Exercise 2.4.8

Question

For each series, find an explicit formula for the sequence of partial sums and determine if the series converges.
  \begin{enumerate}[label=(\alph*)] 
  \item $\sum_{n=1}^{\infty} \frac{1}{2^{n}}$
  \item $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$
  \item $\sum_{n=1}^{\infty} \log \left(\frac{n+1}{n}\right)$
   \end{enumerate}
  (In (c), $\log (x)$ refers to the natural logarithm function from calculus.)

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item This is a geometric series, we can use the usual trick to derive $s_n$. Let $r = 1/2$ for convenience
    $$
    \begin{aligned}
      s_n  &= 1 + r + r^2 + \dots + r^n \
      rs_n &= r + r^2 + \dots + r^{n+1} \
      rs_n - s_n &= r^{n+1} - 1 \implies s_n = \frac{r^{n+1} - 1}{r - 1}
    \end{aligned}
    $$
    This is the formula when $n$ starts at zero, but the sum in question starts at one so we subtract the first term to correct this
    $$
    \sum_{n=1}^\infty \frac{1}{2^n}
    = -1 + \sum_{n=0}^\infty \frac{1}{2^n}
    = -1 + \lim_{n	o \infty} \frac{(1/2)^{n+1} - 1}{1/2 - 1}
    = -1 + \frac{-1}{-1/2} = 1
    $$
  \item We can use partial fractions to get
    $$
    \frac{1}{n(n+1)} = \frac{1}{n} -\frac{1}{n+1}
    $$
    Which gives us a telescoping series, most of the terms cancel and we get
    $$
    s_n = 1 - \frac{1}{n+1}
    $$
    Therefor
    $$
    \sum_{n=1}^\infty \frac{1}{n(n+1)} = \lim_{n 	o \infty} \left(1 - \frac{1}{n+1}ight) = 1
    $$
  \item Another telescoping series, since
    $$
    \log\left(\frac{n+1}{n}ight) = \log(n+1) - \log(n)
    $$
    therefore most of the terms cancel and we get
    $$
    s_n = \log(n+1)
    $$
    Which doesn't converge.
   \end{enumerate}

Exercise 2.4.8

Answers

Comments

Exercise 2.4.8

Answers

Comments

Add answer