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 am trying to show that the series $$\sum _{n=1}^{\infty }\dfrac {nz^{n-1}\left( \left( 1+\dfrac {1} {n}\right) ^{n}-1\right) } {\left( z^{n}-1\right) \left( z^{n}-\left( 1+\dfrac {1} {n}\right) ^{n}\right) }$$ converges absolutely for all values of $z$, except the values $$z=\left( 1+\dfrac {a} {m}\right) e^{\dfrac {2k\pi i} {m}}$$ ($a= 0, 1; k = 0,1,\ldots m-1;m=1,2,3,\ldots)$.

Since we are looking for absolute convergence D'Alembert's Ratio Test for absolute convergence and Gamma's convergence criterion come to mind. So if we can show $$\lim _{n\rightarrow \infty }\left| \dfrac {U_{n+1}} {U_{n}}\right| =l < 1$$ or $$\lim _{n\rightarrow \infty }\left| \dfrac {U_{n+1}} {U_{n}}\right| = 1 +\dfrac {A_{1}} {n}+O\left( \dfrac {1} {n^{2}}\right) $$, where $A_{1}$ is independent of n and $A_{1} < -1$, then we'll establish the series is absolutely convergent.

I was hoping to first establish that the series is absolutely convergent and wishfully thinking that i might stumble across an expression while doing this to prove the exception values.

Although this may seem like a good plan while solving the limit i am coming up against undefined expression as $n\rightarrow \infty $

I am unsure if i am going down the right line here any suggestion, alternative approaches , help would be much appreciated.

Edit: Maybe Cauchy's test if $$\lim _{n\rightarrow \infty }\sup \left| U_{n}\right| ^{\dfrac {1} {n}} < 1$$, then the series converges absolutely might be what we need here.

share|improve this question
    
Your formulas are distractingly huge. Instead of abusing \dfrac in exponents, I'd recommend using \exp(...) instead. Especially for that last formula with everything inside an e^{...}. –  Rahul Mar 11 '12 at 1:06
    
@RahulNarain, i could n't agree more with u buddy, but i am new to TEX so i was wondering if you could possibly provide an example which i could follow and rewrite the expression in a much more readable form. –  Hardy Mar 11 '12 at 1:09
    
i removed the expression all together, hopefully it is for the better. –  Hardy Mar 11 '12 at 1:14

2 Answers 2

up vote 1 down vote accepted

First of all, the term $$ \left(1+\frac1n\right)^n-1 $$ approaches $e-1$ eventually, so it doesn't play any role in the convergence. The same can be said of $$ z^n-\left(1+\frac1n\right)^n $$ when $|z|\leq1$, since $4\geq|z^n-(1+\frac1n)^n|\geq1$ eventually. So, for $|z|<1$, the general term in your series of absolute values is comparable with $n|z|^{n-1}$, and the series converges.

For $|z|>1$, $$ \left|\dfrac {nz^{n-1} } {\left( z^{n}-1\right) \left( z^{n}-\left( 1+\dfrac {1} {n}\right) ^{n}\right) }\right|=\dfrac{n|z|^{n-1}}{|z|^{2n}|1-\frac1{z^n}|\,|1-\frac{(1+1/n)^n}{z^n}|} $$ For $n$ big, the two differences in the denominator approach 1, so the convergence of the series is decided by $$ \frac{n|z|^{n-1}}{|z|^{2n}}=\frac{n}{|z|^{n+1}}, $$ Which shows that the series converges absolutely for every $z$ with $|z|>1$.

The case $|z|=1$: here $z=e^{2\pi i t}$, with $t\in [0,1]$. When $t$ is rational, the series is not defined (as there are infinitely many values of $n$ where $z^n=1$. For $t$ irrational the terms of the series are defined, but they do not go to zero as $n\to\infty$, so the series is not convergent.

share|improve this answer
    
wow, mate u have taught me a new trick in analysing series convergence. I had n't considered it by breaking the ugly expression down. I apologize in advance if this is a dumb question but does $z=\left( 1+\frac {a} {m}\right) e^{\frac {2k\pi i} {m}}$ represents the rationals ? –  Hardy Mar 11 '12 at 20:00
    
No, what I said is that the numbers $e^{\frac{2k\pi i}m}$ (with $m\in\mathbb{N}$, $k=0,1,\ldots,m$) represent all elements in the circle where $t$ is rational (i.e. $t$ is $k/m$). Unless I'm making a mistake, I cannot make the numbers $(1+a/m)$ play a role. –  Martin Argerami Mar 11 '12 at 22:33
    
thanks for the clarification. Yes i was wrong to say do they represent the rationals ? i meant to say if that expression represented complex numbers where both the coefficients for the real and the imaginary parts are rationals. Yeah i am not so sure about the $\left(1 + a/m\right)$ part either, but none the less i found your answer very insightful. –  Hardy Mar 11 '12 at 22:45

In all honesty, the general term of the series as written has no desire to tend to $0$ for $|z|=1$, so what absolute convergence can we talk about?

share|improve this answer
    
Look i did n't make the question up. This is a problem given to me to solve in a problem set, so it has a solution for sure. –  Hardy Mar 11 '12 at 0:35

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.