9.5

1. Show that the sequence \(\{\frac{2n}{1+n}\}_{n=1}^{\infty}\) is bounded

   Proof

   Since \(|\frac{2n}{1+n}|=\frac{2n}{1+n}=2\cdot \frac{n}{1+n}\leq 2\cdot 1=2\) because \(\frac{n}{1+n}\leq 1\)​

   Therefore \(|\frac{2n}{1+n}|\leq 2\) for all \(n\in N\) and thus the sequence is bounded

2. Show that the sequence \(\{\frac{1}{n}\}^{\infty}_{n=1}\) is convergent

   Proof

   Since \(|\frac{1}{n}|=\frac{1}{n}\leq 1\) for all \(n\in N\) the sequence is bounded

   bounded plus monotonic implies convergent

   Note that

   \(\frac{1}{n+1} - \frac{1}{n} = \frac{n - (n - 1)}{(n+1)n} = \frac{-1}{(n+1)n} < 0\)

   Hence the sequence is decreasing.

   Therefore, the sequence \(\left\{ \frac{1}{n} \right\}_{n=1}^{\infty}\) is convergent.

3. Show that the series \(\begin{aligned}\sum^{\infty}_{n=1}\frac{1}{n(n+1)}\end{aligned}\) is convergent and find its sum.

   Solution:

   As \(\begin{aligned} S_n &= \sum_{i=1}^{n} \frac{1}{i(i+1)} = \sum_{i=1}^{n} \left( \frac{1}{i} - \frac{1}{i+1} \right) = 1 - \frac{1}{n+1} \end{aligned}\), then \(\begin{aligned}\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = \lim_{n \to \infty} S_n = \lim_{n \to \infty} ( 1 - \frac{1}{n+1}) = 1 \end{aligned}\)

   Therefore, the series is convergent and its sum is \(1\).

4. Is the series \(\begin{aligned}\sum_{n=1}^{\infty} 2^{2n} 3^{1-n}\end{aligned}\) convergent or divergent?

   \(\begin{aligned}2^{2n} 3^{1-n} &= 4 \cdot \frac{1}{3^{n-1}} = 4 \cdot \frac{4^{n-1}}{3^{n-1}} = 4 \left( \frac{4}{3} \right)^{n-1}\end{aligned}\)

   \(\begin{aligned}\sum_{n=1}^{\infty} 2^{2n} 3^{1-n} &= \sum_{n=1}^{\infty} \left( \frac{4}{3} \right)^{n-1}\end{aligned}\)

   This is a geometric series.

   As \(\left| r \right| = \left| \frac{4}{3} \right| = \frac{4}{3} > 1\), the series is divergent.

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