Series convergence/divergence

Question

I was trying to prove the following question. Part a is intuitive but couldn't give a clear mathematical argument. For parts b and c It seems there is something I am not seeing. Any help ?

If $\sum_{k=1}^{\infty }a_{k}$ diverges, $a_{k}\geq 0$, and $% A_{n}=a_{1}+...+a_{n}$, then ,

(a) $\displaystyle \sum_{n=1}^{\infty }\frac{a_{n}}{1+a_{n}} $ diverges,

(b) $\displaystyle \sum_{k=2}^{\infty }\frac{a_{k}}{A_{k}A_{k-1}}$ diverges, but

(c) $\displaystyle \sum_{k=2}^{\infty }\frac{a_{k}}{A_{k}^{\alpha }}$ converges for each $\alpha >2$.

Answer 1

(a) If $(a_n)$ is divergent sequence then you are done. If $(a_n)$ is convergent sequence, then it is bounded. Let $|a_n|<M$ for some $M$ then $\sum a_n/(1+a_n) >\frac{1}{M+1}\sum a_n$ so the given series is divergent.

(b) $\displaystyle \sum_{k=2}^\infty \frac{a_n}{A_nA_{n-1}}=\sum_{k=2}^\infty \frac{1}{A_{n-1}}-\frac{1}{A_n}=\lim_{N\to\infty}\sum_{k=2}^N \frac{1}{A_{n-1}}-\frac{1}{A_n}=\lim_{N\to\infty}\frac{1}{A_1}-\frac{1}{A_N}$

I think (b) is convergent (and it converge at $a_1$.) Convergence of (c) is easily shown by comparison test. (Because $A_k^{\alpha}\ge A_kA_{k-1}$ if $\alpha>2$.)

Answer 2

for b>

$\displaystyle \sum_{k=n+1}^{n+p }\frac{a_{k}}{A_{k}A_{k-1}}=\displaystyle \sum_{k=n+1}^{n+p }\frac{1}{A_{k-1}}-\frac{1}{A_{k}}=\frac{1}{A_n}-\frac{1}{A_{n+p}}< \frac{1}{A_n}$

using cauchy criteria the given series in b> converges