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Find the Sum of Series

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anemone

MHB POTW Director
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Feb 14, 2012
3,676
Let $x$ be a complex number such that \(\displaystyle x^{2011}=1\) and $x\ne1$.

Compute the sum \(\displaystyle \frac{x^2}{x-1}+\frac{x^4}{x^2-1}+\frac{x^6}{x^3-1}+\cdots+\frac{x^{4020}}{x^{2010}-1}\).
 

PaulRS

Member
Jan 26, 2012
37
Let us first get rid of the doubled exponent on the numerator:

\[ \frac{x^{2n}}{x^n-1} = x^n \cdot \left(\frac{x^n}{x^n-1}\right) = x^n \cdot \left(1 + \frac{1}{x^n-1}\right)\]

So that we have

\[\sum_{k=1}^{2010}\frac{x^{2k}}{x^k-1}=\sum_{k=1}^{2010}x^k + \sum_{k=1}^{2010}\frac{x^k}{x^k-1}\]

and here we have $\sum_{k=1}^{2010}x^k = \frac{x-x^{2011}}{1-x} = \frac{x-1}{1-x}=-1$, hence we will concentrate on the term $\sum_{k=1}^{2010}\frac{x^k}{x^k-1}$.

Let us write $S = \sum_{k=1}^{2010}\frac{x^k}{x^k-1}$.

Note that $x^{2010} = x^{-1}$, thus $\frac{x^{2010} }{x^{2010} - 1} = \frac{x^{-1}}{x^{-1}-1} = \frac{1}{1-x} = - \frac{1}{x-1}$, and, in general we have $x^{2010-k} = x^{-k-1}$ thus
\[\frac{x^{2010-k}}{x^{2010-k}-1} = \frac{x^{-k-1}}{x^{-k-1}-1}=-\frac{1}{x^{k+1}-1}\]

Hence
\[S = \sum_{k=1}^{2010} \frac{x^k}{x^k-1} = \sum_{k=0}^{2009} \frac{x^{2010-k}}{x^{2010-k}-1} = - \sum_{k=0}^{2009}\frac{1}{x^{k+1}-1}= - \sum_{k=1}^{2010}\frac{1}{x^k-1}\]

so that
\[2S = \sum_{k=1}^{2010} \frac{x^k}{x^k - 1} - \sum_{k=1}^{2010}\frac{1}{x^k-1} = \sum_{k=1}^{2010}\frac{x^k-1}{x^k-1}=2010\]

thus $S = 1005$ and so
\[\sum_{k=1}^{2010}\frac{x^{2k}}{x^k-1}=1004\]
 
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anemone

MHB POTW Director
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Feb 14, 2012
3,676
Let $x$ be a complex number such that \(\displaystyle x^{2011}=1\) and $x\ne1$.

Compute the sum \(\displaystyle \frac{x^2}{x-1}+\frac{x^4}{x^2-1}+\frac{x^6}{x^3-1}+\cdots+\frac{x^{4020}}{x^{2010}-1}\).

Hi PaulRS, thanks for participating and yes, your answer is correct.

My solution:

First, let \(\displaystyle S=\frac{x^2}{x-1}+\frac{x^4}{x^2-1}+\frac{x^6}{x^3-1}+\cdots+\frac{x^{4020}}{x^{2010}-1}\).

We're told that \(\displaystyle x^{2011}=1\). This implies

\(\displaystyle x^{4022}=1\;\rightarrow x^2(x^{4020})=1\) or \(\displaystyle x^{4020}=\frac{1}{x^2}\)

From \(\displaystyle x^{4020}=\frac{1}{x^2}\) we get

\(\displaystyle x^{2010}=\frac{1}{x}\) and

\(\displaystyle x^{1005}=\frac{1}{x^{\frac{1}{2}}}\)

Now, if we collect the very first and last term from the given series, we see that
\(\displaystyle
\frac{x^2}{x-1}+\frac{x^{4020}}{x^{2010}-1}=\frac{x^2}{x-1}+\frac{\frac{1}{x^2}}{\frac{1}{x}-1}=\frac{x^2}{x-1}-\frac{1}{x(x-1)}=\frac{x^3-1}{x(x-1)}=\frac{(x-1)(x^2+x+1)}{x(x-1)}=x+1+\frac{1}{x}\)

By continue collecting the terms in this fashion we see that

\(\displaystyle S=\left(\frac{x^2}{x-1}+\frac{x^{4020}}{x^{2010}-1}\right)+\left(\frac{x^4}{x^2-1}+\frac{x^{4018}}{x^{2009}-1}\right)+\cdots+\left(\frac{x^{2010}}{x^{1005}-1}+\frac{x^{2012}}{x^{1006}-1}\right)\)

\(\displaystyle S=\left(x+1+\frac{1}{x}\right)+\left(x^2+1+\frac{1}{x^2}\right)+\cdots+\left(x^{1005}+1+\frac{1}{x^{1005}}\right)\)

\(\displaystyle S=1005+\left(x+x^2+\cdots+x^{1005}\right)+\left( \frac{1}{x}+\frac{1}{x^2}+\cdots+\frac{1}{x^{1005}} \right)\)

And since \(\displaystyle x^{1005}=\frac{1}{x^{\frac{1}{2}}}\), we have \(\displaystyle x^{\frac{1}{2}}=\frac{1}{x^{1005}}\); \(\displaystyle x(x^{\frac{1}{2}})=\frac{1}{x^{1004}}\); \(\displaystyle x^2(x^{\frac{1}{2}})=\frac{1}{x^{1003}}\) and so on and so forth...

\(\displaystyle \frac{1}{x}+\frac{1}{x^2}+\cdots+\frac{1}{x^{1005}}\) becomes \(\displaystyle x^{\frac{1}{2}}(1+x+x^2+\cdots+x^{1004})\) and

\(\displaystyle S=1005+\left(x+x^2+\cdots+x^{1005}\right)+\left( x^{\frac{1}{2}}(1+x+x^2+\cdots+x^{1004}) \right)\)

\(\displaystyle S=1005+\left(1+x+x^2+\cdots+x^{1004}\right)+\left( x^{\frac{1}{2}}(1+x+x^2+\cdots+x^{1004}) \right)-1+x^{1005}\)

\(\displaystyle S=1004+(1+x+x^2+\cdots+x^{1004})(1+ x^{\frac{1}{2}})+\frac{1}{x^{\frac{1}{2}}}\)(*)

I hope I don't confuse you, the reader at this point because I can tell my method is tedious and messy and also, quite confusing...:(

Notice that

\(\displaystyle x^{1005}=\frac{1}{x^{\frac{1}{2}}}\)

\(\displaystyle x^{1005}-1=\frac{1}{x^{\frac{1}{2}}}-1=-\left(\frac{x^{\frac{1}{2}}-1}{x^{\frac{1}{2}}} \right)\)

and since $x \ne 1$,

\(\displaystyle \frac{x^{1005}-1}{x-1}=-\left(\frac{x^{\frac{1}{2}}-1}{x^{\frac{1}{2}}(x-1)} \right)\)

\(\displaystyle \frac{(x-1)(x^{1004}+x^{1003}+\cdots+1)}{x-1}=-\left(\frac{x^{\frac{1}{2}}-1}{x^{\frac{1}{2}}(x^{\frac{1}{2}}-1)(x^{\frac{1}{2}}+1)} \right)\)

\(\displaystyle x^{1004}+x^{1003}+\cdots+1=-\left(\frac{1}{x^{\frac{1}{2}}(x^{\frac{1}{2}}+1)} \right)\)

Now, substitute this into the equation (*) we get

\(\displaystyle S=1004-\left(\frac{1}{x^{\frac{1}{2}}(x^{\frac{1}{2}}+1)} \right)(1+ x^{\frac{1}{2}})+\frac{1}{x^{\frac{1}{2}}}\)

$\therefore S=1004$