At what values of $\beta$ for which $\sum a_n$ converges, where $a_n=\sqrt{1+\frac{(-1)^n}{n^\beta}}-1\,$?

Question

I posted this kind of question before, but didn't get an appropriate answer. However, this is what I'm trying to do. I want to find the values of $\beta$ for which $\sum a_n$ converges, where $a_n=\sqrt{1+\dfrac{(-1)^n}{n^\beta}}-1\,$. Here is what I have done after the hints below, were provided:

$$a_n=\sqrt{1+\dfrac{(-1)^n}{n^\beta}}-1\,=\dfrac{(-1)^n}{2\,n^\beta}+O\left(\frac{1}{n^{2\beta}}\right)$$

$$\sum a_n=\sum \dfrac{(-1)^n}{2\,n^\beta}$$

It seems as if my former Taylor estimate was wrong. So, can anyone help me from here?

Answer 1

Note that for $\beta \le 0$ it suffices to observe that the series is not well defined or

• $|a_n|\not \to 0$

Thus, for $\beta > 0$ by binomial expansion we have that

$$a_n=\sqrt{1+\dfrac{(-1)^n}{n^\beta}}-1= \dfrac{(-1)^n}{2n^\beta} -\dfrac{1}{8n^{2\beta}}+\dfrac{(-1)^n}{16n^{3\beta}}-\dfrac{5}{128n^{4\beta}}+\dots$$

and since

• for the alternating term, by alternating series test, for convergence we need $\beta >0$
• for not alternating terms we need $2\beta>1,\,4\beta>1,\,\dots2k\beta>1\implies \beta>\frac 1 {2k}\, \forall k\in \mathbb{N}$

then the given series converges for $\beta>0$.

Answer 2

If $ \beta \le 0$, the sequence $(a_n)$ does not converge to $0$, hence the series is divergent.

If $ \beta >0$, then, by Leibniz, the series is convergent.

Answer 3

We know that $$\sum_{n=1}^{\infty} \frac{(-1)^n}{2n^{\beta}} = \left(\frac{1}{2^{\beta}}-\frac{1}{2}\right)\zeta(\beta)$$ which is finite iff $\beta > 0$. Let's take $N\in \mathbb{N}$ such that $\frac{1}{n^{\beta}} < \frac{1}{2}$ for $n\geq N$. Then, the first-order Taylor error estimate gives us $$R_1(x) = \max_{\lvert z\rvert < 1/2} \frac{x^2}{8(1+z)^{3/2}}\leq \frac{x^2}{2\sqrt{2}}$$ for all $\lvert x\rvert\leq 1/2$, so we choose $M\in \mathbb{N}$, $M\geq N$, such that $\sup_{k,l\geq M} \left\lvert\sum_{n=k}^l \frac{(-1)^n}{2n^{\beta}}\right\rvert\leq \frac{\epsilon}{2}$ and $\sup_{k,l\geq M} \sum_{n=k}^l \frac{1}{2\sqrt{2}n^{2\beta}}\leq \frac{\epsilon}{2}$ (again, only works iff $\beta > 1/2$). Then, $$\sup_{k,l\geq M} \left\lvert\sum_{n=k}^l a_n\right\rvert\leq \epsilon$$ by the Taylor error estimate, so the sequence of partial sums is Cauchy (i.e. convergent). We can use higher-order estimates to disprove convergence for $\beta\leq 1/2$:

Take the estimate $\sqrt{1+x}-1 = \frac{x}{2}-\frac{x^2}{8}+O(x^3)$, for example. Let $x = \frac{(-1)^n}{n^{\beta}}$. For $\beta\leq 1/2$, we have that $\sum_{n=1}^{\infty} \left[\frac{(-1)^n}{2n^{\beta}}-\frac{1}{8n^{2\beta}}\right]$ diverges, but for $n > 2^{1/\beta}$ the error in any given term is bounded above by $$R_2(x) = \max_{\lvert z\rvert < 1/2} \frac{\lvert x\rvert^3}{16(1+z)^{5/2}} = \frac{\lvert x\rvert^3}{18\sqrt{6}} = \frac{1}{18\sqrt{6}n^{3\beta}}$$ which is summable for $\beta > 1/3$, implying that $\sum_{n=1}^{\infty} a_n$ must diverge for $1/3 < \beta\leq 1/2$. We can extend this argument for all $\beta > 0$ by taking higher-order Taylor expansions.