Series divergent or convergent?

Question

Given that $a_n\rightarrow\frac{1}{3}$ as $n\rightarrow\infty$, does the series $a_{k+1}-a_k$ converge or diverge?

Answer 1

Assume the series $s _k \equiv a_{k+1}-a_k$ starts at $k=0$. Then $$s_0 = a_1 - a_0 \\ s_1 = (a_2-a_1) +(a_1-a_0) = a_2 - a_0 \\ \cdots $$ and in general, $s_k = a_{k+1} - a_0$. So the series $s_k$ converges to $\frac{1}{3} - a_0$.

Answer 2

We try to find the formula for a partial sum:

$$\sum_{n=1}^k(a_{n+1}-a_n)=\require{cancel}a_2-a_1+a_3-a_2+a_4-a_3+...+a_{k+1}-a_k$$ $$=\cancel{a_2}-a_1+\cancel{a_3}-\cancel{a_2}+\cancel{a_4}-\cancel{a_3}+...+a_{k+1}-\cancel{a_k}$$ $$=a_{k+1}-a_1$$

It implies that $$\sum_{n=1}^\infty (a_{n+1}-a_n)=\lim_{k\to\infty}(a_{k+1}-a_1)=\frac13-a_1$$

Answer 3

Hint: $$\sum_{k=0}^n(a_{k+1}-a_k=a_{n+1}-a_0.$$