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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.

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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. – Comic Book Guy Mar 11 '12 at 1:09
i removed the expression all together, hopefully it is for the better. – Comic Book Guy Mar 11 '12 at 1:14
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.

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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 ? – Comic Book Guy 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. – Comic Book Guy 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?

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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. – Comic Book Guy Mar 11 '12 at 0:35

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