Determine the interval of convergence for the following power series

Question

A need in help... in full I guess?

Determine the interval of convergence for the following power series

(a) $$\sum_{n=1}^\infty \operatorname{csch}(n)x^n$$ (b) $$\sum_{n=1}^\infty \bigg(\arctan \bigg({1\over \sqrt{n}}\bigg)\bigg)x^n$$ Qustion (a) need to be fixed because I don't know how to write hyperbolic csc.

Answer 1

• $a_n=\dfrac{2}{e^n-\frac{1}{e^n}}=\dfrac{2e^n}{e^{2n}-1}$

Consider $\dfrac{a_n}{a_{n+1}}=e^{-1}\times\dfrac{e^{2(n+1)}-1}{e^{2n}-1}=e^{-1}\times\dfrac{(e-\frac{1}{e^n})(e+\frac{1}{e^n})}{(1-\frac{1}{e^n})(1+\frac{1}{e^n})}=e$ as $n\to \infty$

So $R=e$

• $a_n=\tan^{-1}(\dfrac{1}{\sqrt n})$

Note that $\tan^{-1}(\dfrac{1}{\sqrt n})<\dfrac{1}{\sqrt n}$

so $a_n^{\frac{1}{n}}=(\tan^{-1}(\dfrac{1}{\sqrt n}))^{\frac{1}{n}}<(\dfrac{1}{\sqrt n})^{\frac{1}{n}}$

$(\dfrac{1}{\sqrt n})^{\frac{1}{n}}=(\dfrac{1}{n^{\frac{1}{n}}})^{\frac{1}{2}}\to 1$

Hence $R=1$

Answer 2

Hints:

a. For large $n$, $\text{csch }n$ is essentially $e^{-n}$, hence radius $e$, $\pm e$ excluded.

b. For large $n$, $\arctan\dfrac1{\sqrt n}$ is essentially $\dfrac1{\sqrt n}$, hence radius $1$, $+1$ excluded.