If $|\frac{c_{n+1}}{c_n}|\leq1+\frac{a}{n}$, where $a<-1$, $a$ does not depend on $n$, then the series $\sum_{n=1}^\infty c_n$ converges absolutely

Question

Question: If $|\frac{c_{n+1}}{c_n}|\leq1+\frac{a}{n}$, where $a<-1$ and $a$ does not depend on $n$, then the series $\sum_{n=1}^\infty c_n$ converges absolutely.

My attempt: Let $\epsilon>0$. To show $\sum c_n$ converges absolutely, we want to show that there exists an $N\in\mathbb{N}$ such that $|\frac{c_{n+1}}{c_n}|$ converges uniformly to some constant $n>N$. Since $a<-1$, there is some $N_0$ such that $||\frac{c_{n+1}}{c_n}|-1|<\epsilon$ whenever $n>N_0$. Thus, $|\frac{c_{n+1}}{c_n}|$ converges uniformly (to $1^-$) hence $\sum c_n$ converges absolutely.

I actually asked this question a little over a year ago here: Showing a series converges absolutely and got a couple neat answers, but I was wondering if this would be a more "direct" (I don't know if that is the right word) of doing it. Or, is there something I messed up? Any help is, as always, greatly appreciated! Thank you.

Answer 1

This is known as Raabe's test. The criteria is as follows:

1. If there is $a>1$ such that $\frac{|a_{n+1}|}{|a_n|}\leq 1-\frac{a}{n}$ for all $n\geq n_0$, then $\sum_na_n$ converges absolutely
2. If there is $a\leq 1$ such that $\frac{|a_{n+1}|}{|a_n|}\geq 1-\frac{a}{n}$ for all $n\geq n_0$, then $\sum_n|a_n|$ diverges.

Another version of the Raabe's test is here

Here is one way to prove this.

Proposition: Suppose $a_n, b_n>0$, and that there is $n_0\in\mathbb{N}$ such that $\frac{a_{n+1}}{a_n}\leq \frac{b_{n+1}}{b_n}$ for all $n\geq n_0$.

1. If $\sum_nb_n$ converges, so does $\sum_na_n$.
2. If $\sum_na_n$ diverges, so does $\sum_nb_n$.

Proof: The hypothesis may be rewritten as $\frac{a_{n+1}}{b_{n+1}}\leq \frac{a_n}{b_n}$ for all $n\geq n_0$. Thus, the sequence of ratios $\frac{a_n}{b_n}$ is monotone non increasing for $n\geq n_0$ and so it bounded above. Hence, there is $c>0$ such that $a_n\leq c b_n$ for all $n\geq n_0$. The conclusions (1) and (2) follow immediately.

Lemma: If $a>1$ or $a<0$, then $1-ax\leq (1-x)^a$ for all $0\leq x\leq 1$. If $0\leq a\leq1$, then $(1-x)^a\leq 1-ax$ for all $0\leq x\leq 1$.

Proof: Define $g(x)=ax + (1-x)^a$. Then, $g'(x)=a\big(1-(1-x)^{a-1}\big)$. If $a>1$, $g$ is monotone nondecreasing on $[0,1]$ and so $g(x)\geq g(0)=1$ for all $0\leq x\leq 1$. If $0\leq a<1$, then $g$ is monotone non increasing on $[0,1]$ and $g(x)\leq g(0)=1$ for all $0\leq x\leq 1$. Finally, if $a<0$, then $g$ is monotone nonincreasing on $[0,1]$ and $g(x)\leq g(0)=1$ for all $0\leq x\leq 1$.

Proof of Raabe's: If $a>1$ and $\frac{a_{n+1}}{a_n}\leq 1-\frac{a}{n}$ for all $n\geq n_0$, then $$\frac{a_{n+1}}{a_n}\leq 1-\frac{a}{n}\leq\Big(1-\frac1n\Big)^a=\frac{(n-1)^a}{n^a},\qquad n\geq n_0$$ Apply the Proposition above with $b_n=\frac{1}{(n-1)^a}$.

If $a\leq0$ and $\frac{a_{n+1}}{a_n}\geq 1-\frac{a}{n}$ for all $n\geq n_0$, then $a_{n+1}\geq a_{n}$ for all $n\geq n_0$ and so, $a_n\not\rightarrow0$ as $n\rightarrow\infty$

If $0< a\leq1$ and $\frac{a_{n+1}}{a_n}\geq 1-\frac{a}{n}$ for all $n\geq n_0$, then $$\frac{a_{n+1}}{a_n}\geq 1-\frac{a}{n}\geq\Big(1-\frac1n\Big)^a=\frac{(n-1)^a}{n^a},\qquad n\geq n_0$$ Since $\sum_n\frac{1}{(n-1)^a}$ diverges, so does $\sum_na_n$ by the Proposition above.

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Reference: Courant, R. and Fritz, J. Introduction to Calculus and Analysis, Vol. 1, pp. 567 (Classics in Mathematics) 1999th Edition.

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Edit: the method presented in his solution is more about insight than ingenuity, for it is based on the fact that many ratio type tests of convergence can be reduced to the setting the proposition above. This was observed by some clever people in the late 1800's, and came up with the systematic treatment I presented above.

The answer provided by Jack D'Aurizio for example is more ingenious once one makes the connection to the harmonic series: $\frac1n\leq H_n=\sum^n_{k=1}\frac1k\sim\log n$.

Here is another more pedestrian solution building things from the ground up. For simplicity assume $a_n>0$ for all $n$.

If $a>1$, and $\frac{a_{n+1}}{a_n}\leq 1-\frac{a}{n}$ for all $n\geq1$, then $$na_{n+1}\leq na_n-aa_n=(n-1)a_n-(a-1)a_n$$ and so, $$(n-1)a_n-na_{n+1}\geq (a-1)a_n>0$$ This means that the sequence $(n-1)a_n$ is monotone non increasing, and adding terms telescopically gives $$a_2\geq a_2-na_{n+1}\geq (a-1)(a_1+\ldots a_n)$$ This shows that the sequence $\sum^n_{k=1}a_n$ is bounded and so, the series converges.

If $a\leq 1$ and $\frac{a_{n+1}}{a_n}\geq 1-\frac{a}{n}$ for all $n\geq n_0$ then $$\frac{a_{n+1}}{a_n}\geq\frac{n-a}{n}\geq \frac{n-1}{n}$$ Since $\sum_n\frac{1}{n-1}$ diverges, so does $\sum_na_n$.

Answer 2

If $$ \left|\frac{c_{n+1}}{c_n}\right| \leq 1-\frac{k}{n}\qquad\text{with }k>1 $$ then $$|c_{n+1}|\leq |c_n| e^{-k/n}\leq |c_1|\exp\left(-k H_n\right)\leq |c_1|\exp(-k\log n)=\frac{|c_1|}{n^k}$$ and $\sum_{n\geq 1}\frac{1}{n^k} = \zeta(k)< +\infty$ for any $k>1$.

Answer 3

Let $|a|<n_0\in\Bbb N.$ For $n>n_0$ we have $$|\;\ln |c_{n+1}|\;|=|\;\ln |c_{n_0}\prod_{j=n_0}^n|c_{j+1}/c_j|\;|\ge$$ $$\ge |\;\ln |c_{n_0}\prod_{j=n_0}^n(1+a/j)|\;|\ge$$ $$\ge -|\;\ln |c_{n_0}|\;|+\sum_{j=n_0}^n|a|/j \quad (\bullet)$$ because if $-1<a/j<0$ then $\ln (1+a/j)=(a/j)-(a/j)^2/2+(a/j)^3/3-...<(a/j).$

Now $(-\ln n)+\sum_{j=1}^n(1/j)$ converges to the Euler-Macheroni constant $\gamma\approx 0.577$ (a. k.a. Euler's constant) as $n\to\infty$. So $\lim_{n\to\infty}[-\ln n+\sum_{j=n_0}^n(1/j)\,]=K$ exists in $\Bbb R.$

So by $(\bullet)$, for all but finitely many $n>n_0$ we have $|\ln |c_{n+1}|\;|>-|\ln |c_{n_0}|\;|+|a|\ln n+|a|K-1.$ For convenience let $L=-|\ln |c_{n_0}|\;|+|a|K-1.$ So for all but finitely many $n$ we have $$(\bullet\bullet)\quad |\ln |c_{n+1}|\;|>L+|a|\ln n.$$ So $(|c_n|)_{n>n_0}$ is a decreasing positive sequence whose log's tend to $\infty,$ so by $(\bullet\bullet)$ we have $$|c_{n+1}|<\frac {e^{-L}}{n^{|a|}}$$ for all but finitely many $n.$