show that $\frac {a_{N+1}}{s_{N+1}} +...+\frac {a_{N+k}}{s_{N+k}}\geq 1 -\frac {a_{N}}{s_{N+k}} $

Question

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.

Please.thank

The right-hand side of the inequality is different in the title and the question body. — Martin R, Oct 12 '18 at 06:53
Also: https://math.stackexchange.com/q/388898/42969, https://math.stackexchange.com/q/1865038/42969 — Martin R, Oct 12 '18 at 06:56

score 1 · Accepted Answer · answered Oct 12 '18 at 05:59

1

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

answered Oct 12 '18 at 05:59

Michael Rozenberg

194,933

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 06:04
@VERA Yes, of course. See my solution. – Michael Rozenberg Oct 12 '18 at 06:05

show that $\frac {a_{N+1}}{s_{N+1}} +...+\frac {a_{N+k}}{s_{N+k}}\geq 1 -\frac {a_{N}}{s_{N+k}} $

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