# show that $\frac {a_{N+1}}{s_{N+1}} +…+\frac {a_{N+k}}{s_{N+k}}\geq 1 -\frac {a_{N}}{s_{N+k}}$ [duplicate]

Suggestion of how to do it, please.

Suppose $$\{a_n\}$$ is a succession in $$\mathbb R ^+$$ such that $$\sum a_n$$ diverges, and if $$s_n = \sum\limits_{k=1}^n{a_k}$$. show that $$\frac {a_{N+1}}{s_{N+1}} +...+\frac {a_{N+k}}{s_{N+k}}\geq 1 -\frac {s_{N}}{s_{N+k}}$$

and infer that $$\sum\limits_{n=1}^\infty{\frac{a_n}{s_n}}$$ diverge.

## marked as duplicate by Martin R, Namaste, Christopher, hardmath, NosratiOct 12 '18 at 18:27

• The right-hand side of the inequality is different in the title and the question body. – Martin R Oct 12 '18 at 6:53
• – Martin R Oct 12 '18 at 6:56

$$\frac {a_{N+1}}{s_{N+1}} +...+\frac {a_{N+k}}{s_{N+k}}\geq\frac {a_{N+1}}{s_{N+k}} +...+\frac {a_{N+k}}{s_{N+k}}=$$ $$=\frac{s_{N+k}-s_N}{s_{N+k}}= 1 -\frac {s_{N}}{s_{N+k}}$$

• is $\frac {a_{N+1}}{s_{N+1}} +...+\frac {a_{N+k}}{s_{N+k}}\geq 1 -\frac {s_{N}}{s_{N+k}}$ – VERA Oct 12 '18 at 6:04
• @VERA Yes, of course. See my solution. – Michael Rozenberg Oct 12 '18 at 6:05