Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have to figure out if the following series converge or diverge:

(1) $\sum_{n=1}^{\infty}\frac{(-i)^{n+1}}{n^{2}+1}$. This has the $N$-th partial sum $s_N=\sum^{N}\frac{(-i)^{N+1}}{N^{2}+1}$.

My attempt: To show whether or not the series converges, I need to show whether {$S_{N}$} is Cauchy. So I have to see that $\forall\epsilon>0$ $\exists N_{\epsilon}$ s.t. for all $n,m>N_{\epsilon}$, $|\sum_{k=m+1}^{n}\frac{(-i)^{n+1}}{n^{2}+1}|<\epsilon$. Where do I go from here?

(2) $\sum_{n=1}^{\infty}e^{in\theta}/n$ for $0<\theta<2\pi$, $\theta$ fixed.

Not sure how to approach this.

share|improve this question

2 Answers 2

When you try to determine the convergence/divergence of a series, always remember that there is not only one trick to do it. So if your attempt failed it can either mean that you didn't do it right or that maybe it's just not a right way to go.

For the first one ; using comparison, you can show that it is absolutely convergent since $$ \sum_{n=1}^{\infty} \left| \frac{(-i)^{n+1}}{n^2+1} \right| = \sum_{n=1}^{\infty} \frac 1{n^2+1} \le \sum_{n=1}^{\infty} \frac 1{n^2} $$ and you know that the latter is a convergent series, hence your series converges absolutely.

For the second one, to know what you're looking for, you can informally do this : \begin{align} \frac 1i \sum_{n=1}^{\infty} \frac{e^{in\theta}}{n} = \sum_{n=1}^{\infty} \frac{e^{in\theta}}{in} = \sum_{n=1}^{\infty} \int e^{in\theta} \, d \theta = \int \sum_{n=1}^{\infty} (e^{i\theta})^n \, d\theta = \int \frac 1{1-e^{i\theta}} \, d\theta \end{align} and from here you can compute this integral : by letting $u=1-e^{i\theta}$, you have \begin{align} \int \frac 1{1-e^{i\theta}} d\theta & = \int \frac{i}{u(1-u)} du = -i \int \frac 1{u(u-1)} du = -i \int \left( \frac 1{u-1} - \frac 1u \right) \\ du \\\ & = -i \log(u-1) + i \log(u) \\\ & = -i \log(-e^{i\theta}) + i \log(1-e^{i\theta}) \\\ & = i \log \left( \frac{1-e^{i\theta}}{-e^{i \theta}} \right) \\\ & = i \log(1 - e^{-i\theta}). \end{align} This made me noticed what I wanted ; the function we're looking for is a logarithm. If you compute the power series for $\log$, you get $$ \log(1-z) = \sum_{n=1}^{\infty} \frac {z^n}{n} $$ which means that your series sums to $-\log(1-e^{-i\theta})$, as long as $\theta \neq 0$ (or an integer multiple of $2\pi$). The trickier part comes because $|e^{i\theta}| = 1$, which means it is on the boundary of the disc of convergence of the function $\log(1-z)$. I must say that at the moment I don't know how to deal with it, but I'll try thinking a little more.

Hope that helps,

share|improve this answer
1  
Patrick: It's "series", whether in singular or plural form. :) –  Srivatsan Jan 23 '12 at 5:17
    
@Srivatsan : bleh, my french language got to me. Thanks –  Patrick Da Silva Jan 23 '12 at 5:35
1  
thanks patrick! –  Emir Jan 23 '12 at 5:43
    
@srivatsan. Patrick gave a good mathematical answer to a mathematical question and you feel the need to criticize his grammar or spelling? Amazing. –  CogitoErgoCogitoSum Feb 12 '13 at 2:25
    
@CogitoErgoCogitoSum : It's okay, I considered his comment to be relevant. –  Patrick Da Silva Feb 19 '13 at 17:36

$$\sum_{n=1}^{\infty}\left|\frac{(-i)^{n+1}}{n^{2}+1}\right|<\sum_{n=1}^{\infty}\frac{1}{n^{2}+1}<929.3111872344565678798794$$

share|improve this answer
3  
A little explanation could be useful here. –  Patrick Da Silva Jan 23 '12 at 4:54
4  
Downvoting this answer: the first inequality is wrong and the numerical value on the RHS is (at best) distracting. –  Did Jan 23 '12 at 7:26
2  
I have removed all off-topic comments. –  Mariano Suárez-Alvarez Jan 24 '12 at 8:06
1  
Just an abuse of this forum from someone that isnt qualified to give an answer of any sort. –  CogitoErgoCogitoSum Feb 12 '13 at 2:26

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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