1. ## Convergence interval

Hello, I have this serie:

$\displaystyle \displaystyle\sum_{i=1}^\infty{(\displaystyle\frac {1}{n}-tg(\displaystyle\frac{1}{n}))}$

I have tried to apply ratio test but it is indefinite. Any ideas?

Thank you!

2. Originally Posted by Bop
Hello, I have this serie:

$\displaystyle \displaystyle\sum_{i=1}^\infty{(\displaystyle\frac {1}{n}-tg(\displaystyle\frac{1}{n}))}$

I have tried to apply ratio test but it is indefinite. Any ideas?

Thank you!
Is this asking to determine the convergence/divergence for $\displaystyle \sum_{n\in\mathbb{N}}\left\{\frac{1}{n}-\tan\left(\frac{1}{n}\right)\right\}$. Note that $\displaystyle \tan\left(\frac{1}{n}\right)=\frac{1}{n}+\frac{1}{ n^3}+\cdots$. Thus, we can see that $\displaystyle \frac{1}{n}-\tan\left(\frac{1}{n}\right)\underset{n\to\infty}{ \sim}\frac{-1}{3n^3}$. So, your series converges.

3. How do you know that

$\displaystyle tan(\frac{1}{x})=\frac{1}{x}+\frac{1}{x^{3}}+..$?

4. Originally Posted by Bop
How do you know that

$\displaystyle tan(\frac{1}{x})=\frac{1}{x}+\frac{1}{x^{3}}+..$?
It should be $\displaystyle \frac{1}{x}+\frac{1}{3x^3}+\cdots$. It's the Maclaurin series for $\displaystyle \tan(x)$.

Alternatively, suppose that $\displaystyle \frac{1}{n}-\tan\left(\frac{1}{n}\right)$ shared convergence with some series $\displaystyle \frac{1}{n^\lambda}$. Then, we get that $\displaystyle \lim_{\infty}\frac{\frac{1}{n}-\tan\left(\frac{1}{n}\right)}{n^\lambda}=C$ for some constant $\displaystyle C$. Letting $\displaystyle z=\frac{1}{n}$ gives $\displaystyle \lim_{z\to0}\frac{z-\tan(z)}{z^\lambda}=C$. Try to find a $\displaystyle \lambda$ that works.

5. Originally Posted by Drexel28
It should be $\displaystyle \frac{1}{x}+\frac{1}{3x^3}+\cdots$. It's the Maclaurin series for $\displaystyle \tan(x)$.

Alternatively, suppose that $\displaystyle \frac{1}{n}-\tan\left(\frac{1}{n}\right)$ shared convergence with some series $\displaystyle \frac{1}{n^\lambda}$. Then, we get that $\displaystyle \lim_{\infty}\frac{\frac{1}{n}-\tan\left(\frac{1}{n}\right)}{n^\lambda}=C$ for some constant $\displaystyle C$. Letting $\displaystyle z=\frac{1}{n}$ gives $\displaystyle \lim_{z\to0}\frac{z-\tan(z)}{{\color{red} z^\lambda}}=C$. Try to find a $\displaystyle \lambda$ that works.
I think it is typo, Right?

6. Originally Posted by TWiX
I think it is typo, Right?
No. that part is correct. It should have been $\displaystyle \frac{\frac{1}{n}-\tan\left(\frac{1}{n}\right)}{\frac{1}{n^\lambda}}$ originally.

7. With $\displaystyle \lambda=2$ the limits is $\displaystyle \frac{1}{2}$, so it converges.

Thank you very much!

8. Hi, I was thinking, I thought we could only apply this method with positive terms series, and this has negative terms, so, is it right?

9. Hi, I have taken a look at my book and it says this test is for postive series, is it wrong?