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- Feb 14, 2012

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Determine the sum \(\displaystyle \sum_{k=1}^n \dfrac{4k}{4k^4+1}\).

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- Thread starter
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- #1

- Feb 14, 2012

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Determine the sum \(\displaystyle \sum_{k=1}^n \dfrac{4k}{4k^4+1}\).

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\(\displaystyle S_n=\sum_{k=1}^n\left(\frac{4k}{4k^4+1} \right)\)

Partial fraction decomposition on the summand allows us to write:

\(\displaystyle S_n=\sum_{k=1}^n\left(\frac{1}{2k^2-2k+1}-\frac{1}{2k^2+2k+1} \right)\)

Observing that:

\(\displaystyle 2(k+1)^2-2(k+1)+1=2k^2+2k+1\)

and using the rule of linearity of the summand and re-indexing the first sum, we obtain:

\(\displaystyle S_n=\sum_{k=0}^{n-1}\left(\frac{1}{2k^2+2k+1} \right)-\sum_{k=1}^n\left(\frac{1}{2k^2+2k+1} \right)\)

Pulling the first term from the first sum and the last term from the second sum, we may write:

\(\displaystyle S_n=1+\sum_{k=1}^{n-1}\left(\frac{1}{2k^2+2k+1} \right)-\sum_{k=1}^{n-1}\left(\frac{1}{2k^2+2k+1} \right)-\frac{1}{2n^2+2n+1}\)

The two sums add to zero, and we are left with:

\(\displaystyle S_n=1-\frac{1}{2n^2+2n+1}=\frac{2n(n+1)}{2n^2+2n+1}\)

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- Feb 14, 2012

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Awesome,

\(\displaystyle S_n=\sum_{k=1}^n\left(\frac{4k}{4k^4+1} \right)\)

Partial fraction decomposition on the summand allows us to write:

\(\displaystyle S_n=\sum_{k=1}^n\left(\frac{1}{2k^2-2k+1}-\frac{1}{2k^2+2k+1} \right)\)

Observing that:

\(\displaystyle 2(k+1)^2-2(k+1)+1=2k^2+2k+1\)

and using the rule of linearity of the summand and re-indexing the first sum, we obtain:

\(\displaystyle S_n=\sum_{k=0}^{n-1}\left(\frac{1}{2k^2+2k+1} \right)-\sum_{k=1}^n\left(\frac{1}{2k^2+2k+1} \right)\)

Pulling the first term from the first sum and the last term from the second sum, we may write:

\(\displaystyle S_n=1+\sum_{k=1}^{n-1}\left(\frac{1}{2k^2+2k+1} \right)-\sum_{k=1}^{n-1}\left(\frac{1}{2k^2+2k+1} \right)-\frac{1}{2n^2+2n+1}\)

The two sums add to zero, and we are left with:

\(\displaystyle S_n=1-\frac{1}{2n^2+2n+1}=\frac{2n(n+1)}{2n^2+2n+1}\)

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- Feb 14, 2012

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\(\displaystyle \begin{align*}\sum_{k=1}^n \dfrac{4k}{4k^4+1}&=\sum_{k=1}^n \dfrac{(2k^2+2k+1)-(2k^2-2k+1)}{(2k^2+2k+1)(2k^2-2k+1)}\\&=\sum_{k=1}^n \left(\dfrac{1}{2k^2-2k+1} -\dfrac{1}{2(k+1)^2-2(k+1)+1} \right)\\&=1-\dfrac{1}{2n^2+2n+1}\end{align*}\)

and we're done.