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Let $\{a_n\}$ be a sequence of positive real numbers such that, for some $N \geq 1$, some $s>1$, and some $M>0$, we have $$ \frac{a_{n+1}}{a_n} = 1 - \frac{A}{n} + \frac{f(n)}{n^s} $$ for all $n \geq N$, where $|f(n)| \leq M$ for all $n$. Then how to prove that the series $\sum a_n$ converges if $A > 1$ and diverges if $A \leq 1$?

I know that the following holds:

Let $\{a_n\}$ and $\{b_n\}$ be two sequences of real numbers such that $a_n > 0$ and $b_n > 0$ for all $n \geq N$, for some fixed $N \geq 1$.

Then if there is a positive constant $r$ such that $$b_n - \frac{a_{n+1} \ b_{n+1}}{a_n} \geq r$$ for all $n \geq N$, then $\sum a_n$ converges.

If, on the other hand, $$b_n - \frac{a_{n+1} \ b_{n+1}}{a_n} \leq 0$$ for all $n \geq N$ and if $\sum 1/b_n$ diverges, then $\sum a_n$ diverges.

I wonder if there's a way of deducing the former result from the latter.

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  • $\begingroup$ This is nothing but Gauss's test. See this. $\endgroup$
    – Kunnysan
    Commented Aug 10, 2013 at 7:46
  • $\begingroup$ Please do not modify entirely the purpose of the question after answers are posted. Asking to use Gauss test (one possible approach, perhaps not the most natural) changes the question. $\endgroup$
    – Did
    Commented Aug 10, 2013 at 7:47
  • $\begingroup$ Then do I pose a fresh question? $\endgroup$ Commented Aug 16, 2013 at 9:34

2 Answers 2

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The question when I posted this answer:

Let $\{a_n\}$ be a sequence of positive real numbers such that, for some $N \geq 1$, some $s>1$, and some $M>0$, we have $$ \frac{a_{n+1}}{a_n} = 1 - \frac{A}{n} + \frac{f(n)}{n^s} $$ for all $n \geq N$, where $|f(n)| \leq M$ for all $n$. Then how to prove that the series $\sum a_n$ converges if $A > 1$ and diverges if $A \leq 1$?

If $A\gt1$, choose $1\lt B\lt A$ and consider $$ c_n=\left(\frac{n+1}{n}\right)^B\cdot\frac{a_{n+1}}{a_n}. $$ Limited expansion of $c_n$ shows that $$ c_n=1-\frac{A-B}n+o\left(\frac1n\right), $$ hence $c_n\leqslant1$ for every $n$ large enough, that is, the sequence $(n^Ba_n)$ is ultimately decreasing, in particular, there exists $N'\geqslant N$ such that for every $n\geqslant N'$, one has $n^Ba_n\leqslant N^Ba_N$, that is, $$ a_n\leqslant\frac{\alpha}{n^B}, $$ for some finite positive $\alpha$ hence the series $\sum\limits_na_n$ converges.

Likewise, if $A\leqslant1$, consider $$ c_n=\frac{n+1}{n}\cdot\frac{a_{n+1}}{a_n}. $$ Can you carry on from here?

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  • $\begingroup$ I'm sorry, but I couldn't make sense of the argument. Would you please take time reviewing my question? $\endgroup$ Commented Aug 10, 2013 at 10:44
  • $\begingroup$ I'm sorry, but I couldn't make sense of the problem you have with the argument. If you need explanations (about the answer), please explain (what your trouble with the answer is). // Regarding the revised version of your question, I already said what I thought of the fact that you completely modified the scope of the question AFTER some answers were posted: this is BAD. Hence, if your (unclear) comment asks that I address the revised version of your question, my answer is: No. $\endgroup$
    – Did
    Commented Aug 10, 2013 at 10:52
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One other way to do this is:

$ln( \frac{A_{n+1}}{A_n})= ln( 1 - \frac{A}{n} + O(\frac{1}{n^s}) )$

-->$ ln(A_{n+1}) -ln(A_n) = -\frac{A}{n} + O(\frac{1}{n^s}) $

You can say that partial sums are equal since both are divergent series:

--> $ln(A_{n+1}) -ln(A_o) = -A*H_n +O(G_n)$ where Hn is the partial harmonic sum and Gn the partial sum of $\frac{1}{n^s}$ . We have : $O(G_n) = o(A_n)$, and $H_n = ln(n) + y + o(1)$ so:

--> $ln(A_n) = ln(A_o) -A*y -A*ln(n) + o(1) $

--> $A_n$ ~ $K*n^{-A} $

Conclusion:

If $A \leq 1$ , the series diverges
If $A > 1$, it converges

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