1
$\begingroup$

Finding value of $\displaystyle \sum^{\infty}_{n=1}\int^{2(n+1)\pi}_{2n\pi}\frac{x\sin x+\cos x}{x^2}$

Try:$$\frac{\cos x}{x} = -\bigg(\frac{x\sin x+\cos x}{x^2}\bigg)$$

So $$\sum^{\infty}_{n=1}\bigg(\frac{\cos x}{x}\bigg)\bigg|^{2(n+1)\pi}_{2n\pi}$$

Could some help me to solve it,Thanks

$\endgroup$
  • $\begingroup$ I have tried to straighten this post out. Feel free to edit further if I've changed your intended meaning. $\endgroup$ – Alfred Yerger Jan 16 '18 at 1:53
  • $\begingroup$ Thanks Alfred Yerger. $\endgroup$ – DXT Jan 16 '18 at 1:54
4
$\begingroup$

This can be cleaned up further. You've done almost everything already.

First, you mean that you recognize the expression to be integrated as the derivative of that quotient, not an equality. But after doing this, you can evaluate.

$$\sum_{n=1}^\infty \frac{\cos(2(n+1)\pi)}{2(n+1)\pi} - \frac{\cos 2n \pi}{2n\pi} =\frac{1}{2\pi}\sum_{n=1}^\infty \frac{1}{n+1} - \frac{1}{n}$$

This is a telescoping series with limit $1$, so you get $\frac{1}{2\pi}$.

$\endgroup$

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.