nature of the series $\sum (-1)^{n}n^{-\tan\left(\tfrac{\pi}{4}+\tfrac{1}{n} \right)}$

Question

I would like to study the nature of the following serie:

$$\sum_{n\geq 0}\ (-1)^{n}n^{-\tan\left(\dfrac{\pi}{4}+\dfrac{1}{n} \right)} $$

we can use simply this question :

Show : $(-1)^{n}n^{-\tan\left(\tfrac{\pi}{4}+\tfrac{1}{n} \right)}=\tfrac{(-1)^{n}}{n}+\mathcal{O}\left(\tfrac{\ln(n)}{n^{2}} \right)$

$$\sum_{n\geq 1} (-1)^{n}n^{-\tan\left(\dfrac{\pi}{4}+\dfrac{1}{n} \right)}=\sum_{n\geq 1}\dfrac{(-1)^{n}}{n}+\sum_{n\geq 1} \mathcal{O}\left(\dfrac{\ln(n)}{n^{2}} \right) $$

Since :

• Alternating harmonic series $\sum_{n\geq 1}\dfrac{(-1)^{n}}{n}$ is convergent
• $\sum_{n\geq 1} \mathcal{O}\left(\dfrac{\ln(n)}{n^{2}} \right) $ is convergent since $\left| \mathcal{O}\left(\dfrac{\ln(n)}{n^{2}} \right)\right|\leq \dfrac{\ln(n)}{n^{2}}$

thus $\sum_{n\geq 0}\ (-1)^{n}n^{-\tan\left(\dfrac{\pi}{4}+\dfrac{1}{n} \right)} $ is convergent

My question :

• Why we can't take expansion up to $2$ $\frac{1}{1-z}=1+z+O(z^2)$ instead of $\frac{1}{1-z}=1+z+z^2+O(z^3)$

\begin{align} \tan\left(\frac{\pi}{4}+\frac1n\right)&=\left(1+\tan(1/n)\right)\left(1-\tan(1/n)\right)^{-1}\\ &= \left(1+\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)\right)\left(1-\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)\right) \\ &= \left(1+\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)\right)\left(1-\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)+\mathcal{O}\left(\left(\frac{-1}{n}+\mathcal{O}\left(\frac{1}{n^3}\right)\right)^{2} \right) \right) \\ &= \left(1+\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)\right)\left(1-\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{2}}\right) \right) \\ &=1-\frac{1}{n^2}+\mathcal{O}\left( \frac{1}{n^{2}}\right) \end{align}

then :

\begin{align} e^{-\tan\left(\frac{\pi}{4}+\frac1n\right)\log(n)}&=e^{\left(-1-\frac{1}{n^2}+\mathcal{O}\left( \frac{1}{n^{2}}\right)\right)\log(n)}\\\\ &=\frac1n\,e^{\left(-\frac{\log(n)}n^{2}+O\left(\frac{\log(n)}{n^2}\right)\right)}\\\\ &=\frac1n\,\left(1-\frac{\log(n)}{n^{2}}+O\left(\frac{\log(n)}{n^2}\right)+O\left(\frac{\log^{2}(n)}{n^4}\right) \right)\\\\ &=\frac1n +O\left(\frac{\log(n)}{n^3}\right) \end{align} $$(-1)^{n}e^{-\tan\left(\frac{\pi}{4}+\frac1n\right)\log(n)}=(-1)^{n}\frac1n +O\left(\frac{(-1)^{n}\log(n)}{n^3}\right) $$ and since $\sum O\left(\frac{\log(n)}{n^3}\right)$ and $(-1)^{n}\frac1n$ are convergent

i prove that the series is conevrgent only by using expansion up to 2 why we can't take it

Answer 1

You made a mistake in

\begin{align} \tan\left(\frac{\pi}{4}+\frac1n\right)&=\left(1+\tan(1/n)\right)\left(1-\tan(1/n)\right)^{-1}\\ &= \left(1+\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)\right)\left(1-\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)\right) \\ &= \left(1+\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)\right)\left(1-\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)+\mathcal{O}\left(\left(\frac{-1}{n}+\mathcal{O}\left(\frac{1}{n^3}\right)\right)^{2} \right) \right) \\ &= \left(1+\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)\right)\left(1-\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{2}}\right) \right) \\ &=1-\frac{1}{n^2}+\mathcal{O}\left( \frac{1}{n^{2}}\right) \end{align}

It should be

\begin{align} \tan\left(\frac{\pi}{4}+\frac1n\right)&=\left(1+\tan(1/n)\right)\left(1-\tan(1/n)\right)^{-1}\\ &= \left(1+\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)\right)\left(1-\left(\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)\right)\right)^{-1} \\ &= \left(1+\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)\right)\left(1+\left(\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)\right)+\mathcal{O}\left(\left(\frac{1}{n}+\mathcal{O}\left(\frac{1}{n^3}\right)\right)^{2} \right) \right) \\ &= \left(1+\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)\right)\left(1+\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{2}}\right) \right) \\ &=1+\frac{2}{n}+\mathcal{O}\left( \frac{1}{n^{2}}\right) \end{align}

That is, you effectively wrote

\begin{align} {1+x\over1-x}&=(1+x)(1-x)^{-1}\\ &=(1+x)(1-x+x^2+\cdots) \end{align}

instead of

\begin{align} {1+x\over1-x}&=(1+x)(1-x)^{-1}\\ &=(1+x)(1+x+x^2+\cdots) \end{align}

(It looks like you also omitted the exponent $-1$ when changing from $(1-\tan(1/n))$ to its Taylor expansion, but I think that was just a typo.)

Answer 2

Without digging into asymptotics, the original series is simply convergent by Dirichlet's test, since $\{(-1)^n\}$ has bounded partial sums and

$$ n^{-\tan\left(\frac{\pi}{4}+\frac{1}{n}\right)} = \frac{1}{n^{\frac{1+\tan(1/n)}{1-\tan(1/n)}}}$$ is decreasing towards zero for any $n\geq 3$.