Find the values of a variable for which a series converges

Question

I'm practising at solving sequence-series related problems to improve my skills at solving such problems and currently I'm tackling the following problem:

1) Find all values of the real variable $x$ for which the series $\sum_{n=1}^{\infty}{\frac{(x+1)^n}{\sqrt{4^n n}}}$ converges.
2) Find all values of the positive constant $a$ for which the series $\sum_{n=1}^{\infty}{(-1)^n\frac{\sqrt{n^4+1}}{n^a}}$ converges. For which out of these values does the series converge absolutely?

My Attempt:

1. Using Cauchy's Criterion — if $\sqrt[n]a_n \to r < 1$, $a_n$ converges. $$ \sqrt[n]a_n = \sqrt[n]{\frac{(x+1)^n}{\sqrt{4^n n}}} = \frac{x+1}{\sqrt[2n]{4^n n}} = \frac{x+1}{2\sqrt[2n]{n}} $$

2. I tried solving this problem in the following ways but both times I got stuck:

   • Using Cauchy's Criterion — if $\sqrt[n]a_n \to r < 1$, $a_n$ converges. $$ \sqrt[n]a_n =\sqrt[n]{\frac{(-1)^n\sqrt{n^4+1}}{n^a}} =\frac{-\sqrt[2n]{n^4+1}}{n^{a/n}} $$
   • Using d'Alembert's Criterion — if $\frac{a_{n+1}}{a_n} \to r < 1$, $a_n$ converges. $$ \frac{a_{n+1}}{a_n} =\frac{(-1)^{n+1}\sqrt{(n+1)^4+1}}{(n+1)^a}\frac{{n^a}}{(-1)^n\sqrt{(n^4+1)}} =\frac{-n^a\sqrt{(n+1)^4+1}}{(n+1)^a\sqrt{n^4+1}} $$

Question:

In all my attempts regarding both sub-problems, I got stuck at some point, which makes me doubt whether I'm using the correct approach of not. How can I solve successfully the above problems?

Answer 1

Question 1:

Using Cauchy's Criterion: $$ \sqrt[n]{|a_n|} = \sqrt[n]{\frac{|x+1|^n}{\sqrt{|4^n n|}}} = \frac{|x+1|}{\sqrt[2n]{|4^n n|}} = \frac{|x+1|}{2\sqrt[2n]{n}} = \frac{|x+1|}{2e^{\frac{ln(n)}{2n}}} = \frac{|x+1|}{2e^{\frac{1}{2n}}} = \frac{|x+1|}{2} $$

• In order for $\sum_{n=1}^{\infty}a_n$ to be convergent, $\sqrt[n]{|a_n|} \to r < 1$. Therefore: $$ \frac{|x+1|}{2} < 1\Leftrightarrow |x+1|<2\Leftrightarrow x+1<2 ~~~ \land ~ -x-1<2\Leftrightarrow -3<x<1 $$

• For $x=-3$, $\sum_{n=1}^{\infty}a_n$ converges conditionally, because it's an alternating series: $$ \sum_{n=1}^{\infty}a_n =\sum_{n=1}^{\infty}\frac{(-2)^n}{\sqrt{4^n n}} =\sum_{n=1}^{\infty}\frac{(-2)^n}{2^n\sqrt{n}} =\sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt{n}} $$

• For $x=1$, $\sum_{n=1}^{\infty}a_n$ diverges, because: $$ a_n =\frac{2^n}{\sqrt{4^n n}} =\frac{2^n}{2^n\sqrt{n}} =\frac{1}{\sqrt{n}} =ζ\left(p=\frac{1}{2}\right),~~~ 0<p\le 1 $$

So, based on the above calculations, $\sum_{n=1}^{\infty}a_n$ converges:

• absolutely $\forall~(-3, 1)$.
• conditionally when $x=-3$.

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Question 2:

Using Leibniz's Criterion:

Since the series $\sum_{n=1}^{\infty}{(-1)^n\frac{\sqrt{n^4+1}}{n^a}}$ is alternating, it converges, if according to Leibniz's Criterion the sequence $a_n=\frac{\sqrt{n^4+1}}{n^a}$ is positive, decreasing and its limit is equal to $0$.

• $a_n$ is positive, since both the numerator and denominator are positive $\forall\ n \in \mathbb{N}$.
• $a_n$ is decreasing $\forall\ a$ that make the denominator greater than the numerator: $$ n^a>\sqrt{n^4+1}\Rightarrow n^a>\sqrt{n^4},\ \forall\ n\gg 1\Rightarrow n^a>n^2\Rightarrow a>2 $$
• $a_n\to 0~~ \forall~a$ that make the denominator $\to\infty$: $$ a_n =\frac{\sqrt{n^4+1}}{n^a} =\frac{n^2\sqrt{1+n^{-4}}}{n^a} =\frac{\sqrt{1+n^{-4}}}{n^{a-2}}\to 0\Rightarrow n^{a-2}\to\infty\Rightarrow a>2 $$

Because all prerequisites of Leibniz's Criterion are fulfilled, $\sum_{n=1}^{\infty}{(-1)^n\frac{\sqrt{n^4+1}}{n^a}}$ converges $\forall~a>2$. Following that, we observe that:

• $\sum_{n=1}^{\infty}{(-1)^n a_n}$ converges absolutely when $\sum_{n=1}^{\infty}{a_n}$ does, because: $$ \sum_{n=1}^{\infty}{|(-1)^n\frac{\sqrt{n^4+1}}{n^a}|} =\sum_{n=1}^{\infty}{\frac{\sqrt{n^4+1}}{n^a}} =\sum_{n=1}^{\infty}{a_n} $$
• $a_n$ is a multiple of Riemann's $ζ$ function: $$ a_n =\frac{\sqrt{n^4+1}}{n^a} =\frac{\sqrt{1+n^{-4}}}{n^{a-2}} =\sqrt{1+n^{-4}}\cdot\frac{1}{n^{a-2}} =\sqrt{1+n^{-4}}\cdot ζ(a-2) $$ Since, Riemann's $ζ$ function converges only when $p>1$, then, in order for $\sum_{n=1}^{\infty}{a_n}$ to converge $p=a-2>1\Rightarrow a>3$.

Based on the above calculations, the series $\sum_{n=1}^{\infty}{(-1)^n\frac{\sqrt{n^4+1}}{n^a}}$ converges:

• conditionally $\forall~a\in(2, 3]$ and
• absolutely $\forall~a\in(3, +\infty)$.

Answer 2

Let follow the following hint

• for 1 let use ratio test
• for 2 note that $\frac{\sqrt{n^4+1}}{n^a}\sim \frac{1}{n^{a-2}}$

Notably for the first by ratio test we obtain the following convergence interval

$$\frac {|x+1|}{\sqrt {4} }\implies |x+1|<2 \implies -3 <x <1$$

and studing a part the limit cases $x=-3$ and $x=1$ we finally find $-3\le x<1$.

For the second by limit comparison test with $\sum \frac {1}{n^{a-2}}$ we find that the series converges for $a-2>1$ that is $a>3$ and diverges otherwise.

Answer 3

For the first, the ratio test gives the limit $$\frac {|x+1|}{\sqrt {4} }$$

it will converge if $$|x+1|<2$$ or $$-3 <x <1$$

For $x=1$ it becomes $$\sum \frac {2^n}{2^n\sqrt {n}} $$ which diverges. For $x=-3$, it is $$\sum\frac {(-1)^n}{\sqrt {n}} $$ which converges.

The series converges $\iff -3\le x <1$.

For the second, as said by Gimusi, the general term is positive and equivalent to $\frac {1}{n^{a-2}} $, so the series converges if $a-2>1$ or $a>3$.