Consider the power series $$\sum_{n=1}^{\infty}\left[x^n(1-\cos\left(1 \over n\right))\right].$$

(I) Determine its radius of convergence $R$

(II) Examine the convergence/divergence of the power series when $x = +R$ and $x = -R$

Help will be much appreciated. Im not sure how to proceed

  • $\begingroup$ Start with $|1-\cos\left(1 \over n\right)| \le 2$. $\endgroup$ – lhf Nov 14 '16 at 16:56
  • 1
    $\begingroup$ $$1- \cos \left( \frac{1}{n} \right) \sim \frac{1}{2n^2}$$ so that the radius of convergence is $1$. $\endgroup$ – Crostul Nov 14 '16 at 17:02

If we use the well known equivalence

$$1-\cos(X)\sim \frac{X^2}{2}\;\;\;(X\to 0)$$

with $X=\frac 1n$ and $n\to +\infty$,

we get

$$1-\cos(\frac 1 n)\sim \frac{1}{2n^2}$$

the two series have the same radius $R=1,$ by ratio test

at $x=\pm1, $

$\sum \frac{1}{n^2}$ and $\sum (1-\cos(\frac{1}{n})) (-1)^n $ converge absolutly by the limit comparison test.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.