-L+L|\leq |a_n-L|+|L|\leq \max\{1,R\}+|L|,\] hence we may take \(M= \max\{1,R\}+|L|\).

Assume that \(\varepsilon > 0\) is given. Since \(\lim_{n\to\infty}a_n=L\) there exists a positive integer \(n_1\) for which \(n\geq n_1\) implies that \(|a_n-L| < \varepsilon\). Similarly, there exists \(n_2\) for which \(n\geq n_2\) implies that \(|c_n-L| < \varepsilon\). Then if we define \(n_3=\max\{n_0,n_1,n_2\}\) we have that \(n\geq n_3\) implies that \(-\varepsilon < a_n-L\); \(a_n < b_n < c_n\), and \(c_n-L < \varepsilon\). This means that \[L-\varepsilon < a_n < b_n < c_n < L+\varepsilon\quad\quad\quad\Rightarrow \quad\quad\quad |b_n-L| < \varepsilon.\] The last inequality holds for every \(n\geq n_3\) hence \(\displaystyle\lim_{n\to\infty} b_n=L\).

Assume that \(\varepsilon > 0\) is given. Since \(\displaystyle \lim_{n\to\infty}a_n=\alpha\) there exists \(n_1\in\mathbb N\) such that if \(n\geq n_1\) the following inequality holds: \(|a_n-\alpha| < \frac{\varepsilon}2\). Similarly, there exists \(n_2\in\mathbb N\) such that \(n\geq n_2\) implies \(|b_n-\beta|\leq \frac{\varepsilon}2\). Let us denote \(n_0=\max\{n_1,n_2\}\). If \(n\geq n_0\) then \(|a_n-\alpha| < \frac{\varepsilon}2\) and \(|b_n-\beta|\leq \frac{\varepsilon}2\) hence: \[\left|a_n+b_n-(\alpha+\beta)\right|\leq \left|a_n-\alpha\right|+\left|b_n-\beta\right|\leq \frac{\varepsilon}2+\frac{\varepsilon}2=\varepsilon.\] The proof is complete.

Assume that \(\varepsilon > 0\) is given. Since \((a_n)_{n=1}^{\infty}\) and \((b_n)_{n=1}^{\infty}\) are convergent, these two sequences are bounded and there exist numbers \(M_a\) and \(M_b\) such that \(|a_n|\leq M_a\) and \(|b_n|\leq M_b\) for all \(n\in\mathbb N\). Let \(M=\max\{M_a,M_b\}+1\).

Since \(\displaystyle \lim_{n\to\infty}a_n=\alpha\) there exists \(n_1\in\mathbb N\) such that if \(n\geq n_1\) the following inequality holds: \(|a_n-\alpha| < \frac{\varepsilon}{2M}\). Similarly, there exists \(n_2\in\mathbb N\) such that \(n\geq n_2\) implies \(|b_n-\beta|\leq \frac{\varepsilon}{2M}\). Let us denote \(n_0=\max\{n_1,n_2\}\). If \(n\geq n_0\) then \begin{eqnarray*} |a_n\cdot b_n-\alpha\cdot \beta|&=&|a_nb_n-\alpha b_n+\alpha b_n-\alpha\beta|\\&\leq& |a_nb_n-\alpha b_n|+|\alpha b_n-\alpha\beta|= |b_n|\cdot |a_n-\alpha|+\alpha\cdot |b_n-\beta| \\ &\leq& M\cdot \frac{\varepsilon}{2M}+M\cdot \frac{\varepsilon}{2M}=\varepsilon. \end{eqnarray*} Thus \(\displaystyle \lim_{n\to\infty} a_nb_n=\alpha\beta\).

Assume that \(\varepsilon > 0\) is given. Since \(\displaystyle \lim_{n\to\infty}b_n=\beta\) and \(\beta\neq 0\), we may assume that \(\beta > 0\). Let us fix \(\beta^{\prime}\in(0,\beta)\). There exists \(n_1\in\mathbb N\) such that \(n\geq n_1\) implies that \(b_n > \beta^{\prime}\). Since \((a_n)_{n=1}^{\infty}\) is convergent there exist a number \(M_a\) such that \(|a_n|\leq M_a\). Let \(M=\max\left\{\frac{M_a}{\beta^{\prime}\,^2},\frac1{\beta^{\prime}}\right\}+1\).

Since \(\displaystyle \lim_{n\to\infty}a_n=\alpha\) there exists \(n_2\in\mathbb N\) such that if \(n\geq n_2\) the following inequality holds: \(|a_n-\alpha| < \frac{\varepsilon}{2M}\). Similarly, there exists \(n_3\in\mathbb N\) such that \(n\geq n_3\) implies \(|b_n-\beta|\leq \frac{\varepsilon}{2M}\). Let us denote \(n_0=\max\{n_1,n_2,n_3\}\). If \(n\geq n_0\) then \begin{eqnarray*} \left|\frac{a_n}{b_n}-\frac{\alpha}{\beta}\right|&=&\left|\frac{a_n}{b_n}-\frac{a_n}{\beta}+\frac{a_n}{\beta}-\frac{\alpha}{\beta}\right|\\ &\leq& \left|\frac{a_n}{b_n}-\frac{a_n}{\beta}\right|+\left|\frac{a_n}{\beta}-\frac{\alpha}{\beta}\right|= \left|\frac{a_n}{b_n\cdot \beta}\right|\cdot|b_n-\beta|+\frac1{\beta}\cdot |a_n-\alpha|\\ &< &M\cdot \frac{\varepsilon}{2M}+M\cdot \frac{\varepsilon}{2M}=\varepsilon. \end{eqnarray*} Thus \(\displaystyle \lim_{n\to\infty} \frac{a_n}{b_n}=\frac{\alpha}{\beta}\).

Assume that \((b_n)_{n=1}^{\infty}\) is a subsequence of \((a_n)_{n=1}^{\infty}\). Assume that \(i_1\), \(i_2\), \(\dots\) is an increasing sequence of positive integers such that \(b_n=a_{i_n}\).

Given \(\varepsilon > 0\), let us choose \(n_0\) such that \(n\geq n_0\) implies \(|a_n-L| < \varepsilon\). Then for \(n\geq n_0\) we are sure \(i_n\geq n\geq n_0\) since \((i_n)_{n=1}^{\infty}\) is increasing. Therefore \(|b_n-L|=|a_{i_n}-L| < \varepsilon\) hence \(\displaystyle \lim_{n\to\infty}b_n=L\).

• Assume first that \(L > 0\).

  Assume that \(\varepsilon > 0\) is given. There exists \(n_0\in\mathbb N\) such that \(n > n_0\) implies that \(|a_n-L| < \varepsilon \cdot \sqrt L\). We now have that for \(n\geq n_0\) the following relations hold: \begin{eqnarray*} |b_n-\sqrt L|&=&|\sqrt{a_n}-\sqrt L|=\frac{|a_n-L|}{\sqrt{a_n}+\sqrt L}\leq\frac{|a_n-L|}{\sqrt L}< \varepsilon. \end{eqnarray*}

• Assume now that \(L=0\). There exists \(n_0\in\mathbb N\) such that \(n\geq n_0\) implies \(a_n < \varepsilon^2\). This in turn implies that \(b_n=\sqrt{a_n} < \varepsilon\) which means that \(\displaystyle \lim_{n\to\infty} b_n=0=\sqrt L\).

Assume that \(a_n\) is non-decreasing. That means \(a_{n+1}\geq a_n\). The case of non-increasing sequence is the same.

Assume that \(M\) is a real number such that \(a_n\leq M\) for all \(n\in\mathbb N\). Consider the set \(A=\{a_1,a_2, a_3,\dots\}\). This set is bounded from above.

Among all upper bounds of \(A\) there exists the smallest one - let’s call it \(L\). Don’t ask for proof of this - this is an axiom. This smallest upper bound \(L\) is called the supremum of \(A\). Existence of the supremum for any bounded set is an axiom for real numbers!

We will now prove that \(L\) is the limit of the sequence \((a_n)_{n=1}^{\infty}\). Assume that \(\varepsilon > 0\) is given. The number \(L-\varepsilon\) is smaller than the smallest upper bound. Therefore it can’t be an upper bound, and there exists an \(n_0\) such that \(a_{n_0} > L-\varepsilon\). Since \((a_n)_{n=1}^{\infty}\) is non-decreasing we have that for every \(n\geq n_0\) the following inequality holds: \(L-\varepsilon < a_{n_0}\leq a_n\). Since \(L\) is an upper bound we also have \(a_n < L\) therefore \(|a_n-L| < \varepsilon\).

• (a) The statement follows immediately from the previous theorem and the fact that \((\overline{a}_n)_{n=1}^{\infty}\) is decreasing and \((\underline{a}_n)_{n=1}^{\infty}\) is increasing.

• (b) Let us denote \(\overline{a}=\limsup_{n\to\infty}a_n\) and \(\underline{a}=\liminf_{n\to\infty}a_n\). Let us prove the existence of the sequence \((b_n)_{n=1}^{\infty}\). The existence of \((c_n)_{n=1}^{\infty}\) is established in an analogous way. For each \(n\in\mathbb N\) there exists an element \(a_{k_n}\) of the set \(\{a_n, a_{n+1},\dots\}\) such that \(a_{k_n}\geq \overline{a}_n-\frac1{n}\). We will set \(b_n=a_{k_n}\). We will now prove that \(\displaystyle \lim_{n\to\infty} b_n=\overline{a}\).

  Let \(\varepsilon > 0\). There exists \(n_1\in\mathbb N\) such that \(n\geq n_1\) implies that \(|\overline{a}_n-\overline a| < \frac{\varepsilon}2\). Let us define \(n_0=\max\left\{n_1,\left\lceil\frac2{\varepsilon}\right\rceil\right\}\). For \(n\geq n_0\) we have that \[|\overline a-b_n|\leq |\overline a-\overline{a}_n|+|\overline{a}_n-b_n|\leq\frac{\varepsilon}2+\frac1n\leq \varepsilon.\]

• (c) This part follows from Theorem 6 and part (b).

We prove this theorem in three steps:

• The sequenze \((a_n)_{n=1}^{\infty}\) is bounded:

  Take \(\varepsilon=1\). There exists an integer \(n_0\) such that \(m,n\geq n_0\) implies that \(|a_n-a_m| < 1\). Let \[M=\max\left\{|a_1|,\dots, |a_{n_0}|\right\}.\] The last maximum is finite because we are taking the maximum of finitely many real number. Therefore for all \(n\geq n_0\) we have \[|a_n|=\left|a_n-a_{n_0}+a_{n_0}\right|\leq \left|a_n-a_{n_0}\right|+\left|a_{n_0}\right|\leq 1+\left|a_{n_0}\right|\leq 1+M.\]

• The sequence \((a_n)_{n=1}^{\infty}\) has a convergent subsequence:

  This follows immediately from the part (b) of Theorem 9.

• Consider a convergent subsequence \((b_n)_{n=1}^{\infty}\) of \((a_n)_{n=1}^{\infty}\). Let \(L\) be the limit of \((b_n)_{n=1}^{\infty}\). We will prove that \(\lim_{n\to \infty} a_n=L\). Let \(\varepsilon > 0\).

  There exists an integer \(n_1\) such that \(|a_n-a_m| < \frac{\varepsilon}2\) for \(n,m\geq n_1\). Let \(n^{\prime}\) be an integer bigger than \(n_1\) such that there exists \(k^{\prime}\) for which \(b_{k^{\prime}}=a_{n^{\prime}}\). There exists a natural number \(k > k^{\prime}\) such that \(|b_k-L| < \frac{\varepsilon}2\). Let \(n_0\) be an integer such that \(a_{n_0}=b_k\). Then for \(n\geq n_0\) we have \[|a_n-L|\leq|a_n-a_{n_0}|+ |a_{n_0}-L| < \frac{\varepsilon}2+|b_k-L| < \frac{\varepsilon}2+\frac{\varepsilon}2=\varepsilon.\]

We will use the following inequality (Bernoulli’s inequality): For all \(a > -1\) and \(n\in\mathbb N\) we have \[(1+a)^n > 1+na.\] The inequality has an easy proof by induction.

We will consider another sequence \((b_n)_{n=1}^{\infty}\) defined by \(b_n=\left(1+\frac1n\right)^{n+1}\) and prove that:

• (a) The sequence \((a_n)_{n=1}^{\infty}\) is increasing.
• (b) The sequence \((b_n)_{n=1}^{\infty}\) is decreasing.
• (c) For all \(n\in\mathbb N\) we have \(a_n\leq b_n\).

• (a) We need to prove that \[\left(1+\frac1n\right)^n\leq \left(1+\frac1{n+1}\right)^{n+1}.\] The last inequality is equivalent to \(\frac{n+1}{n+2}\leq \left(\frac{n(n+2)}{(n+1)^2}\right)^n\). Using Bernoulli’s inequality we get: \[ \left(\frac{n(n+2)}{(n+1)^2}\right)^n=\left(1-\frac1{(n+1)^2}\right)^n\geq 1-\frac{n}{(n+1)^2}.\] Now it remains to prove \(1-\frac{n}{(n+1)^2}\geq\frac{n+1}{n+2}=1-\frac1{n+2}\) and this follows from \(\frac1{n+2} > \frac{n}{(n+1)^2}\) which holds because \((n+1)^2=n(n+2)+1\).

• (b) We need to prove the inequality \[\left(1+\frac1n\right)^{n+1}\geq \left(1+\frac1{n+1}\right)^{n+2}.\] The inequality is equivalent to \((n+1)^{2n+3}\geq\left(n(n+2)\right)^{n+1}\cdot(n+2)\), which is equivalent to \[\left(\frac{(n+1)^2}{n(n+2)}\right)^{n+1}\geq\frac{n+2}{n+1}.\] Bernoulli’s inequality implies: \[\left(\frac{(n+1)^2}{n(n+2)}\right)^{n+1}=\left(1+\frac1{n(n+2)}\right)^{n+1}\geq 1 +\frac{n+1}{n(n+2)}.\] It remains to establish \(\frac{n+1}{n(n+2)}\geq \frac{1}{n+1}\). The last inequality immediately follows from \((n+1)^2 > n(n+2)\).

• (c) The inequality \(a_n\leq b_n\) is obvious because \(\frac{b_n}{a_n}=1+\frac1n > 1\).

Now we have that the sequence \(\left(a_n\right)_{n=1}^{\infty}\) is increasing and bounded above. Each element of the sequence \(\left(b_n\right)_{n=1}^{\infty}\) is an upper bound for \(\left(a_n\right)_{n=1}^{\infty}\). Similarly, the sequence \(\left(b_n\right)_{n=1}^{\infty}\) is decreasing and bounded below. Each element of \(\left(a_n\right)_{n=1}^{\infty}\) is a lower bound for the sequence \(\left(b_n\right)_{n=1}^{\infty}\). Thus both sequences \(\left(a_n\right)_{n=1}^{\infty}\) and \(\left(b_n\right)_{n=1}^{\infty}\) are convergent and the following inequality holds \[\lim_{n\to\infty}a_n\leq \lim_{n\to\infty}b_n.\]

We will now prove that the limits are equal. We will prove that \(\lim_{n\to\infty}\left(b_n-a_n\right)=0\). The difference between the terms \(b_n\) and \(a_n\) is \begin{eqnarray*} b_n-a_n&=&\left(1+\frac1n\right)^{n+1}-\left(1+\frac1n\right)^n=\left(1+\frac1n\right)^n\cdot \frac1n\\ &=&a_n\cdot\frac1n. \end{eqnarray*} Since \(a_n\) is bounded above by \(b_0\), we have \[\left|b_n-a_n\right|\leq b_0\cdot \frac1n.\] For every \(\varepsilon > 0\), we can choose \(n_0 > \frac{b_0}{\varepsilon}\). If \(n\) is an integer that satisfies \(n\geq n_0\), the it is obvious that \(|b_n-a_n| < \varepsilon\).

We have now established that \(\displaystyle\lim_{n\to\infty}(b_n-a_n)=0\), hence

The sequence is convergent and its limit is \(-\frac{19}{29}\) because: \[\lim_{n\to\infty}\frac{2n+19 n^2+\frac3n}{-4n-29 n^2-12}= \lim_{n\to\infty}\frac{19+\frac{2}{n}+\frac3{n^3}}{-29-\frac4n-\frac{12}{n^2}}. \] Let us denote \(b_n=19+\frac{2}{n^3}+\frac3{n^3}\) and \(c_n=-29-\frac4n-\frac{12}{n^2}\). The sequences \((b_n)_{n=1}^{\infty}\) and \((c_n)_{n=1}^{\infty}\) are convergent and their limits are \(19\) and \(-29\), respectively. Thus \((a_n)_{n=1}^{\infty}=\left(\frac{b_n}{c_n}\right)_{n=1}^{\infty}\) is convergent and its limit is \(-\frac{19}{29}\).

The sequence is convergent and its limit is \(\frac{3}{4}\) because: \[\lim_{n\to\infty}\frac{3n^2+\cos n}{4n^2-25}= \lim_{n\to\infty}\frac{3+\frac{\cos n}{n^2}}{4-\frac{25}{n^2}}. \] Let us denote \(b_n= 3+\frac{\cos n}{n^2} \) and \(c_n=4-\frac{25}{n^2}\). The sequences \((c_n)_{n=1}^{\infty}\) is convergent and its limit is \(4\).

The convergence of the sequence \(b_n\) can be established using the squeeze theorem. Set \(p_n=3-\frac1{n^2}\) and \(q_n=3+\frac1{n^2}\). Since \(\cos n\in[-1,1]\) we have that \(p_n\leq b_n\leq q_n\). Since \(\displaystyle \lim_{n\to\infty}p_n=\lim_{n\to\infty}q_n=3\) we have \(\displaystyle \lim_{n\to\infty}b_n=3\).

Thus \((a_n)_{n=1}^{\infty}=\left(\frac{b_n}{c_n}\right)_{n=1}^{\infty}\) is convergent and its limit is \(\frac{3}{4}\).

We first notice that \begin{eqnarray*}a_n&=&\left[\left(1-\frac1{3n}\right)^{3n}\right]^{\frac1{3}}=\left[\left(\frac{3n-1}{3n}\right)^{3n}\right]^{\frac1{3}}= \frac1{ \left[\left(1+\frac1{3n-1}\right)^{3n-1} \right]^{\frac1{3}} } \cdot \frac1{ \left[1+\frac1{3n-1} \right] ^{\frac1{3} } } . \end{eqnarray*} Let us introduce the following two sequences \(b_n\) and \(c_n\): \begin{eqnarray*} b_n&=&\frac1{ \left[\left(1+\frac1{3n-1}\right)^{3n-1} \right]^{\frac1{3}}}\\ c_n&=&\frac1{ \left[1+\frac1{3n-1} \right] ^{\frac1{3} } }. \end{eqnarray*} We will prove that \(\displaystyle \lim_{n\to\infty} b_n\) and \(\displaystyle \lim_{n\to\infty} c_n\) exist. Then the limit of the sequence \((a_n)_{n=1}^{\infty}\) will be product of the previous two limits. We start by finding \(\displaystyle \lim_{n\to\infty} c_n\): \(\displaystyle\lim_{n\to\infty} \left(1+\frac1{3n-1}\right)=1\), hence \(\displaystyle \lim_{n\to\infty}\left(1+\frac1{3n-1}\right)^{\frac1{3}}=1^{\frac13}=1\) and \(\displaystyle \lim_{n\to\infty}c_n=1\).

Let us now find \(\displaystyle \lim_{n\to\infty} b_n\). We first have that \(\displaystyle \lim_{n\to\infty}\left(1+\frac1{3n-1}\right)^{3n-1}=e\) since this is a subsequence of \(\left(\left(1+\frac1n\right)^n\right)_{n=1}^{\infty}\). Therefore \(\displaystyle \lim_{n\to\infty} \left[\left(1+\frac1{3n-1}\right)^{3n-1}\right]^{\frac1{3}}= e^{\frac1{3}}\) and \(\displaystyle \lim_{n\to\infty} b_n=\frac1{e^{\frac1{3}}}\).

We can now conclude that \(\displaystyle\lim_{n\to\infty} a_n=\frac1{e^{\frac1{3}}}\cdot 1=\frac1{e^{\frac1{3}}}\).

We will prove that the sequence is unbounded, and hence divergent. Assume that \(M > 0\) is arbitrary real number. We will prove that there is a term of the sequence that is greater than \(M\).

Given \(M > 0\), let us pick \(n=\max\{M,6\}\). Then since \(n > 1\) we have \(5n^2+\cos n\geq 5n^2-1 > 4n^2\). On the other hand, \(4n-25\leq 4n\) and \(4n-25 > 0\) hence \[\frac{5n^2+\cos n}{4n-25} > \frac{4n^2}{4n}=n\geq M.\]

The convergence of the sequence \(a_n\) can be established using the squeeze theorem. Set \(p_n=1-\frac1{n^2}\) and \(q_n=1+\frac1{n^2}\). Since \(\sin n\in[-1,1]\) we have that \(p_n\leq a_n\leq q_n\). Since \(\displaystyle \lim_{n\to\infty}p_n=\lim_{n\to\infty}q_n=1\) we have \(\displaystyle \lim_{n\to\infty}a_n=1\).

Thus \((a_n)_{n=1}^{\infty}=\left(\frac{b_n}{c_n}\right)_{n=1}^{\infty}\) is convergent and its limit is \(\frac34\).