Convergence of $\sum_{n=1}^{+\infty}\left(\tan\left(\frac{\pi}{2}\lambda\right)\right)^n\left(n^{2\lambda}+ \sin\left(\frac{(-1)^n}{n}\right)\right)$

Question

Study the convergence of the following series for $|\lambda|\lt1, \lambda\in\mathbb{R}$.

$$\sum_{n=1}^{+\infty} \bigg(\tan\bigg(\frac{\pi}{2}\lambda\bigg)\bigg)^n\bigg(n^{2\lambda}+ \sin\bigg(\frac{(-1)^n}{n}\bigg)\bigg)$$

This series is giving me hard times, I have divided it into cases:

$$\tan\bigg(\frac{\pi}{2}\lambda\bigg) : \begin{cases} \gt0, & \text{if } 0 \lt\lambda\lt1\\[2ex] \lt0, & \text{if } -1\lt\lambda\lt0 \end{cases}$$

$$2\lambda : \begin{cases} \gt1, & \text{if } \frac{1}{2}\lt\lambda\lt1 \\[2ex] \gt0, \lt1, & \text{if } 0\lt\lambda\lt\frac{1}{2} \\[2ex] \lt0,\gt-1 & \text{if } -\frac{1}{2}\lt\lambda\lt0 \\[2ex] \lt-1, & \text{if } -1\lt\lambda\lt-\frac{1}{2} \end{cases}$$

besides we have to observe that $\sin\bigg(\frac{(-1)^n}{n}\bigg)$ converges for $n\to+\infty$

Does anyone have ideas and tips?

Answer 1

Split the series into two parts.

For $\sum \tan^n (\pi \lambda /2) n^{2\lambda}$:

• $\lambda > 1/2$: $\tan(\pi \lambda /2)>1$, hence $$\tan^n (\pi \lambda /2) n ^{2\lambda} > \tan^n (\pi \lambda /2) n \to +\infty [n \to \infty],$$ and the series diverges;

• $\lambda = 1/2$: $\tan (\pi \lambda /2) n^{2\lambda} = 1/n$, and $\sum 1/n$ is the harmonic series, which diverges;

• $0 < |\lambda| < 1/2$: use the Cauchy root test: $$ \left\lvert \tan^n \left( \frac {\pi \lambda} 2\right) n^{2\lambda}\right\rvert^{1/n} = \left\lvert \tan \left(\frac {\pi \lambda}2\right)\right\rvert (n^{1/n})^{2\lambda} \to \left\lvert \tan \left(\frac {\pi\lambda} 2\right) \right\rvert < 1, $$ hence the series converges;

• $\lambda = -1/2$: this is a Leibniz series $\sum (-1)^n n^{-1}$, hence converges;

• $-1 < \lambda < -1/2$: use Cauchy root test again, the root of absolute value tends to $-\tan(\pi \lambda/2) > 1$, hence diverges.

Now for the second part:

Since $\sin (1/n) \sim 1/n$, or $$ \frac {2}{n \pi} \leqslant \sin \left( \frac 1n \right) \leqslant \frac 1n $$ for sufficiently large $n$ and the fact that $(1/n)^{1/n} \to 1$, we have $$ \sin ^{1/n} \left( \frac 1 n \right) \to 1 \quad [n \to \infty]. $$

Now by Cauchy root test, this series diverges when $|\lambda| > 1/2$, converges when $|\lambda| < 1/2$. When $\lambda =1/2$, then the series is Leibniz type, hence converges; when $\lambda = -1/2$, the series is $\sum \sin (1/n)$. Since $$ \sin \left( \frac 1n \right) \geqslant \frac 2 \pi \cdot \frac 1 n, $$ the series diverges.

In conclusion, the original series converges only when $|\boldsymbol\lambda| < \mathbf{1/2}$.