Proving that the series $\sum\limits_{n=1}^\infty a(n)$ diverges, where $a(n) = (1/2)^k$ for $2^{k-1} \le n < 2^k$

Question

Let $a(n)=(1/2)^k$ when $2^{k-1} \le n < 2^k$ and $k$ is a natural number. How do you prove that $\sum\limits_{n=1}^\infty a(n)$ diverges?

Answer 1

compute the sum of $a_n$s for n in such a $k$-block. you have $2^{k-1}$ $a_n$s with the same value, hence it will be easy. then just conclude using Cauchy criterion

Answer 2

Define $b_1=a_1$ and $\displaystyle b_n=\sum_{t=2^{n-1}}^{2^n-1}{a_t}$ for $n\geq 2$ and note that $b_n=(2^n-2^{n-1}).\frac{1}{2^n}=\frac{2^{n-1}}{2^n}=\frac{1}{2}$ for each $n \in \mathbb N$. Hence we construct a block series from the series $\sum a_n$. So that the sum of series must be equal. Hence $\displaystyle \sum_{n=1}^{\infty}a_n=\sum_{n=1}^{\infty}b_n=\sum_{n=1}^{\infty}{\frac{1}{2}}=\infty$.