Exercise 2.4.9

Question

Complete the proof of Theorem 2.4.6 by showing that if the series $\sum_{n=0}^{\infty} 2^{n} b_{2^{n}}$ diverges, then so does $\sum_{n=1}^{\infty} b_{n}$. Example $2.4 .5$ may be a useful reference.

Ulisse Mini · Accepted Answer

Let $s_n = b_1 + b_2 + \dots + b_n$ and $t_k = b_1 + 2b_2 + \dots + 2^kb_{2^k}$.

We want to show $s_n$ is unbounded, first we find a series similar to $t_k$ that is less then $s_n$, then rewrite it in terms of $t_k$.

Let $n = 2^k$ so things match up nicely. We get
  $$
  \begin{aligned}
  s_n
  &=   b_1 + b_2 + (b_3 + b_4) + \dots + (b_{2^{k-1}} + \dots + b_{2^k}) \
  &\le b_1 + b_2 + (b_4 + b_4) + \dots + 2^{k-1}b_{2^k}
  \end{aligned}
  $$
  (Notice there are $2^k - 2^{k-1} = 2^{k-1}$ terms in the last term)

Now define $t_k'$ to be our new series $b_1 + b_2 + 2b_4 + 4b_8 + \dots + 2^{k-1}b_{2^k}$.
  This looks a lot like $t_k$, and in fact some algebra gives
  $$
  t_k'
  = \frac 12 \left(b_1 + 2b_2 + 4b_4 + \dots + 2^kb_kight) + \frac 12 b_1
  = \frac 12 t_k + \frac 12 b_1
  $$
  therefore we are justified in writing
  $$
  s_n \ge t_k' \ge \frac 12 t_k
  $$
  And since $t_k/2$ diverges and $s_n$ is bigger, $s_n$ must also diverge.

extbf{Summary}: $s_n$ converges iff $t_k$ conv since $t_k \ge s_n \ge t_k/2$ for $n = 2^k$.

Exercise 2.4.9

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Exercise 2.4.9

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