[Chapter 1~6] Problems #1

[Problem 0.0]

Let \(a_n=\displaystyle \left(1+\frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots+\right)-2\sqrt{n+1}\).  Prove that \(\{a_n\}\) has a limit.

# Solution ▶

(Proof)

\(\{a_n\}\) is strictly increasing and bounded above.(\(<1\)) (See the figure below.)

 

\(a_n=\text{(area of shaded rectangles)}-\int_{1}^{n+1} \frac{1}{\sqrt x} dx\)

Thus \(\{a_n\}\) is convergent by monotone convergence theorem. 

# ◀ 닫기

 

[Problem 0.1]

Prove that the sequence \(a_n=\displaystyle \frac{1\cdot 3 \cdot 5 \cdots (2n-1)}{2\cdot 4 \cdot 6 \cdots (2n)}\) has a limit.

# Solution ▶

(Proof)

For \(n\in \Bbb N\), \[\begin{align}&1\cdot 3<2^2 \\ &3\cdot 5<4^2 \\ &\quad\cdots \\&(2n-1)(2n+1)<(2n)^2\end{align}\] Multiplying both sides, \[\begin{align}&1^2\cdot 3^2\cdots(2n-1)^2\cdot (2n+1)<2^2\cdot 4^2\cdots (2n)^2 \\[.35em] &\Rightarrow \frac{1^2\cdot 3^2\cdots(2n-1)^2}{2^2\cdot 4^2\cdots (2n)^2}<\frac{1}{2n+1} \\[.35em] &\Rightarrow \frac{1\cdot 3\cdots(2n-1)}{2\cdot 4\cdots (2n)}<\frac{1}{\sqrt{2n+1}} \end{align}\] Since \(a_n>0\) and \(\displaystyle \frac{1}{\sqrt{2n+1}} \rightarrow 0 \; \text{as} \; n \rightarrow \infty \), we have \(a_n \rightarrow 0\) by squeeze theorem.

# ◀ 닫기

 

[Problem 0.2]

Prove that \(n^2 a^n \rightarrow 0\) if \(|a|<1\). 

# Solution ▶

(Proof)

\(n^2 a^n \rightarrow 0\) if and only if \(n^2 |a|^n \rightarrow 0\). Let \(|a|=\displaystyle \frac{1}{1+k}\) where \(k>0\). By binomial theorem, \[(1+k)^n=\displaystyle \sum_{i=0}^n\left(\begin{array}{c}n\\i\end{array}\right)k^i\geq 1+nk+\frac{n(n-1)}{2}k^2+\frac{n(n-1)(n-2)}{3!}k^3\] Hence  \[\begin{align}0<n^2 |a|^n &=n^2/(1+k)^n \\ &\leq \displaystyle \frac{n^2}{1+nk+\frac{n(n-1)}{2}k^2+\frac{n(n-1)(n-2)}{3!}k^3} \rightarrow 0 \;\, \textrm{as} \;\, n \rightarrow \infty \end{align}\]

# ◀ 닫기

 

[Problem 0.3]

Show that \(\displaystyle \lim_{n\rightarrow \infty} \int_0^1 (1-x^2)^n dx=0\,\), without attempting to evaluate the integral explicitly.

# Solution ▶

(Proof)

Let \(f_n(x)=(1-x^2)^n\). By MVT, \[\exists\, c\in (0, 1): \int_0^1(1-x^2)^ndx=f_n(c)=(1-c^2)^n\] Since \(0<1-c^2<1\),  \((1-c^2)^n \rightarrow 0\). Hence \(\displaystyle \lim_{n\rightarrow \infty} \int_0^1 (1-x^2)^n dx=0\). 

# ◀ 닫기

 

[Problem 0.4]

Let \(S\) and \(T\) be non-empty bounded subsets of \(\Bbb R\), and suppose that for all \(s\in S\) and \(t \in T\), we have \(s\leq t\). Prove that \(\sup S \leq \inf T\).

# Solution ▶

(Proof)

Suppose \(\sup S:=\alpha > \beta:=\inf T\) and let \(\epsilon=(\alpha-\beta)/2\). By definition of supremum and infimum, \[\begin{align}&\exists \, s\in S: \alpha-\epsilon=(\alpha+\beta)/2<s\leq\alpha\,,\\[.5em]&\exists \, t\in T: \beta\leq t<\beta+\epsilon=(\alpha+\beta)/2\ \\[.5em]&\Rightarrow t<(\alpha+\beta)/2<s\end{align}\] This contrary to the assumption. (\(s\leq t\))

# ◀ 닫기

 

[Problem 0.5]

Let \(\{a_n\}\) and \(\{b_n\}\) be sequences such that \(|a_{n+1}-a_n|\leq b_n\) for all \(n\in \Bbb N\). Define \(B_n=\displaystyle \sum_{k=1}^n b_k\) for all \(n\in \Bbb N\). If \(\{B_n\}\) is convergent, then prove that \(\{a_n\}\) is convergent. 

# Solution ▶

(Proof)

Since \(\{B_n\}\) is convergent, it is also a Cauchy sequence. i,e, \[\exists \, N\in \Bbb N: \; m>n>N \, \Rightarrow \, |B_m-B_{n-1}|=\left|\sum_{k=n}^{m} b_k\right|<\epsilon\] Hence, \[\begin{align} |a_m-a_n|&\leq|a_{n+1}-a_n|+|a_{n+2}-a_{n+1}|+\cdots+|a_m-a_{m-1}| \\[.5em] &\leq b_n+b_{n+1}+\cdots+b_m \\[.5em] &\leq \left|\sum_{k=n}^{m}a_k\right|<\epsilon\end{align}\] This shows \(\{a_n\}\) is a Cauchy sequence, so that is convergent.

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Reference: <Introduction to Analysis> Arthur Mattuck, 1999.